Related
A set of integers is given as input. You have to return the subset of that set so that the mean - median is maximum for that subset.
Example 1
Input
{1,2,3,4}
Output
{1,2,4}
Example 2
Input
{1,2,2,3,3}
Output
{2,2,3}
package subsetMean_Median;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class MySolution {
public static void main(String[] args) {
int[] arr=
{2,3,2,1,3};
// {1,3,2,4};
Arrays.sort(arr);
int[] outp=meanMedian(arr);
for(int e:outp) {
System.out.print(e+"\t");
}
}
protected static int[] meanMedian(int[] arr) {
double median=findMedian(arr);
double mean=findMean(arr);
double diff=median-mean;
int MAXINDEX=0;
int n=arr.length;
double sets=(1<<n);
System.out.println("sets:"+sets);
for(int i=1;i<=sets;i++) {
int[] subset=findSubset(i,arr);
mean=findMean(subset);
median=findMedian(subset);
if(mean -median>diff) {
diff=mean-median;MAXINDEX=i;
}
}
System.out.println("mean: "+mean+"\tmedian: "+median+"\tdiff: "+diff);
return findSubset(MAXINDEX,arr);
}
protected static int[] findSubset(int counter, int[] arr) {
int n=arr.length;
List<Integer> ls=new ArrayList<Integer>();
for(int j=0;j<n;j++) {
if((counter & (1<<j))>0) {
ls.add(arr[j]);
}
}
int[] output= new int[ls.size()];
for(int j=0;j<ls.size();j++) {
output[j]=ls.get(j);
}
return output;
}
protected static double findMean(int[] arr) {
int n=arr.length;
double sum=0;
if(n==0) return 0;
for(int i=0;i<n;i++)
sum +=arr[i];
return (sum/n);
}
protected static double findMedian(int[] arr) {
int n=arr.length;
if(n%2==1)
return arr[(n/2)];
else if(n>=2)
return 0.5*(arr[((n-2)/2)]+arr[n/2]);
else return 0;
}
}
For every possible median:
lllllmrrrrr
Sort both parts L and R, then start choosing in pair lr maximal elements from both parts and with addition of every next element recompute mean, store arrangement with the best difference. Then the same for minimal elements.
There are about N possible medians, sorting takes O(N*lgN), on every iteration you need to compute up to N means, you can do it in O(N). So, overall complexity is O(N^3*LgN), but most likely you can avoid sorting on every iteration, instead sort whole array only once and update parts in O(1) on every iteration. With such an improvements it is O(N^2).
The most important thing in this problem is to find the Subset.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class MeanMedian {
public static void main(String[] args) {
int[] arr = { 1, 2, 3 };// { 1, 2, 2, 3, 3 };// { 1, 2, 3, 4 };
returnMaxMeanMedian(arr);
}
private static void returnMaxMeanMedian(int[] arr) {
double max = -999.9;
List<Integer[]> subArr = subSet(arr);
Integer[] maxArr = new Integer[1];
for (Integer[] sub : subArr) {
double newMax = calcDiff(sub);
if (max <= newMax) {
max = newMax;
maxArr = sub;
}
}
System.out.println(Arrays.toString(maxArr));
}
private static double calcDiff(Integer[] sub) {
// calc. mean
double sum = 0;
for (int i = 0; i < sub.length; i++) {
sum += sub[i];
}
sum = sum / sub.length;
// calc. median
double median = 0;
if (sub.length % 2 == 0)
median = (double) (sub[(sub.length / 2) - 1] + sub[sub.length / 2]) / 2;
else
median = sub[sub.length / 2];
double diff = sum - median;
return diff;
}
private static List<Integer[]> subSet(int[] arr) {
List<Integer[]> subArr = new ArrayList<Integer[]>();
int n = arr.length;
// Run a loop until 2^n
// subsets one by one
for (int i = 0; i < (1 << n); i++) {
String subSet = "";
// Print current subset
for (int j = 0; j < n; j++)
if ((i & (1 << j)) > 0)
subSet += arr[j] + " ";
subArr.add(convertToInt(subSet.trim().split(" ")));
}
return subArr;
}
private static Integer[] convertToInt(String[] arr) {
if (arr[0] == "")
return new Integer[] { 0 };
Integer[] intArr = new Integer[arr.length];
for (int i = 0; i < arr.length; i++) {
intArr[i] = Integer.parseInt(arr[i].trim());
}
return intArr;
}
}
Sort the list in O(n log n).
Deleting any element to the left of the median (center element or pair) has the same effect on the median, but affect the mean differently. Ditto for elements to the right.
That means that if anything will improve (mean - median), one of these will improve it the most:
the smallest element in the array
the smallest element to the right of the median
one of the element(s) that comprises the median
I.e., for each possible new median, how can we achieve the largest mean?
Repeatedly check these 3-4 for improving mean-median, deleting whatever improves the most. Each operation is O(1), as is recalculating the mean and median. You have to do this at most O(n) times.
The running time is O(n log n) if the list is unsorted, otherwise O(n).
Is this question only for a positive sequence of numbers? If yes, there's this efficient piece of code I wrote:
import java.util.Scanner;
public class MeanMedian {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
int i;
int j;
int k;
int in_length;
int mid_loc;
int sum_arr;
float median = 0.0f;
float mean = 0.0f;
float delta = 0.0f;
float incremental_delta = 0.0f;
float MEDIAN_FOR_MAX_DELTA = 0.0f;
float MEAN_FOR_MAX_DELTA = 0.0f;
float MAX_DELTA = -1.0f;
int MAX_SEQ_LENGTH = 0;
System.out.print("Enter the length of input: ");
in_length = sc.nextInt();
int in_arr[]= new int [in_length+1];
int out_arr[] = new int [in_length+1]; //This is the maximum size of the output array.
int MAX_DELTA_ARR[] = new int [in_length+1];
// STAGE-1: Accept the input sequence
for (i = 1; i <= in_length; i++) {
System.out.print("Enter the input #" + i + ": ");
in_arr[i] = sc.nextInt();
}
// STAGE-1 completed.
// STAGE-2: Sort the array (Bubble sort in Ascending order)
for (j = 1; j < in_length; j++) {
for (i = in_length; i > j; i--) {
if (in_arr[i-1] > in_arr[i]) {
k = in_arr[i];
in_arr[i] = in_arr[i-1];
in_arr[i-1] = k;
}
}
}
// STAGE-2 completed.
// STAGE-3: Compute Max Delta
MAX_DELTA = -99999; //Store as large -ve number as float data type can hold.
for (i = in_length; i > 2; i--) {
// STAGE-3a: Optional - Clear the out_arr[]
for (j = 1; j <= in_length; j++) {
out_arr [j] = 0;
}
// STAGE-3a completed.
// STAGE-3b: Determine the index of the median for the sequence of length i
if (i % 2 == 1) {
mid_loc = (i + 1)/2;
}
else {
mid_loc = (i / 2) + 1;
}
// STAGE-3b completed.
// STAGE-3c: Create the selection that gives the min median and max mean.
// STAGE-3c1: Create left side of mid point.
for (j = mid_loc; j > 0; j--) {
out_arr[j] = in_arr[j];
}
// STAGE-3c1 completed.
// STAGE-3c2: Create right side of mid point.
k = in_length;
for (j = i; j > mid_loc; j--) {
out_arr[j] = in_arr[k];
k = k - 1;
}
// STAGE-3c2 completed.
// STAGE-3c3: Do the SHIFT TEST.
//for (; k <= mid_loc + in_length - i; k++) {
for (k = mid_loc + 1; k <= mid_loc + in_length - i; k++) {
if (i % 2 == 1) {
incremental_delta = ((float)in_arr[k] - (float)out_arr[1])/i - ((float)in_arr[k] - (float)out_arr[mid_loc]);
}
else {
incremental_delta = ((float)in_arr[k] - (float)out_arr[1])/i - (((float)in_arr[k] - (float)out_arr[mid_loc]/2));
}
if (incremental_delta >= 0 ) {
//Insert this new element
for(j = 1; j < mid_loc; j++) {
out_arr[j] = out_arr[j+1];
}
out_arr[mid_loc] = in_arr[k];
}
}
// STAGE-3c3 completed.
// STAGE-3d: Find the median of the present sequence.
if(i % 2 == 1) {
median = out_arr[mid_loc];
}
else {
median = ((float)out_arr[mid_loc] + (float)out_arr[mid_loc - 1])/2;
}
// STAGE-3d completed.
// STAGE-3e: Find the mean of the present sequence.
sum_arr = 0;
for(j=1; j <= i ; j++) {
sum_arr = sum_arr + out_arr[j];
}
mean = (float)sum_arr / i;
// STAGE-3e completed.
// STAGE-3f: Find the delta for the present sequence and compare with previous MAX_DELTA. Store the result.
delta = mean - median;
if(delta > MAX_DELTA) {
MAX_DELTA = delta;
MEAN_FOR_MAX_DELTA = mean;
MEDIAN_FOR_MAX_DELTA = median;
MAX_SEQ_LENGTH = i;
for (j = 1; j <= MAX_SEQ_LENGTH; j++) {
MAX_DELTA_ARR[j] = out_arr[j];
}
}
// STAGE-3f completed.
}
// STAGE-4: Print the result.
System.out.println("--- RESULT ---");
System.out.print("The given input sequence is: ");
System.out.print("{ ");
for(i=1; i <= in_length; i++) {
System.out.print(in_arr[i]);
System.out.print(" ");
}
System.out.print("}");
System.out.println("");
System.out.print("The sequence with maximum difference between mean and median is: ");
System.out.print("{ ");
for(i=1; i <= MAX_SEQ_LENGTH; i++) {
System.out.print(MAX_DELTA_ARR[i]);
System.out.print(" ");
}
System.out.print("}");
System.out.println("");
System.out.println("The mean for this sequence is: " + MEAN_FOR_MAX_DELTA);
System.out.println("The median for this sequence is: " + MEDIAN_FOR_MAX_DELTA);
System.out.println("The maximum difference between mean and median for this sequence is: " + MAX_DELTA);
}
}
This code has order O(n) (if we ignore the necessity to sort the input array).
