Maximum number that can be formed from the given digits - algorithm

Given a Integer, find the maximum number that can be formed from the digits.
Input : 8754365
output : 8765543
I told solution in $O(n logn)$. He asked me optimize further.
We can use Hash Table to optimize further, $\rightarrow$ O(n)
Algorithm:
1. Take a hash table with size 10 (0 to 9).
2. Store the count of every digit from 0 to 9.
3. Print the index of the Hash table with respect to digit count in the reverse direction (from 9 to 0).
Example:
Hash table after digit count for 8754365 $\rightarrow$ (0 0 0 1 1 2 1 1 1 0)
Print the index of the hash table with respect to their count in reverse order $\rightarrow$ 8765543
Time Complexity : O(n)
Correct me if I am wrong.

The following greedy code does this in O(n) time. Where n is the number of digits in the number.
int num = 8756404;
int[] times = new int[10];
while(true){
if(num==0){
break;
}
int val = num%10;
times[val]++;
num /= 10;
}
for(int i=9; i>=0; i--){
for(int j=0; j<times[i]; j++){
System.out.print(i);
}
}
It works by counting the number of occurences of each of the digits in the input number. Then printing each number the number of times it was in the input number in reverse order, ie. starting from 9 to 0.

RunTime: 00:00:00.01
public int Assignment(int number)
{
// Consider that int.MaxValue equals to 2147483647
var siblingString = String.Join("", number.ToString().ToCharArray().OrderByDescending(n => n));
int sibling = -1;
if (!int.TryParse(siblingString, out sibling) || sibling > 100000000)
{
return -1;
}
return sibling;
}
Performances tested with the following code:
static void Main()
{
Stopwatch stopWatch = new Stopwatch();
stopWatch.Start();
var result = AssignmentOne(2147483646);
stopWatch.Stop();
TimeSpan ts = stopWatch.Elapsed;
string elapsedTime = String.Format("{0:00}:{1:00}:{2:00}.{3:00}", ts.Hours, ts.Minutes, ts.Seconds, ts.Milliseconds / 10);
Console.WriteLine("RunTime " + elapsedTime);
}

Related

greatest sum sub array such that each element is less than or equal to X

Given an array A with N integers we need to find the highest sum of sub array such that each element is less than or equal to given integer X
Example : Let N=8 and array be [3 2 2 3 1 1 1 3] . Now if x=2 then answer is 4 by summing A[2] + A[3] if we consider 1 base indexing . How to do this question in O(N) or O(N*logN)
Currently am having O(N^2) approach by checking each possible subarray. How to reduce the complexity ?
You can use the fact that if some array contain integers only less than or equal to X, then all its subarrays also have this property. Lets find for each index i the greatest possible sum of subarray, ending at i (sub_sum).
sub_sum[i] = 0, if array[i] > X
sub_sum[i] = max(array[i], sub_sum[i - 1] + array[i]), otherwise
Initial conditions are:
sub_sum[1] = 0, if array[1] > X
sub_sum[1] = max(array[1], 0), otherwise
You can compute all sub_sum values in one loop using the formulas above. The answer to your question is the maximum in sub_sum array. The computation complexity is O(n).
I am just giving you a simple step by step approach
Time complexity O(n) Space complexity O(n)
1. Input array=A[1..n] and x be the element and ans= -INF
(smallest int value)
2. Take another array B[1..n]={0,0,...0}.
3. For i=1 to n
if(A[i]<=x)
B[i]=1;
sum=0;
4. For i=1 to n
if(B[i])
sum+=A[i];
else
{
ans=maximum of(sum,ans);
sum= 0;
}
5. ans is the output.
Time complexity O(n) Space complexity O(1)
Note ans= -INF;(smallest int value)
sum=0;
1. for(i=1;i<=n;i++)
//get input Ai in variable a(temporary int variable to store the elements)
if(a<=x)
sum+=a
else
{
ans=max of (ans,sum);
sum= 0;
}
2. ans will be the output.
An O(n) C++ code:
const int INF = 2147483647;
int A[] = {3,2,2,3,1,1,1,3};
int ArraySize = 8;
int X = 2;
int max = -INF; //currenly max
int si = -1; //starting index
int ei = -1; //ending index
int tmax = 0; //temp currenly max
int tsi = -1; //temp starting index
int tei = -1; //temp ending index
for (int i = 0;i<ArraySize;i++) {
if (A[i]<=X) {
tmax+=A[i];
if (tsi==-1) tsi = i;
}
else {
tei = i-1;
if (tmax>max) {
max = tmax;
si = tsi;
ei = tei;
}
tsi = -1;
tei = -1;
tmax = 0;
}
}
cout<<"Max is: "<<max<<" starting from "<<si<<" ending to "<<ei<<"\n";

Is it possible to develop a recursive word wrap algorithm?

