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I am a beginner in programming, especially in Scilab.
I need your help to write this short algorithm:
We have 2, sorted, non-decreasing tables A=[1,2,2,2,3,4,4,4,5,8,10] and B=[2,2,3,4,4,4].
Table A will always be bigger then Table B.
I need your help constructing The algorytm for checking whether the table B is contained entirely in the table A (without violating the order of the elements of the B array).
So for tables A=[1,2,2,2,3,4,4,4,5,8,10] and B=[2,2,3,4,4,4] the condition is met.
i have this for now
function sprawdz (A,B, n, m)
A=[2 3 0 5 1 1 2]
B=[3 0 5 1]
n=length(A)
m=length(B)
i=1
j=1
while (i<n && j<m)
if A(i)==B(j) then
i=i+1
j=j+1
if (j==m) then
return %t
break
end
else
i=i+1
j=1
end
end
return %f
if sprawdz (A,B, n, m) == %t then
disp("spoko")
else
disp("nie")
end
endfunction
In Scilab language, we can simply write vectorfind(A,B) that returns all positions in A where B is met and starts:
--> A = [1,2,2,2,3,4,4,4,5,8,10], B = [2,2,3,4,4,4]
A =
1. 2. 2. 2. 3. 4. 4. 4. 5. 8. 10.
B =
2. 2. 3. 4. 4. 4.
--> vectorfind(A, B)
ans =
3.
vectorfind() does not require from A or/and B to be sorted.
In addition, it can work with some wildcards. A is an array with any number of dimensions: 1D (vector), 2D (matrix), 3D, .. ND
If you just need a boolean answer:
vectorfind(A,B)<>[]
For illustrated examples or more information: https://help.scilab.org/docs/6.1.1/en_US/vectorfind.html
below code checks if an array contains all elements in another array:
public static boolean linearIn(Integer[] outer, Integer[] inner) {
return Arrays.asList(outer).containsAll(Arrays.asList(inner));
}
In your case there will be your two array
Another approch if you want to check contiguous subsequence
Simple Approach:
A simple approach is to run two nested loops and generate all subarrays of the array A[] and use one more loop to check if any of the subarray of A[] is equal to the array B[].
Efficient Approach :
An efficient approach is to use two pointers to traverse both the array simultaneously. Keep the pointer of array B[] still and if any element of A[] matches with the first element of B[] then increase the pointer of both the array else set the pointer of A to the next element of the previous starting point and reset the pointer of B to 0. If all the elements of B are matched then print YES otherwise print NO.
Below is the implementation of the above approach:
class SubTable
{
// Function to check if an array is subarray of another array
static boolean isSubArray(int A[], int B[],
int n, int m)
{
// Two pointers to traverse the arrays
int i = 0, j = 0;
// Traverse both arrays simultaneously
while (i < n && j < m)
{
// If element matches
// increment both pointers
if (A[i] == B[j])
{
i++;
j++;
// If array B is completely
// traversed
if (j == m)
return true;
}
// If not,
// increment i and reset j
else
{
i = i - j + 1;
j = 0;
}
}
return false;
}
public static void main(String arr[])
{
int A[] = { 2, 3, 0, 5, 1, 1, 2 };
int n = A.length;
int B[] = { 3, 0, 5, 1 };
int m = B.length;
if (isSubArray(A, B, n, m))
System.out.println("YES");
else
System.out.println("NO");
}
}
Question link: http://codeforces.com/contest/2/problem/B
There is a square matrix n × n, consisting of non-negative integer numbers. You should find such a way on it that
starts in the upper left cell of the matrix;
each following cell is to the right or down from the current cell;
the way ends in the bottom right cell.
Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros.
Input
The first line contains an integer number n (2 ≤ n ≤ 1000), n is the size of the matrix. Then follow n lines containing the matrix elements (non-negative integer numbers not exceeding 10^9).
Output
In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.