In case, -ve inputs are also expected - the only way out is by evaluating each subset. The downside to this approach is that the algorithm has exponential order: O(2^n).
As a compromise you could use both types of algorithm in your code and switch between the two by evaluating the input sequence. By the way, where did you come across this question?
from itertools import combinations
[Verfication of the code][1]
# function to generate all subsets possible, there will be 2^n - 1 subsets(combinations)
def subsets(arr):
temp = []
for i in range(1, len(arr)+1):
comb = combinations(arr, i)
for j in comb:
temp.append(j)
return temp
# function to calculate median
def median(arr):
mid = len(arr)//2
if(len(arr)%2==0):
median = (arr[mid] + arr[mid-1])/2
else:`
median = arr[mid]
return median
# function to calculate median
def mean(arr):
temp = 0
for i in arr:
temp = temp + i
return temp/len(arr)
# function to solve given problem
def meanMedian(arr):
sets = subsets(arr)
max_value = 0
for i in sets:
mean_median = mean(i)-median(i)
if(mean_median>max_value):
max_value = mean_median
needed_set = i
return needed_set
[1]: https://i.stack.imgur.com/Mx4pc.png
So I tried a little on the problem and here is a code that might help you. Its written in a way that should be easy to read, and if not, do let me know. Maybe you need to take array input from the user as I have taken a fixed array. That shouldn't be much of a problem I am sure.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
class MeanMinusMedian
{
private static float mean = 0;
private static float median = 0;
private static float meanMinusMedian = 0;
private static List<Integer> meanMinusMedianList = null;
private static void formMeanMinusMedianArr(int data[], int sumOfData)
{
findMean(data, sumOfData);
findMedian(data);
if ((mean - median) > meanMinusMedian) {
meanMinusMedian = mean - median;
meanMinusMedianList = new ArrayList<Integer>();
Arrays.stream(data)
.forEach(e->meanMinusMedianList.add(e));
}
}
/**
* #param data
*/
private static void findMedian(int[] data) {
int dataLen = data.length;
median = data.length % 2 == 0 ? ((float)data[dataLen / 2] + (float)data[dataLen / 2 - 1]) / 2 : data[dataLen / 2];
}
/**
* #param data
* #param sumOfData
*/
private static void findMean(int[] data, int sumOfData) {
mean = ((float)sumOfData /(float) data.length);
}
/**
*
* #param arr
* #param data
* #param start
* #param end
* #param index
* #param runningVal
*/
private static void combinationUtil(int arr[], int data[], int start, int end, int index, int runningVal) {
// Current combination is ready to be printed, print it
if (index == runningVal) {
formMeanMinusMedianArr(data, Arrays.stream(data) // Step 1
.sum());
return;
}
// replace index with all possible elements. The condition
// "end-i+1 >= r-index" makes sure that including one element
// at index will make a combination with remaining elements
// at remaining positions
for (int i = start; i <= end && end - i + 1 >= runningVal - index; i++) {
data[index] = arr[i];
combinationUtil(arr, data, i + 1, end, index + 1, runningVal);
}
}
/**
*
* #param arr
* #param n
* #param runningVal
*/
private static void printCombination(int arr[], int n, int runningVal) {
int data[] = new int[runningVal];
// Print all combination using temporary array 'data[]'
combinationUtil(arr, data, 0, n - 1, 0, runningVal);
}
public static void main(String[] args) {
int arr[] = { 1, 2, 2, 3, 3 };
int runningVal = 1;//Running value
int len = arr.length;
for (int i = 1; i < arr.length; i++) {
printCombination(arr, len, runningVal + i);
}
System.out.println(meanMinusMedianList);
}
}
Taking reference of answer of Bhaskar13 https://stackoverflow.com/a/59386801/3509609 , I solved it without using the bit shift operators, to add more readability.
package array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
public class MeanMinusMedianMax {
public static void main(String[] args) {
System.out.println(Arrays.toString(maxDiffrenceSubSet(4, new int[] { 4, 2, 3, 1 })));
System.out.println(Arrays.toString(maxDiffrenceSubSet(4, new int[] { 1, 2, 2, 3, 3 })));
}
public static int[] maxDiffrenceSubSet(int n, int[] input2) {
int totalSubsets = (int) Math.pow(2, n);
Map<Integer, ArrayList<Integer>> subsetsMap = new HashMap<Integer, ArrayList<Integer>>();
Integer maxKey = null;
double maxDiff = 0;
for (int i = 0; i < totalSubsets; i++) {
String binaryString = Integer.toBinaryString(i);
while (binaryString.length() < 4) {
binaryString = "0" + binaryString;
}
char[] currentPick = binaryString.toCharArray();
ArrayList<Integer> currentList = new ArrayList<Integer>();
for (int x = 0; x < currentPick.length; x++) {
if ((currentPick[x]) == '1') {
currentList.add(input2[x]);
}
}
Collections.sort(currentList);
subsetsMap.put(i, currentList);
double mean = findMean(currentList);
double median = findMedian(currentList);
double currentDifference = mean - median;
if (currentDifference > maxDiff) {
maxDiff = currentDifference;
maxKey = i;
}
}
return subsetsMap.get(maxKey).stream().mapToInt(i -> i).toArray();
}
static double findMean(ArrayList<Integer> arr) {
int n = arr.size();
double sum = 0;
if (n == 0)
return 0;
for (int i = 0; i < n; i++)
sum += arr.get(i);
return (sum / n);
}
static double findMedian(ArrayList<Integer> arr) {
int n = arr.size();
if (n % 2 == 1)
return arr.get((n / 2));
else if (n >= 2)
return 0.5 * (arr.get(((n - 2) / 2)) + arr.get(n / 2));
else
return 0;
}
}
class UserMainCode (object):
def meanmedian(cls,ip1,ip2=[]):
s = []
s = ip2
lst = []
final = []
op = []
max_val = 0
diff = 0
for i in range(1,ip1+1):
n=i
lst = list(itertools.combinations(s,n))
final = final +lst
for i in range(len(final)):
men = statistics.mean(final[i])
med = statistics.median(final[i])
diff = men - med
if max_val < diff:
op = final[i]
max_val = diff
return op
Find minimum cost of tickets required to buy for traveling on known days of the month (1...30). Three types of tickets are available : 1-day ticket valid for 1 days and costs 2 units, 7-days ticket valid for 7 days and costs 7 units, 30-days ticket valid for 30 days and costs 25 units.
For eg: I want to travel on [1,4,6,7,28,30] days of the month i.e. 1st, 4th, 6th ... day of the month. How to buy tickets so that the cost is minimum.
I tried to use dynamic programming to solve this but the solution is not giving me the correct answer for all cases. Here is my solution in Java :
public class TicketsCost {
public static void main(String args[]){
int[] arr = {1,5,6,9,28,30};
System.out.println(findMinCost(arr));
}
public static int findMinCost(int[] arr) {
int[][] dp = new int[arr.length][3];
int[] tDays = {1,7,30};
int[] tCost = {2,7,25};
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < 3; j++) {
if (j==0){
dp[i][j]= (i+1)*tCost[j];
}
else{
int c = arr[i]-tDays[j];
int tempCost = tCost[j];
int k;
if (c>=arr[0] && i>0){
for (k = i-1; k >= 0; k--) {
if (arr[k]<=c){
c = arr[k];
}
}
tempCost += dp[c][j];
int tempCostX = dp[i-1][j] + tCost[0];
tempCost = Math.min(tempCost,tempCostX);
}
dp[i][j] = Math.min(tempCost,dp[i][j-1]);
}
}
}
return dp[arr.length-1][2];
}
}
The solution doesn't work for {1,7,8,9,10} input, it gives 10 but the correct answer should be 9. Also, for {1,7,8,9,10,15} it give 13 but the correct is 11.
I have posted my solution not for other to debug it for me but just for reference. I was taken a bottom-up dynamic programming approach for this problem. Is this approach correct?
Let MC(d) denote the minimum cost that will pay for all trips on days 1 through d. The desired answer is then MC(30).
To calculate MC(d), observe the following:
If there's no trip on day d, then MC(d) = MC(d − 1).
As a special case, MC(d) = 0 for all d ≤ 0.
Otherwise, the minimum cost involves one of the following:
A 1-day pass on day d. In this case, MC(d) = MC(d − 1) + 2.
A 7-day pass ending on or after day d. In this case, MC(d) = min(MC(d − 7), MC(d − 6), …, MC(d − 1)) + 7.
And since MC is nondecreasing (adding a day never reduces the minimum cost), this can be simplified to MC(d) = MC(d − 7) + 7. (Hat-tip to Ravi for pointing this out.)
A 30-day pass covering the whole period. In this case, MC(d) = 25.
As you've realized, dynamic programming (bottom-up recursion) is well-suited to this.
For ease of coding, I suggest we start by converting the list of days into a lookup table for "is this a trip day?":
boolean[] isDayWithTrip = new boolean[31]; // note: initializes to false
for (final int dayWithTrip : arr) {
isDayWithTrip[dayWithTrip] = true;
}
We can then create an array to track the minimum costs, and populate it starting from index 0:
int[] minCostUpThroughDay = new int[31];
minCostUpThroughDay[0] = 0; // technically redundant
for (int d = 1; d <= 30; ++d) {
if (! isDayWithTrip[d]) {
minCostUpThroughDay[d] = minCostUpThroughDay[d-1];
continue;
}
int minCost;
// Possibility #1: one-day pass on day d:
minCost = minCostUpThroughDay[d-1] + 2;
// Possibility #2: seven-day pass ending on or after day d:
minCost =
Math.min(minCost, minCostUpThroughDay[Math.max(0, d-7)] + 7);
// Possibility #3: 30-day pass for the whole period:
minCost = Math.min(minCost, 25);
minCostUpThroughDay[d] = minCost;
}
And minCostUpThroughDay[30] is the result.