I want to develop a recursive word wrap algorithm that takes a specified string and wrap length (the maximum number of characters on one line) to return a wrapped output at the input length. I don't want it to break apart words. So for example, This is the first paragraph that you need to input with length 20 returns as:
This is the first
paragraph that you
need to input
I already have a dynamic programming (bottom-up) solution implemented, but I was wondering if it's possible to write an algorithm to do this using just recursion (top-down) instead? I'd also like to memoize it if I can. Please don't give me any runnable code... I"m just wondering about ideas/pseudocode.
Something like the pseudocode below should work. (I'm sure we'll get comments if I made a mistake!)
function Wrap(the_text,line_len)
if length(the_text) > line_len then
text_bit = the first few words of the_text, keeping their length shorter than line_len
remove text_bit from the beginning of the_text
return text_bit + linefeed + Wrap(the_text, line_len)
else
return the_text
end if
end function
import java.lang.Math;
public int RCS(int[] l , int n , int m , int index) {
// first base condition - if index gets beyond the array 'l' , then return 0;
if (index > n - 1) return 0;
// second base condition - if index is the last word i.e there is only one word left in the
// array to be inserted in the line then return the cost if added in that line.
if (index == n - 1) return (m - l[n - 1]) * (m - l[n - 1]) * (m - l[n - 1]);
// make a global cost variable to be returned
int cost = Integer.MAX_VALUE;
// Here , we try to select words from the array and apply RCS on the rest of the array.
// From index to last element , we iteratvely select first , or first two and so on.
for (int i = index ; i < n ; i++) {
int current_space_sum = 0 ;
// we add the length of the selected word. We have selected words in array from index to i.
for (int k = index ; k <= i ; k++) {
current_space_sum = current_space_sum + l[k] ;
}
// Adding the space between the words choses. If 2 words are chosen , there is one space and so on
current_space_sum = current_space_sum + i - index;
// If the length of the chosen words is greater than the line can accept , no need of looking beyond.
if (current_space_sum > m) break;
// Iteratively find the minimum cost
cost = Math.min(cost , (m - current_space_sum) * (m - current_space_sum) * (m - current_space_sum) + RCS(l , n , m , i + 1));
}
return cost;
}
public static void main(String[] args) {
WordWrap w = new WordWrap();
int[] l = {3, 2 , 2 , 5};
int n = l.length;
int m = 6;
int result = w.RCS(l , n , m , 0);
System.out.println(result);
}
The below code will help you to get the optimal cost for that problem.
#include<bits/stdc++.h>
using namespace std;
// method to get the optimal cost
int findOptimalCost(int *arr, int s, int e,int lineLength,map<pair<int,int>,int>dp) {
if(s>=e) // return 0 for the last line because we are not calculating the last line space
return 0;
if(dp.find({s,e}) != dp.end()) { // return cost if we already calculate
return dp[{s,e}];
}
int minCost = INT_MAX;
for(int i=s;i<=e;i++) {
int sum = 0,space=i-s;
for(int j =s; j<=i; j++)
sum += arr[j]; // add the word length
sum += space; // add the space for words (if 2 word then we will count 1 space )
int cost;
if(sum<=lineLength)
cost = (lineLength-sum)*(lineLength-sum) + findOptimalCost(arr,s+1+space,e,lineLength,dp); // calculate the cost for perticular line and call for rest line
if(minCost > cost) {
minCost = cost; // update the minCost variable if the latest cost is less then the previous calculated cost
}
}
return dp[{s,e}] = minCost; // store the minimum cost for particular line and return
}
int main()
{
//code
int len = 4; // total word in the list
int arr[] = {3,2,2,5}; // let us assume the length of word
int lineLength = 6; // size of max line length
map<pair<int,int>,int> dp;
cout<<findOptimalCost(arr,0,len-1,lineLength,dp)<<endl;
return 0;
}