I thought of the following: In the end, whatever the answer will be, it should contain minimum powers of 2's and 5's. Therefore, what I did was, for each entry in the input matrix, I calculated the powers of 2's and 5's and stored them in separate matrices.
for (i = 0; i < n; i++)
{
for ( j = 0; j < n; j++)
{
cin>>foo;
matrix[i][j] = foo;
int n1 = calctwo(foo); // calculates the number of 2's in factorisation of that number
int n2 = calcfive(foo); // calculates number of 5's
two[i][j] = n1;
five[i][j] = n2;
}
}
After that, I did this:
for (i = 0; i < n; i++)
{
for ( j = 0; j < n; j++ )
{
dp[i][j] = min(two[i][j],five[i][j]); // Here, dp[i][j] will store minimum number of 2's and 5's.
}
}
But the above doesn't really a valid answer, I don't know why? Have I implemented the correct approach? Or, is this the correct way of solving this question?
Edit: Here are my functions of calculating the number of two's and number of five's in a number.
int calctwo (int foo)
{
int counter = 0;
while (foo%2 == 0)
{
if (foo%2 == 0)
{
counter++;
foo = foo/2;
}
else
break;
}
return counter;
}
int calcfive (int foo)
{
int counter = 0;
while (foo%5 == 0)
{
if (foo%5 == 0)
{
counter++;
foo = foo/5;
}
else
break;
}
return counter;
}
Edit2: I/O Example as given in the link:
Input:
3
1 2 3
4 5 6
7 8 9
Output:
0
DDRR
Since you are interested only in the number of trailing zeroes you need only to consider the powers of 2, 5 which you could keep in two separate nxn arrays. So for the array
1 2 3
4 5 6
7 8 9
you just keep the arrays
the powers of 2 the powers of 5
0 1 0 0 0 0
2 0 1 0 1 0
0 3 0 0 0 0
The insight for the problem is the following. Notice that if you find a path which minimizes the sum of the powers of 2 and a path which minimizes the number sum of the powers of 5 then the answer is the one with lower value of those two paths. So you reduce your problem to the two times application of the following classical dp problem: find a path, starting from the top-left corner and ending at the bottom-right, such that the sum of its elements is minimum. Again, following the example, we have:
minimal path for the
powers of 2 value
* * - 2
- * *
- - *
minimal path for the
powers of 5 value
* - - 0
* - -
* * *
so your answer is
* - -
* - -
* * *
with value 0
Note 1
It might seem that taking the minimum of the both optimal paths gives only an upper bound so a question that may rise is: is this bound actually achieved? The answer is yes. For convenience, let the number of 2's along the 2's optimal path is a and the number of 5's along the 5's optimal path is b. Without loss of generality assume that the minimum of the both optimal paths is the one for the power of 2's (that is a < b). Let the number of 5's along the minimal path is c. Now the question is: are there as much as 5's as there are 2's along this path (i.e. is c >= a?). Assume that the answer is no. That means that there are less 5's than 2's along the minimal path (that is c < a). Since the optimal value of 5's paths is b we have that every 5's path has at least b 5's in it. This should also be true for the minimal path. That means that c > b. We have that c < a so a > b but the initial assumption was that a < b. Contradiction.
Note 2
You might also want consider the case in which there is an element 0 in your matrix. I'd assume that number of trailing zeroes when the product is 1. In this case, if the algorithm has produced a result with a value more than 1 you should output 1 and print a path that goes through the element 0.
Here is the code. I've used pair<int,int> to store factor of 2 and 5 in the matrix.
#include<vector>
#include<iostream>
using namespace std;
#define pii pair<int,int>
#define F first
#define S second
#define MP make_pair
int calc2(int a){
int c=0;
while(a%2==0){
c++;
a/=2;
}
return c;
}
int calc5(int a){
int c=0;
while(a%5==0){
c++;
a/=5;
}
return c;
}
int mini(int a,int b){
return a<b?a:b;
}
pii min(pii a, pii b){
if(mini(a.F,a.S) < mini(b.F,b.S))
return a;
return b;
}
int main(){
int n;
cin>>n;
vector<vector<pii > > v;
vector<vector<int> > path;
int i,j;
for(i=0;i<n;i++){
vector<pii > x;
vector<int> q(n,0);
for(j=0;j<n;j++){
int y;cin>>y;
x.push_back(MP(calc2(y),calc5(y))); //I store factors of 2,5 in the vector to calculate
}
x.push_back(MP(100000,100000)); //padding each row to n+1 elements (to handle overflow in code)
v.push_back(x);
path.push_back(q); //initialize path matrix to 0
}
vector<pii > x(n+1,MP(100000,100000));
v.push_back(x); //pad 1 more row to handle index overflow
for(i=n-1;i>=0;i--){
for(j=n-1;j>=0;j--){ //move from destination to source grid
if(i==n-1 && j==n-1)
continue;
//here, the LHS of condition in if block is the condition which determines minimum number of trailing 0's. This is the same condition that is used to manipulate "v" for getting the same result.