You can see the above code in action at: https://ideone.com/1Xx1fd.
One recursive solution in Python3.
from typing import List
def solution(A: List[int]) -> int:
if not any(A):
return 0
tickets = {
1: 2,
7: 7,
30: 25,
}
import sys
min_cost = sys.maxsize
size = len(A)
for length, price in tickets.items():
current_cost = price
idx = 0
last_day = A[idx] + length
while idx < size and A[idx] < last_day:
idx += 1
if current_cost > min_cost:
continue
current_cost += solution(A[idx:])
if current_cost < min_cost:
min_cost = current_cost
return min_cost
if __name__ == '__main__':
cases = {
11: [1, 4, 6, 7, 28, 30],
9: [1, 7, 8, 9, 10],
}
for expect, parameters in cases.items():
status = (expect == solution(parameters))
print("case pass status: %s, detail: %s == solution(%s)" %
(status, expect, parameters))
public class Main03v3
{
public static void main(String[] args)
{
int[] A = {1,7,8,9,10,15,16,17,18,21,25};
System.out.println("Traveling days:\r\n "+Arrays.toString(A));
int cost = solution(A);
System.out.println("\r\nMinimum cost is " + cost);
System.out.println("\r\n" + new String(new char[40]).replace("\0", "-"));
}
public static int solution(int[] A)
{
if (A == null) return -1;
int sevenDays = 7;
int dayCost = 2, weekCost = 7, monthCost = 25;
int ratio_WeekAndDays = weekCost / dayCost;
int len = A.length;
if (len == 0) return -1;
if (len <= 3) return len * dayCost;
int cost[] = new int[len];
int i = 0;
while (i < len)
{
int startIdx = i, endIdx = i + 1;
while (endIdx < len && A[endIdx]-A[startIdx] < sevenDays)
endIdx++;
if (endIdx-startIdx > ratio_WeekAndDays)
{
if (endIdx >= startIdx + sevenDays)
endIdx = startIdx + sevenDays;
int j = startIdx;
cost[j] = ((j == 0) ? 0 : cost[j-1]) + weekCost;
while (++j < endIdx) {
cost[j] = cost[j-1];
}
i = j;
}
else
{
cost[i] = ((i == 0) ? 0 : cost[i-1]) + dayCost;
i++;
}
}
int finalCost = Math.min(cost[len-1], monthCost);
return finalCost;
}
}
Find minimum cost of tickets in JavaScript
case 1 : if input is [1,7,8,9,10] then the required output is 9
case 2 : if input is [1,7,8,9,10,15] then the required output is 11
function calMinCosts(arr){
if(!arr || arr.length===0)
return 0;
var len = arr.length;
var costsOfDateArr = Array.apply(null,{length:arr[len-1]+1}).map(()=>0);
var price1=2,price2=7,price3=25;
var days=7;
var index=0,n=costsOfDateArr.length;
for(var i=1;i<n;i++){
if(i===arr[index]){
if(i>=days+1){
costsOfDateArr[i] = Math.min(costsOfDateArr[i-days-1]+price2, costsOfDateArr[i-1]+price1);
}else{
costsOfDateArr[i] = Math.min(costsOfDateArr[0]+price2, costsOfDateArr[i-1]+price1);
}
index+=1;
}else{
costsOfDateArr[i] = costsOfDateArr[i-1];
}
}
return Math.min(price3,costsOfDateArr[n-1]);
}
console.log(calMinCosts([1,7,8,9,10]))
console.log(calMinCosts([1,7,8,9,10,15]))
Here is the C++ solution including print outs
#include <vector>
#include <iostream>
#include <cmath>
#include <algorithm>
int compute(std::vector<int> &A)
{
int sum[A.size()][A.size()+1];
for (int i = 0; i < A.size(); i++)
{
for(int j =0; j < A.size(); j++)
{
sum[i][j]=2;
}
}
for (int k = 0; k < A.size();k++)
{
sum[k][A.size()]=0;
}
for (int i = 0; i < A.size(); i++)
{
for(int j = 0; j < A.size(); j++)
{
if (i!=j)
{
if (sum[i][i] != 7)
{
int temp = abs(A[j]-A[i]);
if (temp<7 && abs(j-i)>=3)
{
sum[i][i]=7;
sum[i][j]=7;
if (i>j)
{
for(int k = j;k < i;k++)
sum[i][k]=7;
}
else
{
for(int k = i;k < j;k++)
sum[i][k]=7;
}
}
}
}
}
}
for (int i = 0; i < A.size(); ++i)
{
for(int j = 0; j < A.size(); ++j)
{
if (sum[i][j]==7)
{
sum[i][A.size()]+=1;
}
}
}
for (int i = 0; i < A.size(); ++i)
{
for (int j = 0; j < A.size()+1; ++j)
std::cout<<sum[i][j]<<" ";
std::cout<<std::endl;
}
int result = 0;
int row = A.size()-1;
int column = A.size()-1;
while(1)
{
int value = sum[row][A.size()];
if (value == 0)
value=1;
int temp = sum[row][column];
result += temp;
row = row-value;
column = column-value;
while (sum[row][column+1]==7 && row>=0)
{
row-=1;
column-=1;
result+=2;
}
if (row < 0)
break;
}
return result;
}
int solution(std::vector<int> &A) {
if (A.size() > 24)
return 25;
if (A.size() <= 3)
return A.size() * 2;
return std::min(25,compute(A));
}
int main()
{
std::vector<int> AA={1,2,3,4,5,29,30};
std::vector<int> B={1,2,3,4,5};
std::vector<int> A={1,2,3,4,5,9,10,11,12,13,14,17,18,20,21};
std::vector<int> C={1,2,3,12};
std::vector<int> D={1,2,3,4,12,13,14,15,29,30};
std::vector<int> DD={1,2,3,4,5,14,17,18,19,20,23,28,29,30};
std::vector<int> CC={1,2,3,4,5,6,7,9,14,17,18,19,20,23,28,29,30};
std::cout<<solution(AA)<<std::endl;
std::cout<<solution(D)<<std::endl;
std::cout<<solution(B)<<std::endl;
std::cout<<solution(A)<<std::endl;
std::cout<<solution(C)<<std::endl;
std::cout<<solution(DD)<<std::endl;
std::cout<<solution(CC)<<std::endl;
return 0;
}
Solved using the same approach of bottom-up dynamic programming. Here is the full solution :
public class PublicTicketCost {
public static void main(String args[]){
int[] arr = {1,7,8,9,10,15,16,17,18,21,25};
int[] tDays = {1,7,30};
int[] tCost = {2,7,25};
System.out.println(minCost(arr, tDays, tCost));
}
public static int minCost(int[] arr, int[] tDays, int[] tCost) {
int[][] dp = new int[arr.length][tDays.length];
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < tDays.length; j++) {
int prevDayIndex = findPrevDayIndex(arr,i,tDays,j);
int prevCost = prevDayIndex>=0 ? dp[prevDayIndex][tDays.length-1] : 0;
int currCost = prevCost + tCost[j];
if(j-1>=0){
currCost = Math.min(currCost, dp[i][j-1]);
}
dp[i][j] = currCost;
}
}
//print(dp);
return dp[arr.length-1][tDays.length-1];
}
private static void print(int arr[][]){
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[0].length; j++) {
System.out.print(arr[i][j]+" ");
}
System.out.println();
}
}
private static int findPrevDayIndex(int[] arr, int i, int[] days, int j){
int validAfterDate = arr[i] - days[j];
if (validAfterDate<1){
return -1;
}
for (int k = i-1; k >= 0; k--) {
if (arr[k]<=validAfterDate){
return k;
}
}
return -1;
}
}
http://ideone.com/sfgxGo
I was wondering if there is an efficient premade algorithm for determining if the sum/difference of a group of numbers can equal a different number. Example:
5, 8, 10, 2, using + or -, to equal 9.
5 - 8 = -3 + 10 = 7 + 2 = 9
If there is a preexisting algorithm, what is it called. If not, I can figure out how to program it, though it may not be efficient.
Thank you!
Yeah, this is basically knapsack problem, but it can be computed in pseudopolynomial time using dynamic programming.
I did it few month ago, so maybe this java code can help you, if you want to implement it :
public void solve() {
while (this.isEnd() == false) {
int priceSum = this.getItemsInstance().getTotalPrice()/divide;
int numOfItems = this.getItemsInstance().itemCount();
int maxWeight = this.getItemsInstance().getMaxWeight();
int[][] array = new int[numOfItems + 1][priceSum + 1];
boolean[][] arrayCounted = new boolean[numOfItems + 1][priceSum + 1];
for (int i = 0; i < numOfItems + 1; i++) {
array[i][0] = 0;
arrayCounted[i][0] = true;
}
int max = 0;
int price = 0;
for (int j = 1; j < priceSum + 1; j++) {
for (int i = 1; i < numOfItems + 1; i++) {
int temp = W(i, j, array, arrayCounted);
if (temp <= maxWeight) {
max = temp;
price = j;
}
}
}
}
}
private int W(int i, int c, int[][] array, boolean[][] arrayCounted) {
if (c < 0) {
return MAX_PRICE / divide;
}
if (i == 0) {
if (c == 0) {
return 0;
} else {
return MAX_PRICE / divide;
}
}
if (arrayCounted[i][c]) {
return array[i][c];
}
arrayCounted[i][c] = true;
array[i][c] = Math.min(W(i - 1, c, array, arrayCounted), W(i - 1, c - this.items[i - 1].price/divide, array, arrayCounted) + this.items[i - 1].weight);
return array[i][c];
}
its not an NP problem, if the problem is to find a given number from adding or subtracting each elements of a list/array. if you think about AP. here is a sample code in C++
int Np( int mn, list<int>a, int c )
{
int size = a.size(), rst = 0, maxI = 0;
std::list<int>::iterator it;
while( size > c )
{
a.sort();
maxI += a.back();
a.pop_back();
rst = 0;
for( auto ele : a )
{
rst += ele;
cout << rst << endl;
}
if( (rst - maxI) == mn or (maxI - rst) == mn or (maxI + rst) == mn )
{
return mn;
}
size--;
}
return rst;
}
this should help. i think.