How to perform K-swap operations on an N-digit integer to get maximum possible number

I recently went through an interview and was asked this question. Let me explain the question properly:
Given a number M (N-digit integer) and K number of swap operations(a swap
operation can swap 2 digits), devise an algorithm to get the maximum
possible integer?
Examples:
M = 132 K = 1 output = 312
M = 132 K = 2 output = 321
M = 7899 k = 2 output = 9987
My solution ( algorithm in pseudo-code). I used a max-heap to get the maximum digit out of N-digits in each of the K-operations and then suitably swapping it.
for(int i = 0; i<K; i++)
{
int max_digit_currently = GetMaxFromHeap();
// The above function GetMaxFromHeap() pops out the maximum currently and deletes it from heap
int index_to_swap_with = GetRightMostOccurenceOfTheDigitObtainedAbove();
// This returns me the index of the digit obtained in the previous function
// .e.g If I have 436659 and K=2 given,
// then after K=1 I'll have 936654 and after K=2, I should have 966354 and not 963654.
// Now, the swap part comes. Here the gotcha is, say with the same above example, I have K=3.
// If I do GetMaxFromHeap() I'll get 6 when K=3, but I should not swap it,
// rather I should continue for next iteration and
// get GetMaxFromHeap() to give me 5 and then get 966534 from 966354.
if (Value_at_index_to_swap == max_digit_currently)
continue;
else
DoSwap();
}
Time complexity: O(K*( N + log_2(N) ))
// K-times [log_2(N) for popping out number from heap & N to get the rightmost index to swap with]
The above strategy fails in this example:
M = 8799 and K = 2
Following my strategy, I'll get M = 9798 after K=1 and M = 9978 after K=2. However, the maximum I can get is M = 9987 after K=2.
What did I miss?
Also suggest other ways to solve the problem & ways to optimize my solution.
I think the missing part is that, after you've performed the K swaps as in the algorithm described by the OP, you're left with some numbers that you can swap between themselves. For example, for the number 87949, after the initial algorithm we would get 99748. However, after that we can swap 7 and 8 "for free", i.e. not consuming any of the K swaps. This would mean "I'd rather not swap the 7 with the second 9 but with the first".
So, to get the max number, one would perform the algorithm described by the OP and remember the numbers which were moved to the right, and the positions to which they were moved. Then, sort these numbers in decreasing order and put them in the positions from left to right.
This is something like a separation of the algorithm in two phases - in the first one, you choose which numbers should go in the front to maximize the first K positions. Then you determine the order in which you would have swapped them with the numbers whose positions they took, so that the rest of the number is maximized as well.
Not all the details are clear, and I'm not 100% sure it handles all cases correctly, so if anyone can break it - go ahead.
This is a recursive function, which sorts the possible swap values for each (current-max) digit:
function swap2max(string, K) {
// the recursion end:
if (string.length==0 || K==0)
return string
m = getMaxDigit(string)
// an array of indices of the maxdigits to swap in the string
indices = []
// a counter for the length of that array, to determine how many chars
// from the front will be swapped
len = 0
// an array of digits to be swapped
front = []
// and the index of the last of those:
right = 0
// get those indices, in a loop with 2 conditions:
// * just run backwards through the string, until we meet the swapped range