I actually wrote a simple java program, I was not actually aware of knapsack strategies. This is my own solution. Hope this helps
import java.util.ArrayList;
import java.util.List;
public class Puzzle {
public static void main(String[] args) {
int targetNumber = 0;
int min = 2147483647;
int[] numbers = {-10, -30, -20, -50};
//int[] numbers = {0,0,0,0};
//int[] numbers = {7, 2, 10};
//int[] numbers = {1, 2, 3, 4, 5};
//int[] numbers = {1000, 2, 3, 4, 100};
char set[] = {'+', '-'};
min = getNumberClosestToTarget(numbers, set, min, targetNumber);
System.out.println(String.format(" %d is closest to %d", min, targetNumber));
}
private static int getNumberClosestToTarget(int[] numbers, char[] set, int min, int targetNumber) {
List<String> operators = new ArrayList<>();
computeAllOperatorsCombination(set, "", set.length, numbers.length - 1, operators);
for (String operatorString : operators) {
String[] ops = operatorString.split("");
int sum = computeSum(numbers, ops, numbers.length - 1);
min = getClosestToTarget(min, targetNumber, sum);
}
return min;
}
static int computeSum(int[] numbers, String[] operators, int index) {
int result = numbers[index];
if (index == 0) {
return result;
} else {
switch (operators[index - 1]) {
case "+":
return computeSum(numbers, operators, index - 1) + result;
case "-":
return computeSum(numbers, operators, index - 1) - result;
}
return result;
}
}
static void computeAllOperatorsCombination(char set[], String prefix, int n, int k, List<String> result) {
if (k == 0) {
result.add(prefix);
return;
}
for (int i = 0; i < n; i++) {
String newPrefix;
newPrefix = prefix + set[i];
computeAllOperatorsCombination(set, newPrefix, n, k - 1, result);
}
}
private static int getClosestToTarget(int min, int targetNumber, int r) {
int distance = Math.abs(targetNumber - r) < Math.abs(r - targetNumber) ? Math.abs(targetNumber - r) : Math.abs(r - targetNumber);
if (distance < Math.abs(min)) {
min = distance;
if (r < 0) {
min = -distance;
}
}
return min;
}
}
Given a set of numbers, divide the numbers into two subsets such that difference between the sum of numbers in two subsets is minimal.
This is the idea that I have, but I am not sure if this is a correct solution:
Sort the array
Take the first 2 elements. Consider them as 2 sets (each having 1 element)
Take the next element from the array.
Decide in which set should this element go (by computing the sum => it should be minimum)
Repeat
Is this the correct solution? Can we do better?
The decision version of the problem you are describing is an NP-complete problem and it is called the partition problem. There are a number of approximations which provide, in many cases, optimal or, at least, good enough solutions.
The simple algorithm you described is a way playground kids would pick teams. This greedy algorithm performs remarkably well if the numbers in the set are of similar orders of magnitude.
The article The Easiest Hardest Problem, by American Scientist, gives an excellent analysis of the problem. You should go through and read it!
No, that doesn't work. There is no polynomial time solution (unless P=NP). The best you can do is just look at all different subsets. Have a look at the subset sum problem.
Consider the list [0, 1, 5, 6]. You will claim {0, 5} and {1, 6}, when the best answer is actually {0, 1, 5} and {6}.
No, Your algorithm is wrong. Your algo follows a greedy approach.
I implemented your approach and it failed over this test case:
(You may try here)
A greedy algorithm:
#include<bits/stdc++.h>
#define rep(i,_n) for(int i=0;i<_n;i++)
using namespace std;
#define MXN 55
int a[MXN];
int main() {
//code
int t,n,c;
cin>>t;
while(t--){
cin>>n;
rep(i,n) cin>>a[i];
sort(a, a+n);
reverse(a, a+n);
ll sum1 = 0, sum2 = 0;
rep(i,n){
cout<<a[i]<<endl;
if(sum1<=sum2)
sum1 += a[i];
else
sum2 += a[i];
}
cout<<abs(sum1-sum2)<<endl;
}
return 0;
}
Test case:
1
8
16 14 13 13 12 10 9 3
Wrong Ans: 6
16 13 10 9
14 13 12 3
Correct Ans: 0
16 13 13 3
14 12 10 9
The reason greedy algorithm fails is that it does not consider cases when taking a larger element in current larger sum set and later a much smaller in the larger sum set may result much better results. It always try to minimize current difference without exploring or knowing further possibilities, while in a correct solution you might include an element in a larger set and include a much smaller element later to compensate this difference, same as in above test case.
Correct Solution:
To understand the solution, you will need to understand all below problems in order:
0/1 Knapsack with Dynamic Programming
Partition Equal Subset Sum with DP
Solution
My Code (Same logic as this):
#include<bits/stdc++.h>
#define rep(i,_n) for(int i=0;i<_n;i++)
using namespace std;
#define MXN 55
int arr[MXN];
int dp[MXN][MXN*MXN];
int main() {
//code
int t,N,c;
cin>>t;
while(t--){
rep(i,MXN) fill(dp[i], dp[i]+MXN*MXN, 0);
cin>>N;
rep(i,N) cin>>arr[i];
int sum = accumulate(arr, arr+N, 0);
dp[0][0] = 1;
for(int i=1; i<=N; i++)
for(int j=sum; j>=0; j--)
dp[i][j] |= (dp[i-1][j] | (j>=arr[i-1] ? dp[i-1][j-arr[i-1]] : 0));
int res = sum;
for(int i=0; i<=sum/2; i++)
if(dp[N][i]) res = min(res, abs(i - (sum-i)));
cout<<res<<endl;
}
return 0;
}
Combinations over combinations approach:
import itertools as it
def min_diff_sets(data):
"""
Parameters:
- `data`: input list.
Return:
- min diff between sum of numbers in two sets
"""
if len(data) == 1:
return data[0]
s = sum(data)
# `a` is list of all possible combinations of all possible lengths (from 1
# to len(data) )
a = []
for i in range(1, len(data)):
a.extend(list(it.combinations(data, i)))
# `b` is list of all possible pairs (combinations) of all elements from `a`
b = it.combinations(a, 2)
# `c` is going to be final correct list of combinations.
# Let's apply 2 filters:
# 1. leave only pairs where: sum of all elements == sum(data)
# 2. leave only pairs where: flat list from pairs == data
c = filter(lambda x: sum(x[0])+sum(x[1])==s, b)
c = filter(lambda x: sorted([i for sub in x for i in sub])==sorted(data), c)
# `res` = [min_diff_between_sum_of_numbers_in_two_sets,
# ((set_1), (set_2))
# ]
res = sorted([(abs(sum(i[0]) - sum(i[1])), i) for i in c],
key=lambda x: x[0])
return min([i[0] for i in res])
if __name__ == '__main__':
assert min_diff_sets([10, 10]) == 0, "1st example"
assert min_diff_sets([10]) == 10, "2nd example"
assert min_diff_sets([5, 8, 13, 27, 14]) == 3, "3rd example"
assert min_diff_sets([5, 5, 6, 5]) == 1, "4th example"
assert min_diff_sets([12, 30, 30, 32, 42, 49]) == 9, "5th example"
assert min_diff_sets([1, 1, 1, 3]) == 0, "6th example"
The recursive approach is to generate all possible sums from all the values of array and to check
which solution is the most optimal one.
To generate sums we either include the i’th item in set 1 or don’t include, i.e., include in
set 2.
The time complexity is O(n*sum) for both time and space.T
public class MinimumSubsetSum {
static int dp[][];
public static int minDiffSubsets(int arr[], int i, int calculatedSum, int totalSum) {
if(dp[i][calculatedSum] != -1) return dp[i][calculatedSum];
/**
* If i=0, then the sum of one subset has been calculated as we have reached the last
* element. The sum of another subset is totalSum - calculated sum. We need to return the
* difference between them.
*/
if(i == 0) {
return Math.abs((totalSum - calculatedSum) - calculatedSum);
}
//Including the ith element
int iElementIncluded = minDiffSubsets(arr, i-1, arr[i-1] + calculatedSum,
totalSum);
//Excluding the ith element
int iElementExcluded = minDiffSubsets(arr, i-1, calculatedSum, totalSum);
int res = Math.min(iElementIncluded, iElementExcluded);
dp[i][calculatedSum] = res;
return res;
}
public static void util(int arr[]) {
int totalSum = 0;
int n = arr.length;
for(Integer e : arr) totalSum += e;
dp = new int[n+1][totalSum+1];
for(int i=0; i <= n; i++)
for(int j=0; j <= totalSum; j++)
dp[i][j] = -1;
int res = minDiffSubsets(arr, n, 0, totalSum);
System.out.println("The min difference between two subset is " + res);
}
public static void main(String[] args) {
util(new int[]{3, 1, 4, 2, 2, 1});
}
}
We can use Dynamic Programming (similar to the way we find if a set can be partitioned into two equal sum subsets). Then we find the max possible sum, which will be our first partition.
Second partition will be the difference of the total sum and firstSum.
Answer will be the difference of the first and second partitions.
public int minDiffernce(int set[]) {
int sum = 0;
int n = set.length;
for(int i=0; i<n; i++)
sum+=set[i];
//finding half of total sum, because min difference can be at max 0, if one subset reaches half
int target = sum/2;
boolean[][] dp = new boolean[n+1][target+1];//2
for(int i = 0; i<=n; i++)
dp[i][0] = true;
for(int i= 1; i<=n; i++){
for(int j = 1; j<=target;j++){
if(set[i-1]>j) dp[i][j] = dp[i-1][j];
else dp[i][j] = dp[i-1][j] || dp[i-1][j-set[i-1]];
}
}
// we now find the max sum possible starting from target
int firstPart = 0;
for(int j = target; j>=0; j--){
if(dp[n][j] == true) {
firstPart = j; break;
}
}
int secondPart = sum - firstPart;
return Math.abs(firstPart - secondPart);
}
One small change: reverse the order - start with the largest number and work down. This will minimize the error.
Are you sorting your subset into decending order or ascending order?
Think about it like this, the array {1, 3, 5, 8, 9, 25}
if you were to divide, you would have {1,8,9} =18 {3,5,25} =33
If it were sorted into descending order it would work out a lot better
{25,1}=26 {9,8,5,3}=25
So your solution is basically correct, it just needs to make sure to take the largest values first.
EDIT: Read tskuzzy's post. Mine does not work
This is a variation of the knapsack and subset sum problem.
In subset sum problem, given n positive integers and a value k and we have to find the sum of subset whose value is less than or equal to k.
In the above problem we have given an array, here we have to find the subset whose sum is less than or equal to total_sum(sum of array values).
So the
subset sum can be found using a variation in knapsack algorithm,by
taking profits as given array values. And the final answer is
total_sum-dp[n][total_sum/2]. Have a look at the below code for clear
understanding.
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int n;
cin>>n;
int arr[n],sum=0;
for(int i=1;i<=n;i++)
cin>>arr[i],sum+=arr[i];
int temp=sum/2;
int dp[n+1][temp+2];
for(int i=0;i<=n;i++)
{
for(int j=0;j<=temp;j++)
{
if(i==0 || j==0)
dp[i][j]=0;
else if(arr[i]<=j)
dp[i][j]=max(dp[i-1][j],dp[i-1][j-arr[i]]+arr[i]);
else
{
dp[i][j]=dp[i-1][j];
}
}
}
cout<<sum-2*dp[n][temp]<<endl;
}
This can be solve using BST.
First sort the array say arr1
To start create another arr2 with the last element of arr1 (remove this ele from arr1)
Now:Repeat the steps till no swap happens.
Check arr1 for an element which can be moved to arr2 using BST such that the diff is less MIN diff found till now.
if we find an element move this element to arr2 and go to step1 again.
if we don't find any element in above steps do steps 1 & 2 for arr2 & arr1.
i.e. now check if we have any element in arr2 which can be moved to arr1
continue steps 1-4 till we don't need any swap..
we get the solution.
Sample Java Code:
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
/**
* Divide an array so that the difference between these 2 is min
*
* #author shaikhjamir
*
*/
public class DivideArrayForMinDiff {
/**
* Create 2 arrays and try to find the element from 2nd one so that diff is
* min than the current one
*/
private static int sum(List<Integer> arr) {
int total = 0;
for (int i = 0; i < arr.size(); i++) {
total += arr.get(i);
}
return total;
}
private static int diff(ArrayList<Integer> arr, ArrayList<Integer> arr2) {
int diff = sum(arr) - sum(arr2);
if (diff < 0)
diff = diff * -1;
return diff;
}
private static int MIN = Integer.MAX_VALUE;
private static int binarySearch(int low, int high, ArrayList<Integer> arr1, int arr2sum) {
if (low > high || low < 0)
return -1;
int mid = (low + high) / 2;
int midVal = arr1.get(mid);
int sum1 = sum(arr1);
int resultOfMoveOrg = (sum1 - midVal) - (arr2sum + midVal);
int resultOfMove = (sum1 - midVal) - (arr2sum + midVal);
if (resultOfMove < 0)
resultOfMove = resultOfMove * -1;
if (resultOfMove < MIN) {
// lets do the swap
return mid;
}
// this is positive number greater than min
// which mean we should move left
if (resultOfMoveOrg < 0) {
// 1,10, 19 ==> 30
// 100
// 20, 110 = -90
// 29, 111 = -83
return binarySearch(low, mid - 1, arr1, arr2sum);
} else {
// resultOfMoveOrg > 0
// 1,5,10, 15, 19, 20 => 70
// 21
// For 10
// 60, 31 it will be 29
// now if we move 1
// 71, 22 ==> 49
// but now if we move 20
// 50, 41 ==> 9
return binarySearch(mid + 1, high, arr1, arr2sum);
}
}
private static int findMin(ArrayList<Integer> arr1) {
ArrayList<Integer> list2 = new ArrayList<>(arr1.subList(arr1.size() - 1, arr1.size()));
arr1.remove(arr1.size() - 1);
while (true) {
int index = binarySearch(0, arr1.size(), arr1, sum(list2));
if (index != -1) {
int val = arr1.get(index);
arr1.remove(index);
list2.add(val);
Collections.sort(list2);
MIN = diff(arr1, list2);
} else {
// now try for arr2
int index2 = binarySearch(0, list2.size(), list2, sum(arr1));
if (index2 != -1) {
int val = list2.get(index2);
list2.remove(index2);
arr1.add(val);
Collections.sort(arr1);
MIN = diff(arr1, list2);
} else {
// no switch in both the cases
break;
}
}
}
System.out.println("MIN==>" + MIN);
System.out.println("arr1==>" + arr1 + ":" + sum(arr1));
System.out.println("list2==>" + list2 + ":" + sum(list2));
return 0;
}
public static void main(String args[]) {
ArrayList<Integer> org = new ArrayList<>();
org = new ArrayList<>();
org.add(1);
org.add(2);
org.add(3);
org.add(7);
org.add(8);
org.add(10);
findMin(org);
}
}
you can use bits to solve this problem by looping over all the possible combinations using bits:
main algorithm:
for(int i = 0; i < 1<<n; i++) {
int s = 0;
for(int j = 0; j < n; j++) {
if(i & 1<<j) s += arr[j];
}
int curr = abs((total-s)-s);
ans = min(ans, curr);
}
use long long for greater inputs.
but here I found a recursive and dynamic programming solution and I used both the approaches to solve the question and both worked for greater inputs perfectly fine. Hope this helps :) link to solution
Please check this logic which I have written for this problem. It worked for few scenarios I checked. Please comment on the solution,
Approach :
Sort the main array and divide it into 2 teams.
Then start making the team equal by shift and swapping elements from one array to other, based on the conditions mentioned in the code.
If the difference is difference of sum is less than the minimum number of the larger array(array with bigger sum), then shift the elements from the bigger array to smaller array.Shifting happens with the condition, that element from the bigger array with value less than or equal to the difference.When all the elements from the bigger array is greater than the difference, the shifting stops and swapping happens. I m just swapping the last elements of the array (It can be made more efficient by finding which two elements to swap), but still this worked. Let me know if this logic failed in any scenario.
public class SmallestDifference {
static int sum1 = 0, sum2 = 0, diff, minDiff;
private static List<Integer> minArr1;
private static List<Integer> minArr2;
private static List<Integer> biggerArr;
/**
* #param args
*/
public static void main(String[] args) {
SmallestDifference sm = new SmallestDifference();
Integer[] array1 = { 2, 7, 1, 4, 5, 9, 10, 11 };
List<Integer> array = new ArrayList<Integer>();
for (Integer val : array1) {
array.add(val);
}
Collections.sort(array);
CopyOnWriteArrayList<Integer> arr1 = new CopyOnWriteArrayList<>(array.subList(0, array.size() / 2));
CopyOnWriteArrayList<Integer> arr2 = new CopyOnWriteArrayList<>(array.subList(array.size() / 2, array.size()));
diff = Math.abs(sm.getSum(arr1) - sm.getSum(arr2));
minDiff = array.get(0);
sm.updateSum(arr1, arr2);
System.out.println(arr1 + " : " + arr2);
System.out.println(sum1 + " - " + sum2 + " = " + diff + " : minDiff = " + minDiff);
int k = arr2.size();
biggerArr = arr2;
while (diff != 0 && k >= 0) {
while (diff != 0 && sm.findMin(biggerArr) < diff) {
sm.swich(arr1, arr2);
int sum1 = sm.getSum(arr1), sum2 = sm.getSum(arr2);
diff = Math.abs(sum1 - sum2);
if (sum1 > sum2) {
biggerArr = arr1;
} else {
biggerArr = arr2;
}
if (minDiff > diff || sm.findMin(biggerArr) > diff) {
minDiff = diff;
minArr1 = new CopyOnWriteArrayList<>(arr1);
minArr2 = new CopyOnWriteArrayList<>(arr2);
}
sm.updateSum(arr1, arr2);
System.out.println("Shifting : " + sum1 + " - " + sum2 + " = " + diff + " : minDiff = " + minDiff);
}
while (k >= 0 && minDiff > array.get(0) && minDiff != 0) {
sm.swap(arr1, arr2);
diff = Math.abs(sm.getSum(arr1) - sm.getSum(arr2));
if (minDiff > diff) {
minDiff = diff;
minArr1 = new CopyOnWriteArrayList<>(arr1);
minArr2 = new CopyOnWriteArrayList<>(arr2);
}
sm.updateSum(arr1, arr2);
System.out.println("Swapping : " + sum1 + " - " + sum2 + " = " + diff + " : minDiff = " + minDiff);
k--;
}
k--;
}
System.out.println(minArr1 + " : " + minArr2 + " = " + minDiff);
}
private void updateSum(CopyOnWriteArrayList<Integer> arr1, CopyOnWriteArrayList<Integer> arr2) {
SmallestDifference sm1 = new SmallestDifference();
sum1 = sm1.getSum(arr1);
sum2 = sm1.getSum(arr2);
}
private int findMin(List<Integer> biggerArr2) {
Integer min = biggerArr2.get(0);
for (Integer integer : biggerArr2) {
if(min > integer) {
min = integer;
}
}
return min;
}
private int getSum(CopyOnWriteArrayList<Integer> arr) {
int sum = 0;
for (Integer val : arr) {
sum += val;
}
return sum;
}
private void swap(CopyOnWriteArrayList<Integer> arr1, CopyOnWriteArrayList<Integer> arr2) {
int l1 = arr1.size(), l2 = arr2.size(), temp2 = arr2.get(l2 - 1), temp1 = arr1.get(l1 - 1);
arr1.remove(l1 - 1);
arr1.add(temp2);
arr2.remove(l2 - 1);
arr2.add(temp1);
System.out.println(arr1 + " : " + arr2);
}
private void swich(CopyOnWriteArrayList<Integer> arr1, CopyOnWriteArrayList<Integer> arr2) {
Integer e;
if (sum1 > sum2) {
e = this.findElementJustLessThanMinDiff(arr1);
arr1.remove(e);
arr2.add(e);
} else {
e = this.findElementJustLessThanMinDiff(arr2);
arr2.remove(e);
arr1.add(e);
}
System.out.println(arr1 + " : " + arr2);
}
private Integer findElementJustLessThanMinDiff(CopyOnWriteArrayList<Integer> arr1) {
Integer e = arr1.get(0);
int tempDiff = diff - e;
for (Integer integer : arr1) {
if (diff > integer && (diff - integer) < tempDiff) {
e = integer;
tempDiff = diff - e;
}
}
return e;
}
}
A possible solution here- https://stackoverflow.com/a/31228461/4955513
This Java program seems to solve this problem, provided one condition is fulfilled- that there is one and only one solution to the problem.