// * no more swaps than left (K)
for (i=string.length; i-->right && len<K;)
if (m == string[i])
// omit digits that are already in the right place
while (right<=i && string[right] == m)
right++
// and when they need to be swapped
if (i>=right)
front.push(string[right++])
indices.push(i)
len++
// sort the digits to swap with
front.sort()
// and swap them
for (i=0; i<len; i++)
string.setCharAt(indices[i], front[i])
// the first len digits are the max ones
// the rest the result of calling the function on the rest of the string
return m.repeat(right) + swap2max(string.substr(right), K-len)
}
This is all pseudocode, but converts fairly easy to other languages. This solution is nonrecursive and operates in linear worst case and average case time.
You are provided with the following functions:
function k_swap(n, k1, k2):
temp = n[k1]
n[k1] = n[k2]
n[k2] = temp
int : operator[k]
// gets or sets the kth digit of an integer
property int : magnitude
// the number of digits in an integer
You could do something like the following:
int input = [some integer] // input value
int digitcounts[10] = {0, ...} // all zeroes
int digitpositions[10] = {0, ...) // all zeroes
bool filled[input.magnitude] = {false, ...) // all falses
for d = input[i = 0 => input.magnitude]:
digitcounts[d]++ // count number of occurrences of each digit
digitpositions[0] = 0;
for i = 1 => input.magnitude:
digitpositions[i] = digitpositions[i - 1] + digitcounts[i - 1] // output positions
for i = 0 => input.magnitude:
digit = input[i]
if filled[i] == true:
continue
k_swap(input, i, digitpositions[digit])
filled[digitpositions[digit]] = true
digitpositions[digit]++
I'll walk through it with the number input = 724886771
computed digitcounts:
{0, 1, 1, 0, 1, 0, 1, 3, 2, 0}
computed digitpositions:
{0, 0, 1, 2, 2, 3, 3, 4, 7, 9}
swap steps:
swap 0 with 0: 724886771, mark 0 visited
swap 1 with 4: 724876781, mark 4 visited
swap 2 with 5: 724778881, mark 5 visited
swap 3 with 3: 724778881, mark 3 visited
skip 4 (already visited)
skip 5 (already visited)
swap 6 with 2: 728776481, mark 2 visited
swap 7 with 1: 788776421, mark 1 visited
swap 8 with 6: 887776421, mark 6 visited
output number: 887776421
Edit:
This doesn't address the question correctly. If I have time later, I'll fix it but I don't right now.
How I would do it (in pseudo-c -- nothing fancy), assuming a fantasy integer array is passed where each element represents one decimal digit:
int[] sortToMaxInt(int[] M, int K) {
for (int i = 0; K > 0 && i < M.size() - 1; i++) {
if (swapDec(M, i)) K--;
}
return M;
}
bool swapDec(int[]& M, int i) {
/* no need to try and swap the value 9 as it is the
* highest possible value anyway. */
if (M[i] == 9) return false;
int max_dec = 0;
int max_idx = 0;
for (int j = i+1; j < M.size(); j++) {
if (M[j] >= max_dec) {
max_idx = j;
max_dec = M[j];
}
}
if (max_dec > M[i]) {
M.swapElements(i, max_idx);
return true;
}
return false;
}
From the top of my head so if anyone spots some fatal flaw please let me know.
Edit: based on the other answers posted here, I probably grossly misunderstood the problem. Anyone care to elaborate?
You start with max-number(M, N, 1, K).
max-number(M, N, pos, k)
{
if k == 0
return M
max-digit = 0
for i = pos to N
if M[i] > max-digit
max-digit = M[i]
if M[pos] == max-digit
return max-number(M, N, pos + 1, k)
for i = (pos + 1) to N
maxs.add(M)
if M[i] == max-digit
M2 = new M
swap(M2, i, pos)
maxs.add(max-number(M2, N, pos + 1, k - 1))
return maxs.max()
}
Here's my approach (It's not fool-proof, but covers the basic cases). First we'll need a function that extracts each DIGIT of an INT into a container:
std::shared_ptr<std::deque<int>> getDigitsOfInt(const int N)
{
int number(N);
std::shared_ptr<std::deque<int>> digitsQueue(new std::deque<int>());
while (number != 0)
{
digitsQueue->push_front(number % 10);
number /= 10;
}
return digitsQueue;
}
You obviously want to create the inverse of this, so convert such a container back to an INT:
const int getIntOfDigits(const std::shared_ptr<std::deque<int>>& digitsQueue)
{
int number(0);
for (std::deque<int>::size_type i = 0, iMAX = digitsQueue->size(); i < iMAX; ++i)
{
number = number * 10 + digitsQueue->at(i);
}
return number;
}
You also will need to find the MAX_DIGIT. It would be great to use std::max_element as it returns an iterator to the maximum element of a container, but if there are more you want the last of them. So let's implement our own max algorithm:
int getLastMaxDigitOfN(const std::shared_ptr<std::deque<int>>& digitsQueue, int startPosition)
{
assert(!digitsQueue->empty() && digitsQueue->size() > startPosition);
int maxDigitPosition(0);
int maxDigit(digitsQueue->at(startPosition));
for (std::deque<int>::size_type i = startPosition, iMAX = digitsQueue->size(); i < iMAX; ++i)
{
const int currentDigit(digitsQueue->at(i));
if (maxDigit <= currentDigit)
{
maxDigit = currentDigit;
maxDigitPosition = i;
}
}
return maxDigitPosition;
}
From here on its pretty straight what you have to do, put the right-most (last) MAX DIGITS to their places until you can swap:
const int solution(const int N, const int K)
{
std::shared_ptr<std::deque<int>> digitsOfN = getDigitsOfInt(N);
int pos(0);
int RemainingSwaps(K);
while (RemainingSwaps)
{
int lastHDPosition = getLastMaxDigitOfN(digitsOfN, pos);
if (lastHDPosition != pos)
{
std::swap<int>(digitsOfN->at(lastHDPosition), digitsOfN->at(pos));
++pos;
--RemainingSwaps;
}
}
return getIntOfDigits(digitsOfN);
}
There are unhandled corner-cases but I'll leave that up to you.
I assumed K = 2, but you can change the value!
Java code
public class Solution {
public static void main (String args[]) {
Solution d = new Solution();
System.out.println(d.solve(1234));
System.out.println(d.solve(9812));
System.out.println(d.solve(9876));
}
public int solve(int number) {
int[] array = intToArray(number);
int[] result = solve(array, array.length-1, 2);
return arrayToInt(result);
}
private int arrayToInt(int[] array) {
String s = "";
for (int i = array.length-1 ;i >= 0; i--) {
s = s + array[i]+"";
}
return Integer.parseInt(s);
}
private int[] intToArray(int number){
String s = number+"";
int[] result = new int[s.length()];
for(int i = 0 ;i < s.length() ;i++) {
result[s.length()-1-i] = Integer.parseInt(s.charAt(i)+"");
}
return result;
}
private int[] solve(int[] array, int endIndex, int num) {
if (endIndex == 0)
return array;
int size = num ;
int firstIndex = endIndex - size;
if (firstIndex < 0)
firstIndex = 0;
int biggest = findBiggestIndex(array, endIndex, firstIndex);
if (biggest!= endIndex) {
if (endIndex-biggest==num) {
while(num!=0) {
int temp = array[biggest];
array[biggest] = array[biggest+1];
array[biggest+1] = temp;
biggest++;
num--;
}
return array;
}else{
int n = endIndex-biggest;
for (int i = 0 ;i < n;i++) {
int temp = array[biggest];
array[biggest] = array[biggest+1];
array[biggest+1] = temp;
biggest++;
}
return solve(array, --biggest, firstIndex);
}
}else{
return solve(array, --endIndex, num);
}
}
private int findBiggestIndex(int[] array, int endIndex, int firstIndex) {
int result = firstIndex;
int max = array[firstIndex];
for (int i = firstIndex; i <= endIndex; i++){
if (array[i] > max){
max = array[i];
result = i;
}
}
return result;
}
}