I'll convert this problem to subset sum problem
let's take array int[] A = { 10,20,15,5,25,33 };
it should be divided into {25 20 10} and { 33 20 } and answer is 55-53=2
Notations : SUM == sum of whole array
sum1 == sum of subset1
sum2 == sum of subset1
step 1: get sum of whole array SUM=108
step 2: whichever way we divide our array into two part one thing will remain true
sum1+ sum2= SUM
step 3: if our intention is to get minimum sum difference then sum1 and sum2 should be near SUM/2 (example sum1=54 and sum2=54 then diff=0 )
steon 4: let's try combinations
sum1 = 54 AND sum2 = 54 (not possible to divide like this)
sum1 = 55 AND sum2 = 53 (possible and our solution, should break here)
sum1 = 56 AND sum2 = 52
sum1 = 57 AND sum2 = 51 .......so on
pseudo code
SUM=Array.sum();
sum1 = SUM/2;
sum2 = SUM-sum1;
while(true){
if(subSetSuMProblem(A,sum1) && subSetSuMProblem(A,sum2){
print "possible"
break;
}
else{
sum1++;
sum2--;
}
}
Java code for the same
import java.util.ArrayList;
import java.util.List;
public class MinimumSumSubsetPrint {
public static void main(String[] args) {
int[] A = {10, 20, 15, 5, 25, 32};
int sum = 0;
for (int i = 0; i < A.length; i++) {
sum += A[i];
}
subsetSumDynamic(A, sum);
}
private static boolean subsetSumDynamic(int[] A, int sum) {
int n = A.length;
boolean[][] T = new boolean[n + 1][sum + 1];
// sum2[0][0]=true;
for (int i = 0; i <= n; i++) {
T[i][0] = true;
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= sum; j++) {
if (A[i - 1] > j) {
T[i][j] = T[i - 1][j];
} else {
T[i][j] = T[i - 1][j] || T[i - 1][j - A[i - 1]];
}
}
}
int sum1 = sum / 2;
int sum2 = sum - sum1;
while (true) {
if (T[n][sum1] && T[n][sum2]) {
printSubsets(T, sum1, n, A);
printSubsets(T, sum2, n, A);
break;
} else {
sum1 = sum1 - 1;
sum2 = sum - sum1;
System.out.println(sum1 + ":" + sum2);
}
}
return T[n][sum];
}
private static void printSubsets(boolean[][] T, int sum, int n, int[] A) {
List<Integer> sumvals = new ArrayList<Integer>();
int i = n;
int j = sum;
while (i > 0 && j > 0) {
if (T[i][j] == T[i - 1][j]) {
i--;
} else {
sumvals.add(A[i - 1]);
j = j - A[i - 1];
i--;
}
}
System.out.println();
for (int p : sumvals) {
System.out.print(p + " ");
}
System.out.println();
}
}
Here is recursive approach
def helper(arr,sumCal,sumTot,n):
if n==0:
return abs(abs(sumCal-sumTot)-sumCal)
return min(helper(arr,sumCal+arr[n-1],sumTot,n-1),helper(arr,sumCal,sumTot,n-1))
def minimum_subset_diff(arr,n):
sum=0
for i in range(n):
sum+=arr[i]
return helper(arr,0,sum,n)
Here is a Top down Dynamic approach to reduce the time complexity
dp=[[-1]*100 for i in range(100)]
def helper_dp(arr,sumCal,sumTot,n):
if n==0:
return abs(abs(sumCal-sumTot)-sumCal)
if dp[n][sumTot]!=-1:
return dp[n][sumTot]
return min(helper_dp(arr,sumCal+arr[n-1],sumTot,n-1),helper_dp(arr,sumCal,sumTot,n-1))
def minimum_subset_diff_dp(arr,n):
sum=0
for i in range(n):
sum+=arr[i]
return helper_dp(arr,0,sum,n)
int ModDiff(int a, int b)
{
if(a < b)return b - a;
return a-b;
}
int EqDiv(int *a, int l, int *SumI, int *SumE)
{
static int tc = 0;
int min = ModDiff(*SumI,*SumE);
for(int i = 0; i < l; i++)
{
swap(a,0,i);
a++;
int m1 = EqDiv(a, l-1, SumI,SumE);
a--;
swap(a,0,i);
*SumI = *SumI + a[i];
*SumE = *SumE - a[i];
swap(a,0,i);
a++;
int m2 = EqDiv(a,l-1, SumI,SumE);
a--;
swap(a,0,i);
*SumI = *SumI - a[i];
*SumE = *SumE + a[i];
min = min3(min,m1,m2);
}
return min;
}
call the function with SumI =0 and SumE= sumof all the elements in a.
This O(n!) solution does compute the way we can divide the given array into 2 parts such the difference is minimum.
But definitely not practical due to the n! time complexity looking to improve this using DP.
#include<bits/stdc++.h>
using namespace std;
bool ison(int i,int x)
{
if((i>>x) & 1)return true;
return false;
}
int main()
{
// cout<<"enter the number of elements : ";
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++)
cin>>a[i];
int sumarr1[(1<<n)-1];
int sumarr2[(1<<n)-1];
memset(sumarr1,0,sizeof(sumarr1));
memset(sumarr2,0,sizeof(sumarr2));
int index=0;
vector<int>v1[(1<<n)-1];
vector<int>v2[(1<<n)-1];
for(int i=1;i<(1<<n);i++)
{
for(int j=0;j<n;j++)
{
if(ison(i,j))
{
sumarr1[index]+=a[j];
v1[index].push_back(a[j]);
}
else
{
sumarr2[index]+=a[j];
v2[index].push_back(a[j]);
}
}index++;
}
int ans=INT_MAX;
int ii;
for(int i=0;i<index;i++)
{
if(abs(sumarr1[i]-sumarr2[i])<ans)
{
ii=i;
ans=abs(sumarr1[i]-sumarr2[i]);
}
}
cout<<"first partitioned array : ";
for(int i=0;i<v1[ii].size();i++)
{
cout<<v1[ii][i]<<" ";
}
cout<<endl;
cout<<"2nd partitioned array : ";
for(int i=0;i<v2[ii].size();i++)
{
cout<<v2[ii][i]<<" ";
}
cout<<endl;
cout<<"minimum difference is : "<<ans<<endl;
}
Many answers mentioned about getting an 'approximate' solution in a very acceptable time bound . But since it is asked in an interview , I dont expect they need an approximation algorithm. Also I dont expect they need a naive exponential algorithm either.
Coming to the problem , assuming the maximum value of sum of numbers is known , it can infact be solved in polynomial time using dynamic programming. Refer this link
https://people.cs.clemson.edu/~bcdean/dp_practice/dp_4.swf
HI I think This Problem can be solved in Linear Time on a sorted array , no Polynomial Time is required , rather than Choosing Next Element u can choose nest two Element and decide which side which element to go. in This Way
in this way minimize the difference, let suppose
{0,1,5,6} ,
choose {0,1}
{0} , {1}
choose 5,6
{0,6}, {1,5}
but still that is not exact solution , now at the end there will be difference of sum in 2 array let suppose x
but there can be better solution of difference of (less than x)
for that Find again 1 greedy approach over sorted half sized array
and move x/2(or nearby) element from 1 set to another or exchange element of(difference x/2) so that difference can be minimized***
Given array of n integers and given a number X, find all the unique pairs of elements (a,b), whose summation is equal to X.
The following is my solution, it is O(nLog(n)+n), but I am not sure whether or not it is optimal.
int main(void)
{
int arr [10] = {1,2,3,4,5,6,7,8,9,0};
findpair(arr, 10, 7);
}
void findpair(int arr[], int len, int sum)
{
std::sort(arr, arr+len);
int i = 0;
int j = len -1;
while( i < j){
while((arr[i] + arr[j]) <= sum && i < j)
{
if((arr[i] + arr[j]) == sum)
cout << "(" << arr[i] << "," << arr[j] << ")" << endl;
i++;
}
j--;
while((arr[i] + arr[j]) >= sum && i < j)
{
if((arr[i] + arr[j]) == sum)
cout << "(" << arr[i] << "," << arr[j] << ")" << endl;
j--;
}
}
}
There are 3 approaches to this solution:
Let the sum be T and n be the size of array
Approach 1:
The naive way to do this would be to check all combinations (n choose 2). This exhaustive search is O(n2).
Approach 2:
A better way would be to sort the array. This takes O(n log n)
Then for each x in array A,
use binary search to look for T-x. This will take O(nlogn).
So, overall search is O(n log n)
Approach 3 :
The best way
would be to insert every element into a hash table (without sorting). This takes O(n) as constant time insertion.
Then for every x,
we can just look up its complement, T-x, which is O(1).
Overall the run time of this approach is O(n).
You can refer more here.Thanks.