Count the subsequences of length 4 divisible by 9

To count the subsequences of length 4 of a string of length n which are divisible by 9.
For example if the input string is 9999
then cnt=1
My approach is similar to Brute Force and takes O(n^3).Any better approach than this?
If you want to check if a number is divisible by 9, You better look here.
I will describe the method in short:
checkDividedByNine(String pNum) :
If pNum.length < 1
return false
If pNum.length == 1
return toInt(pNum) == 9;
Sum = 0
For c in pNum:
Sum += toInt(pNum)
return checkDividedByNine(toString(Sum))
So you can reduce the running time to less than O(n^3).
EDIT:
If you need very fast algorithm, you can use pre-processing in order to save for each possible 4-digit number, if it is divisible by 9. (It will cost you 10000 in memory)
EDIT 2:
Better approach: you can use dynamic programming:
For string S in length N:
D[i,j,k] = The number of subsequences of length j in the string S[i..N] that their value modulo 9 == k.
Where 0 <= k <= 8, 1 <= j <= 4, 1 <= i <= N.
D[i,1,k] = simply count the number of elements in S[i..N] that = k(mod 9).
D[N,j,k] = if j==1 and (S[N] modulo 9) == k, return 1. Otherwise, 0.
D[i,j,k] = max{ D[i+1,j,k], D[i+1,j-1, (k-S[i]+9) modulo 9]}.
And you return D[1,4,0].
You get a table in size - N x 9 x 4.
Thus, the overall running time, assuming calculating modulo takes O(1), is O(n).
Assuming that the subsequence has to consist of consecutive digits, you can scan from left to right, keeping track of what order the last 4 digits read are in. That way, you can do a linear scan and just apply divisibility rules.
If the digits are not necessarily consecutive, then you can do some finangling with lookup tables. The idea is that you can create a 3D array named table such that table[i][j][k] is the number of sums of i digits up to index j such that the sum leaves a remainder of k when divided by 9. The table itself has size 45n (i goes from 0 to 4, j goes from 0 to n-1, and k goes from 0 to 8).
For the recursion, each table[i][j][k] entry relies on table[i-1][j-1][x] and table[i][j-1][x] for all x from 0 to 8. Since each entry update takes constant time (at least relative to n), that should get you an O(n) runtime.
How about this one:
/*NOTE: The following holds true, if the subsequences consist of digits in contagious locations */
public int countOccurrences (String s) {
int count=0;
int len = s.length();
String subs = null;
int sum;
if (len < 4)
return 0;
else {
for (int i=0 ; i<len-3 ; i++) {
subs = s.substring(i, i+4);
sum = 0;
for (int j=0; j<=3; j++) {
sum += Integer.parseInt(String.valueOf(subs.charAt(j)));
}
if (sum%9 == 0)
count++;
}
return count;
}
}
Here is the complete working code for the above problem based on the above discussed ways using lookup tables
int fun(int h)
{
return (h/10 + h%10);
}
int main()
{
int t;
scanf("%d",&t);
int i,T;
for(T=0;T<t;T++)
{
char str[10001];
scanf("%s",str);
int len=strlen(str);
int arr[len][5][10];
memset(arr,0,sizeof(int)*(10*5*len));
int j,k,l;
for(j=0;j<len;j++)
{
int y;
y=(str[j]-48)%10;
arr[j][1][y]++;
}
//printarr(arr,len);
for(i=len-2;i>=0;i--) //represents the starting index of the string
{
int temp[5][10];
//COPYING ARRAY
int a,b,c,d;
for(a=0;a<=4;a++)
for(b=0;b<=9;b++)
temp[a][b]=arr[i][a][b]+arr[i+1][a][b];
for(j=1;j<=4;j++) //represents the length of the string
{
for(k=0;k<=9;k++) //represents the no. of ways to make it
{
if(arr[i+1][j][k]!=0)
{
for(c=1;c<=4;c++)
{
for(d=0;d<=9;d++)
{
if(arr[i][c][d]!=0)
{
int h,r;
r=j+c;
if(r>4)
continue;
h=k+d;
h=fun(h);
if(r<=4)
temp[r][h]=( temp[r][h]+(arr[i][c][d]*arr[i+1][j][k]))%1000000007;
}}}
}
//copy back from temp array
}
}
for(a=0;a<=4;a++)
for(b=0;b<=9;b++)
arr[i][a][b]=temp[a][b];
}
printf("%d\n",(arr[0][1][9])%1000000007);
}
return 0;
}