# Let arr be the given array.
# And K be the give sum
for i=0 to arr.length - 1 do
# key is the element and value is its index.
hash(arr[i]) = i
end-for
for i=0 to arr.length - 1 do
# if K-th element exists and it's different then we found a pair
if hash(K - arr[i]) != i
print "pair i , hash(K - arr[i]) has sum K"
end-if
end-for
Implementation in Java : Using codaddict's algorithm (Maybe slightly different)
import java.util.HashMap;
public class ArrayPairSum {
public static void main(String[] args) {
int []a = {2,45,7,3,5,1,8,9};
printSumPairs(a,10);
}
public static void printSumPairs(int []input, int k){
Map<Integer, Integer> pairs = new HashMap<Integer, Integer>();
for(int i=0;i<input.length;i++){
if(pairs.containsKey(input[i]))
System.out.println(input[i] +", "+ pairs.get(input[i]));
else
pairs.put(k-input[i], input[i]);
}
}
}
For input = {2,45,7,3,5,1,8,9} and if Sum is 10
Output pairs:
3,7
8,2
9,1
Some notes about the solution :
We iterate only once through the array --> O(n) time
Insertion and lookup time in Hash is O(1).
Overall time is O(n), although it uses extra space in terms of hash.
Solution in java. You can add all the String elements to an ArrayList of strings and return the list. Here I am just printing it out.
void numberPairsForSum(int[] array, int sum) {
HashSet<Integer> set = new HashSet<Integer>();
for (int num : array) {
if (set.contains(sum - num)) {
String s = num + ", " + (sum - num) + " add up to " + sum;
System.out.println(s);
}
set.add(num);
}
}
Python Implementation:
import itertools
list = [1, 1, 2, 3, 4, 5,]
uniquelist = set(list)
targetsum = 5
for n in itertools.combinations(uniquelist, 2):
if n[0] + n[1] == targetsum:
print str(n[0]) + " + " + str(n[1])
Output:
1 + 4
2 + 3
C++11, run time complexity O(n):
#include <vector>
#include <unordered_map>
#include <utility>
std::vector<std::pair<int, int>> FindPairsForSum(
const std::vector<int>& data, const int& sum)
{
std::unordered_map<int, size_t> umap;
std::vector<std::pair<int, int>> result;
for (size_t i = 0; i < data.size(); ++i)
{
if (0 < umap.count(sum - data[i]))
{
size_t j = umap[sum - data[i]];
result.push_back({data[i], data[j]});
}
else
{
umap[data[i]] = i;
}
}
return result;
}
Here is a solution witch takes into account duplicate entries. It is written in javascript and assumes array is sorted. The solution runs in O(n) time and does not use any extra memory aside from variable.
var count_pairs = function(_arr,x) {
if(!x) x = 0;
var pairs = 0;
var i = 0;
var k = _arr.length-1;
if((k+1)<2) return pairs;
var halfX = x/2;
while(i<k) {
var curK = _arr[k];
var curI = _arr[i];
var pairsThisLoop = 0;
if(curK+curI==x) {
// if midpoint and equal find combinations
if(curK==curI) {
var comb = 1;
while(--k>=i) pairs+=(comb++);
break;
}
// count pair and k duplicates
pairsThisLoop++;
while(_arr[--k]==curK) pairsThisLoop++;
// add k side pairs to running total for every i side pair found
pairs+=pairsThisLoop;
while(_arr[++i]==curI) pairs+=pairsThisLoop;
} else {
// if we are at a mid point
if(curK==curI) break;
var distK = Math.abs(halfX-curK);
var distI = Math.abs(halfX-curI);
if(distI > distK) while(_arr[++i]==curI);
else while(_arr[--k]==curK);
}
}
return pairs;
}
I solved this during an interview for a large corporation. They took it but not me.
So here it is for everyone.
Start at both side of the array and slowly work your way inwards making sure to count duplicates if they exist.
It only counts pairs but can be reworked to
find the pairs
find pairs < x
find pairs > x
Enjoy!
O(n)
def find_pairs(L,sum):
s = set(L)
edgeCase = sum/2
if L.count(edgeCase) ==2:
print edgeCase, edgeCase
s.remove(edgeCase)
for i in s:
diff = sum-i
if diff in s:
print i, diff
L = [2,45,7,3,5,1,8,9]
sum = 10
find_pairs(L,sum)
Methodology: a + b = c, so instead of looking for (a,b) we look for a = c -
b
Implementation in Java : Using codaddict's algorithm:
import java.util.Hashtable;
public class Range {
public static void main(String[] args) {
// TODO Auto-generated method stub
Hashtable mapping = new Hashtable();
int a[]= {80,79,82,81,84,83,85};
int k = 160;
for (int i=0; i < a.length; i++){
mapping.put(a[i], i);
}
for (int i=0; i < a.length; i++){
if (mapping.containsKey(k - a[i]) && (Integer)mapping.get(k-a[i]) != i){
System.out.println(k-a[i]+", "+ a[i]);
}
}
}
}
Output:
81, 79
79, 81
If you want duplicate pairs (eg: 80,80) also then just remove && (Integer)mapping.get(k-a[i]) != i from the if condition and you are good to go.
Just attended this question on HackerRank and here's my 'Objective C' Solution:
-(NSNumber*)sum:(NSArray*) a andK:(NSNumber*)k {
NSMutableDictionary *dict = [NSMutableDictionary dictionary];
long long count = 0;
for(long i=0;i<a.count;i++){
if(dict[a[i]]) {
count++;
NSLog(#"a[i]: %#, dict[array[i]]: %#", a[i], dict[a[i]]);
}
else{
NSNumber *calcNum = #(k.longLongValue-((NSNumber*)a[i]).longLongValue);
dict[calcNum] = a[i];
}
}
return #(count);
}
Hope it helps someone.
this is the implementation of O(n*lg n) using binary search implementation inside a loop.
#include <iostream>
using namespace std;
bool *inMemory;
int pairSum(int arr[], int n, int k)
{
int count = 0;
if(n==0)
return count;
for (int i = 0; i < n; ++i)
{
int start = 0;
int end = n-1;
while(start <= end)
{
int mid = start + (end-start)/2;
if(i == mid)
break;
else if((arr[i] + arr[mid]) == k && !inMemory[i] && !inMemory[mid])
{
count++;
inMemory[i] = true;
inMemory[mid] = true;
}
else if(arr[i] + arr[mid] >= k)
{
end = mid-1;
}
else
start = mid+1;
}
}
return count;
}
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
inMemory = new bool[10];
for (int i = 0; i < 10; ++i)
{
inMemory[i] = false;
}
cout << pairSum(arr, 10, 11) << endl;
return 0;
}
In python
arr = [1, 2, 4, 6, 10]
diff_hash = {}
expected_sum = 3
for i in arr:
if diff_hash.has_key(i):
print i, diff_hash[i]
key = expected_sum - i
diff_hash[key] = i
Nice solution from Codeaddict. I took the liberty of implementing a version of it in Ruby:
def find_sum(arr,sum)
result ={}
h = Hash[arr.map {|i| [i,i]}]
arr.each { |l| result[l] = sum-l if h[sum-l] && !result[sum-l] }
result
end
To allow duplicate pairs (1,5), (5,1) we just have to remove the && !result[sum-l] instruction
Here is Java code for three approaches:
1. Using Map O(n), HashSet can also be used here.
2. Sort array and then use BinarySearch to look for complement O(nLog(n))
3. Traditional BruteForce two loops O(n^2)
public class PairsEqualToSum {
public static void main(String[] args) {
int a[] = {1,10,5,8,2,12,6,4};
findPairs1(a,10);
findPairs2(a,10);
findPairs3(a,10);
}
//Method1 - O(N) use a Map to insert values as keys & check for number's complement in map
static void findPairs1(int[]a, int sum){
Map<Integer, Integer> pairs = new HashMap<Integer, Integer>();
for(int i=0; i<a.length; i++){
if(pairs.containsKey(sum-a[i]))
System.out.println("("+a[i]+","+(sum-a[i])+")");
else
pairs.put(a[i], 0);
}
}
//Method2 - O(nlog(n)) using Sort
static void findPairs2(int[]a, int sum){
Arrays.sort(a);
for(int i=0; i<a.length/2; i++){
int complement = sum - a[i];
int foundAtIndex = Arrays.binarySearch(a,complement);
if(foundAtIndex >0 && foundAtIndex != i) //to avoid situation where binarySearch would find the original and not the complement like "5"
System.out.println("("+a[i]+","+(sum-a[i])+")");
}
}
//Method 3 - Brute Force O(n^2)
static void findPairs3(int[]a, int sum){
for(int i=0; i<a.length; i++){
for(int j=i; j<a.length;j++){
if(a[i]+a[j] == sum)
System.out.println("("+a[i]+","+a[j]+")");
}
}
}
}
A Simple program in java for arrays having unique elements:
import java.util.*;
public class ArrayPairSum {
public static void main(String[] args) {
int []a = {2,4,7,3,5,1,8,9,5};
sumPairs(a,10);
}
public static void sumPairs(int []input, int k){
Set<Integer> set = new HashSet<Integer>();
for(int i=0;i<input.length;i++){
if(set.contains(input[i]))
System.out.println(input[i] +", "+(k-input[i]));
else
set.add(k-input[i]);
}
}
}
A simple Java code snippet for printing the pairs below:
public static void count_all_pairs_with_given_sum(int arr[], int S){
if(arr.length < 2){
return;
}
HashSet values = new HashSet(arr.length);
for(int value : arr)values.add(value);
for(int value : arr){
int difference = S - value;
if(values.contains(difference) && value<difference){
System.out.printf("(%d, %d) %n", value, difference);
}
}
}
Another solution in Swift: the idea is to create an hash that store values of (sum - currentValue) and compare this to the current value of the loop. The complexity is O(n).
func findPair(list: [Int], _ sum: Int) -> [(Int, Int)]? {
var hash = Set<Int>() //save list of value of sum - item.
var dictCount = [Int: Int]() //to avoid the case A*2 = sum where we have only one A in the array
var foundKeys = Set<Int>() //to avoid duplicated pair in the result.
var result = [(Int, Int)]() //this is for the result.
for item in list {
//keep track of count of each element to avoid problem: [2, 3, 5], 10 -> result = (5,5)
if (!dictCount.keys.contains(item)) {
dictCount[item] = 1
} else {
dictCount[item] = dictCount[item]! + 1
}
//if my hash does not contain the (sum - item) value -> insert to hash.
if !hash.contains(sum-item) {
hash.insert(sum-item)
}
//check if current item is the same as another hash value or not, if yes, return the tuple.
if hash.contains(item) &&
(dictCount[item] > 1 || sum != item*2) // check if we have item*2 = sum or not.