Find the top k sums of two sorted arrays

You are given two sorted arrays, of sizes n and m respectively. Your task (should you choose to accept it), is to output the largest k sums of the form a[i]+b[j].
A O(k log k) solution can be found here. There are rumors of a O(k) or O(n) solution. Does one exist?
I found the responses at your link mostly vague and poorly structured. Here's a start with a O(k * log(min(m, n))) O(k * log(m + n)) O(k * log(k)) algorithm.
Suppose they are sorted decreasing. Imagine you computed the m*n matrix of the sums as follows:
for i from 0 to m
for j from 0 to n
sums[i][j] = a[i] + b[j]
In this matrix, values monotonically decrease down and to the right. With that in mind, here is an algorithm which performs a graph search through this matrix in order of decreasing sums.
q : priority queue (decreasing) := empty priority queue
add (0, 0) to q with priority a[0] + b[0]
while k > 0:
k--
x := pop q
output x
(i, j) : tuple of int,int := position of x
if i < m:
add (i + 1, j) to q with priority a[i + 1] + b[j]
if j < n:
add (i, j + 1) to q with priority a[i] + b[j + 1]
Analysis:
The loop is executed k times.
There is one pop operation per iteration.
There are up to two insert operations per iteration.
The maximum size of the priority queue is O(min(m, n)) O(m + n) O(k).
The priority queue can be implemented with a binary heap giving log(size) pop and insert.
Therefore this algorithm is O(k * log(min(m, n))) O(k * log(m + n)) O(k * log(k)).
Note that the general priority queue abstract data type needs to be modified to ignore duplicate entries. Alternately, you could maintain a separate set structure that first checks for membership in the set before adding to the queue, and removes from the set after popping from the queue. Neither of these ideas would worsen the time or space complexity.
I could write this up in Java if there's any interest.
Edit: fixed complexity. There is an algorithm which has the complexity I described, but it is slightly different from this one. You would have to take care to avoid adding certain nodes. My simple solution adds many nodes to the queue prematurely.
private static class FrontierElem implements Comparable<FrontierElem> {
int value;
int aIdx;
int bIdx;
public FrontierElem(int value, int aIdx, int bIdx) {
this.value = value;
this.aIdx = aIdx;
this.bIdx = bIdx;
}
#Override
public int compareTo(FrontierElem o) {
return o.value - value;
}
}
public static void findMaxSum( int [] a, int [] b, int k ) {
Integer [] frontierA = new Integer[ a.length ];
Integer [] frontierB = new Integer[ b.length ];
PriorityQueue<FrontierElem> q = new PriorityQueue<MaxSum.FrontierElem>();
frontierA[0] = frontierB[0]=0;
q.add( new FrontierElem( a[0]+b[0], 0, 0));
while( k > 0 ) {
FrontierElem f = q.poll();
System.out.println( f.value+" "+q.size() );
k--;
frontierA[ f.aIdx ] = frontierB[ f.bIdx ] = null;
int fRight = f.aIdx+1;
int fDown = f.bIdx+1;
if( fRight < a.length && frontierA[ fRight ] == null ) {
q.add( new FrontierElem( a[fRight]+b[f.bIdx], fRight, f.bIdx));
frontierA[ fRight ] = f.bIdx;
frontierB[ f.bIdx ] = fRight;
}
if( fDown < b.length && frontierB[ fDown ] == null ) {
q.add( new FrontierElem( a[f.aIdx]+b[fDown], f.aIdx, fDown));
frontierA[ f.aIdx ] = fDown;
frontierB[ fDown ] = f.aIdx;
}
}
}
The idea is similar to the other solution, but with the observation that as you add to your result set from the matrix, at every step the next element in our set can only come from where the current set is concave. I called these elements frontier elements and I keep track of their position in two arrays and their values in a priority queue. This helps keep the queue size down, but by how much I've yet to figure out. It seems to be about sqrt( k ) but I'm not entirely sure about that.
(Of course the frontierA/B arrays could be simple boolean arrays, but this way they fully define my result set, This isn't used anywhere in this example but might be useful otherwise.)
As the pre-condition is the Array are sorted hence lets consider the following
for N= 5;
A[]={ 1,2,3,4,5}
B[]={ 496,497,498,499,500}
Now since we know Summation of N-1 of A&B would be highest hence just insert this in to heap along with the indexes of A & B element ( why, indexes? we'll come to know in a short while )
H.insert(A[N-1]+B[N-1],N-1,N-1);
now
while(!H.empty()) { // the time heap is not empty
H.pop(); // this will give you the sum you are looking for
The indexes which we got at the time of pop, we shall use them for selecting the next sum element.
Consider the following :
if we have i & j as the indexes in A & B , then the next element would be max ( A[i]+B[j-1], A[i-1]+B[j], A[i+1]+B[j+1] ) ,
So, insert the same if that has not been inserted in the heap
hence
(i,j)= max ( A[i]+B[j-1], A[i-1]+B[j], A[i+1]+B[j+1] ) ;
if(Hash[i,j]){ // not inserted