{
if !foundKeys.contains(item) && !foundKeys.contains(sum-item) {
foundKeys.insert(item) //add to found items in order to not to add duplicated pair.
result.append((item, sum-item))
}
}
}
return result
}
//test:
let a = findPair([2,3,5,4,1,7,6,8,9,5,3,3,3,3,3,3,3,3,3], 14) //will return (8,6) and (9,5)
My Solution - Java - Without duplicates
public static void printAllPairSum(int[] a, int x){
System.out.printf("printAllPairSum(%s,%d)\n", Arrays.toString(a),x);
if(a==null||a.length==0){
return;
}
int length = a.length;
Map<Integer,Integer> reverseMapOfArray = new HashMap<>(length,1.0f);
for (int i = 0; i < length; i++) {
reverseMapOfArray.put(a[i], i);
}
for (int i = 0; i < length; i++) {
Integer j = reverseMapOfArray.get(x - a[i]);
if(j!=null && i<j){
System.out.printf("a[%d] + a[%d] = %d + %d = %d\n",i,j,a[i],a[j],x);
}
}
System.out.println("------------------------------");
}
This prints the pairs and avoids duplicates using bitwise manipulation.
public static void findSumHashMap(int[] arr, int key) {
Map<Integer, Integer> valMap = new HashMap<Integer, Integer>();
for(int i=0;i<arr.length;i++)
valMap.put(arr[i], i);
int indicesVisited = 0;
for(int i=0;i<arr.length;i++) {
if(valMap.containsKey(key - arr[i]) && valMap.get(key - arr[i]) != i) {
if(!((indicesVisited & ((1<<i) | (1<<valMap.get(key - arr[i])))) > 0)) {
int diff = key-arr[i];
System.out.println(arr[i] + " " +diff);
indicesVisited = indicesVisited | (1<<i) | (1<<valMap.get(key - arr[i]));
}
}
}
}
I bypassed the bit manuplation and just compared the index values. This is less than the loop iteration value (i in this case). This will not print the duplicate pairs and duplicate array elements also.
public static void findSumHashMap(int[] arr, int key) {
Map<Integer, Integer> valMap = new HashMap<Integer, Integer>();
for (int i = 0; i < arr.length; i++) {
valMap.put(arr[i], i);
}
for (int i = 0; i < arr.length; i++) {
if (valMap.containsKey(key - arr[i])
&& valMap.get(key - arr[i]) != i) {
if (valMap.get(key - arr[i]) < i) {
int diff = key - arr[i];
System.out.println(arr[i] + " " + diff);
}
}
}
}
in C#:
int[] array = new int[] { 1, 5, 7, 2, 9, 8, 4, 3, 6 }; // given array
int sum = 10; // given sum
for (int i = 0; i <= array.Count() - 1; i++)
if (array.Contains(sum - array[i]))
Console.WriteLine("{0}, {1}", array[i], sum - array[i]);
One Solution can be this, but not optimul (The complexity of this code is O(n^2)):
public class FindPairsEqualToSum {
private static int inputSum = 0;
public static List<String> findPairsForSum(int[] inputArray, int sum) {
List<String> list = new ArrayList<String>();
List<Integer> inputList = new ArrayList<Integer>();
for (int i : inputArray) {
inputList.add(i);
}
for (int i : inputArray) {
int tempInt = sum - i;
if (inputList.contains(tempInt)) {
String pair = String.valueOf(i + ", " + tempInt);
list.add(pair);
}
}
return list;
}
}
A simple python version of the code that find a pair sum of zero and can be modify to find k:
def sumToK(lst):
k = 0 # <- define the k here
d = {} # build a dictionary
# build the hashmap key = val of lst, value = i
for index, val in enumerate(lst):
d[val] = index
# find the key; if a key is in the dict, and not the same index as the current key
for i, val in enumerate(lst):
if (k-val) in d and d[k-val] != i:
return True
return False
The run time complexity of the function is O(n) and Space: O(n) as well.
public static int[] f (final int[] nums, int target) {
int[] r = new int[2];
r[0] = -1;
r[1] = -1;
int[] vIndex = new int[0Xfff];
for (int i = 0; i < nums.length; i++) {
int delta = 0Xff;
int gapIndex = target - nums[i] + delta;
if (vIndex[gapIndex] != 0) {
r[0] = vIndex[gapIndex];
r[1] = i + 1;
return r;
} else {
vIndex[nums[i] + delta] = i + 1;
}
}
return r;
}
less than o(n) solution will be=>
function(array,k)
var map = {};
for element in array
map(element) = true;
if(map(k-element))
return {k,element}
Solution in Python using list comprehension
f= [[i,j] for i in list for j in list if j+i==X];
O(N2)
also gives two ordered pairs- (a,b) and (b,a) as well
I can do it in O(n). Let me know when you want the answer. Note it involves simply traversing the array once with no sorting, etc... I should mention too that it exploits commutativity of addition and doesn't use hashes but wastes memory.
using System;
using System.Collections.Generic;
/*
An O(n) approach exists by using a lookup table. The approach is to store the value in a "bin" that can easily be looked up(e.g., O(1)) if it is a candidate for an appropriate sum.
e.g.,
for each a[k] in the array we simply put the it in another array at the location x - a[k].
Suppose we have [0, 1, 5, 3, 6, 9, 8, 7] and x = 9
We create a new array,
indexes value
9 - 0 = 9 0
9 - 1 = 8 1
9 - 5 = 4 5
9 - 3 = 6 3
9 - 6 = 3 6
9 - 9 = 0 9
9 - 8 = 1 8
9 - 7 = 2 7
THEN the only values that matter are the ones who have an index into the new table.
So, say when we reach 9 or equal we see if our new array has the index 9 - 9 = 0. Since it does we know that all the values it contains will add to 9. (note in this cause it's obvious there is only 1 possible one but it might have multiple index values in it which we need to store).
So effectively what we end up doing is only having to move through the array once. Because addition is commutative we will end up with all the possible results.
For example, when we get to 6 we get the index into our new table as 9 - 6 = 3. Since the table contains that index value we know the values.
This is essentially trading off speed for memory.
*/
namespace sum
{
class Program
{
static void Main(string[] args)
{
int num = 25;
int X = 10;
var arr = new List<int>();
for(int i = 0; i <= num; i++) arr.Add((new Random((int)(DateTime.Now.Ticks + i*num))).Next(0, num*2));
Console.Write("["); for (int i = 0; i < num - 1; i++) Console.Write(arr[i] + ", "); Console.WriteLine(arr[arr.Count-1] + "] - " + X);
var arrbrute = new List<Tuple<int,int>>();
var arrfast = new List<Tuple<int,int>>();
for(int i = 0; i < num; i++)
for(int j = i+1; j < num; j++)
if (arr[i] + arr[j] == X)
arrbrute.Add(new Tuple<int, int>(arr[i], arr[j]));
int M = 500;
var lookup = new List<List<int>>();
for(int i = 0; i < 1000; i++) lookup.Add(new List<int>());
for(int i = 0; i < num; i++)
{
// Check and see if we have any "matches"
if (lookup[M + X - arr[i]].Count != 0)
{
foreach(var j in lookup[M + X - arr[i]])
arrfast.Add(new Tuple<int, int>(arr[i], arr[j]));
}
lookup[M + arr[i]].Add(i);
}
for(int i = 0; i < arrbrute.Count; i++)
Console.WriteLine(arrbrute[i].Item1 + " + " + arrbrute[i].Item2 + " = " + X);
Console.WriteLine("---------");
for(int i = 0; i < arrfast.Count; i++)
Console.WriteLine(arrfast[i].Item1 + " + " + arrfast[i].Item2 + " = " + X);
Console.ReadKey();
}
}
}
I implemented logic in Scala with out a Map. It gives duplicate pairs since the counter loops thru entire elements of the array. If duplicate pairs are needed, you can simply return the value pc
val arr = Array[Int](8, 7, 2, 5, 3, 1, 5)
val num = 10
var pc = 0
for(i <- arr.indices) {
if(arr.contains(Math.abs(arr(i) - num))) pc += 1
}
println(s"Pairs: ${pc/2}")
It is working with duplicates values in the array as well.
GOLANG Implementation
func findPairs(slice1 []int, sum int) [][]int {
pairMap := make(map[int]int)
var SliceOfPairs [][]int
for i, v := range slice1 {
if valuei, ok := pairMap[v]; ok {
//fmt.Println("Pair Found", i, valuei)
SliceOfPairs = append(SliceOfPairs, []int{i, valuei})
} else {
pairMap[sum-v] = i
}
}
return SliceOfPairs
}
function findPairOfNumbers(arr, targetSum) {
arr = arr.sort();
var low = 0, high = arr.length - 1, sum, result = [];
while(low < high) {
sum = arr[low] + arr[high];
if(sum < targetSum)
low++;
else if(sum > targetSum)
high--;
else if(sum === targetSum) {
result.push({val1: arr[low], val2: arr[high]});
high--;
}
}
return (result || false);
}
var pairs = findPairOfNumbers([1,2,3,4,5,6,7,8,9,0], 7);
if(pairs.length) {
console.log(pairs);
} else {
console.log("No pair of numbers found that sums to " + 7);
}