H.insert (i,j);
}else{
get the next max from max ( A[i]+B[j-1], A[i-1]+B[j], A[i+1]+B[j+1] ) ; and insert.
}
K pop-ing them will give you max elements required.
Hope this helps
Many thanks to #rlibby and #xuhdev with such an original idea to solve this kind of problem. I had a similar coding exercise interview require to find N largest sums formed by K elements in K descending sorted arrays - means we must pick 1 element from each sorted arrays to build the largest sum.
Example: List findHighestSums(int[][] lists, int n) {}
[5,4,3,2,1]
[4,1]
[5,0,0]
[6,4,2]
[1]
and a value of 5 for n, your procedure should return a List of size 5:
[21,20,19,19,18]
Below is my code, please take a look carefully for those block comments :D
private class Pair implements Comparable<Pair>{
String state;
int sum;
public Pair(String state, int sum) {
this.state = state;
this.sum = sum;
}
#Override
public int compareTo(Pair o) {
// Max heap
return o.sum - this.sum;
}
}
List<Integer> findHighestSums(int[][] lists, int n) {
int numOfLists = lists.length;
int totalCharacterInState = 0;
/*
* To represent State of combination of largest sum as String
* The number of characters for each list should be Math.ceil(log(list[i].length))
* For example:
* If list1 length contains from 11 to 100 elements
* Then the State represents for list1 will require 2 characters
*/
int[] positionStartingCharacterOfListState = new int[numOfLists + 1];
positionStartingCharacterOfListState[0] = 0;
// the reason to set less or equal here is to get the position starting character of the last list
for(int i = 1; i <= numOfLists; i++) {
int previousListNumOfCharacters = 1;
if(lists[i-1].length > 10) {
previousListNumOfCharacters = (int)Math.ceil(Math.log10(lists[i-1].length));
}
positionStartingCharacterOfListState[i] = positionStartingCharacterOfListState[i-1] + previousListNumOfCharacters;
totalCharacterInState += previousListNumOfCharacters;
}
// Check the state <---> make sure that combination of a sum is new
Set<String> states = new HashSet<>();
List<Integer> result = new ArrayList<>();
StringBuilder sb = new StringBuilder();
// This is a max heap contain <State, largestSum>
PriorityQueue<Pair> pq = new PriorityQueue<>();
char[] stateChars = new char[totalCharacterInState];
Arrays.fill(stateChars, '0');
sb.append(stateChars);
String firstState = sb.toString();
states.add(firstState);
int firstLargestSum = 0;
for(int i = 0; i < numOfLists; i++) firstLargestSum += lists[i][0];
// Imagine this is the initial state in a graph
pq.add(new Pair(firstState, firstLargestSum));
while(n > 0) {
// In case n is larger than the number of combinations of all list entries
if(pq.isEmpty()) break;
Pair top = pq.poll();
String currentState = top.state;
int currentSum = top.sum;
/*
* Loop for all lists and generate new states of which only 1 character is different from the former state
* For example: the initial state (Stage 0) 0 0 0 0 0
* So the next states (Stage 1) should be:
* 1 0 0 0 0
* 0 1 0 0 0 (choose element at index 2 from 2nd array)
* 0 0 1 0 0 (choose element at index 2 from 3rd array)
* 0 0 0 0 1
* But don't forget to check whether index in any lists have exceeded list's length
*/
for(int i = 0; i < numOfLists; i++) {
int indexInList = Integer.parseInt(
currentState.substring(positionStartingCharacterOfListState[i], positionStartingCharacterOfListState[i+1]));
if( indexInList < lists[i].length - 1) {
int numberOfCharacters = positionStartingCharacterOfListState[i+1] - positionStartingCharacterOfListState[i];
sb = new StringBuilder(currentState.substring(0, positionStartingCharacterOfListState[i]));
sb.append(String.format("%0" + numberOfCharacters + "d", indexInList + 1));
sb.append(currentState.substring(positionStartingCharacterOfListState[i+1]));
String newState = sb.toString();
if(!states.contains(newState)) {
// The newSum is always <= currentSum
int newSum = currentSum - lists[i][indexInList] + lists[i][indexInList+1];
states.add(newState);
// Using priority queue, we can immediately retrieve the largest Sum at Stage k and track all other unused states.
// From that Stage k largest Sum's state, then we can generate new states
// Those sums composed by recently generated states don't guarantee to be larger than those sums composed by old unused states.
pq.add(new Pair(newState, newSum));
}
}
}
result.add(currentSum);
n--;
}
return result;
}
Let me explain how I come up with the solution:
The while loop in my answer executes N times, consider the max heap
( priority queue).
Poll operation 1 time with complexity O(log(
sumOfListLength )) because the maximum element Pair in
heap is sumOfListLength.
Insertion operations might up to K times,
the complexity for each insertion is log(sumOfListLength).
Therefore, the complexity is O(N * log(sumOfListLength) ),

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