if(min(MP(v[i][j].F+v[i+1][j].F,v[i][j].S+v[i+1][j].S), MP(v[i][j].F+v[i][j+1].F,v[i][j].S+v[i][j+1].S)) == MP(v[i][j].F+v[i+1][j].F,v[i][j].S+v[i+1][j].S))
path[i][j] = 1; //go down
else
path[i][j] = 2; //go right
v[i][j] = min(MP(v[i][j].F+v[i+1][j].F,v[i][j].S+v[i+1][j].S), MP(v[i][j].F+v[i][j+1].F,v[i][j].S+v[i][j+1].S));
}
}
cout<<mini(v[0][0].F, v[0][0].S)<<endl; //print result
for(i=0,j=0;i<=n-1 && j<=n-1;){ //print path (I don't know o/p format)
cout<<"("<<i<<","<<j<<") -> ";
if(path[i][j]==1)
i++;
else
j++;
}
return 0;
}
This code gives fine results as far as the test cases I checked. If you have any doubts regarding this code, ask in comments.
EDIT:
The basic thought process.
To reach the destination, there are only 2 options. I started with destination to avoid the problem of path ahead calculation, because if 2 have same minimum values, then we chose any one of them. If the path to destination is already calculated, it does not matter which we take.
And minimum is to check which pair is more suitable. If a pair has minimum 2's or 5's than other, it will produce less 0's.
Here is a solution proposal using Javascript and functional programming.
It relies on several functions:
the core function is smallest_trailer that recursively goes through the grid. I have chosen to go in 4 possible direction, left "L", right "R", down "D" and "U". It is not possible to pass twice on the same cell. The direction that is chosen is the one with the smallest number of trailing zeros. The counting of trailing zeros is devoted to another function.
the function zero_trailer(p,n,nbz) assumes that you arrive on a cell with a value p while you already have an accumulator n and met nbz zeros on your way. The function returns an array with two elements, the new number of zeros and the new accumulator. The accumulator will be a power of 2 or 5. The function uses the auxiliary function pow_2_5(n) that returns the powers of 2 and 5 inside n.
Other functions are more anecdotical: deepCopy(arr) makes a standard deep copy of the array arr, out_bound(i,j,n) returns true if the cell (i,j) is out of bound of the grid of size n, myMinIndex(arr) returns the min index of an array of 2 dimensional arrays (each subarray contains the nb of trailing zeros and the path as a string). The min is only taken on the first element of subarrays.
MAX_SAFE_INTEGER is a (large) constant for the maximal number of trailing zeros when the path is wrong (goes out of bound for example).
Here is the code, which works on the example given in the comments above and in the orginal link.
var MAX_SAFE_INTEGER = 9007199254740991;
function pow_2_5(n) {
// returns the power of 2 and 5 inside n
function pow_not_2_5(k) {
if (k%2===0) {
return pow_not_2_5(k/2);
}
else if (k%5===0) {
return pow_not_2_5(k/5);
}
else {
return k;
}
}
return n/pow_not_2_5(n);
}
function zero_trailer(p,n,nbz) {
// takes an input two numbers p and n that should be multiplied and a given initial number of zeros (nbz = nb of zeros)
// n is the accumulator of previous multiplications (a power of 5 or 2)
// returns an array [kbz, k] where kbz is the total new number of zeros (nbz + the trailing zeros from the multiplication of p and n)
// and k is the new accumulator (typically a power of 5 or 2)
function zero_aux(k,kbz) {
if (k===0) {
return [1,0];
}
else if (k%10===0) {
return zero_aux(k/10,kbz+1);
}
else {
return [kbz,k];
}
}
return zero_aux(pow_2_5(p)*n,nbz);
}
function out_bound(i,j,n) {
return !((i>=0)&&(i<n)&&(j>=0)&&(j<n));
}
function deepCopy(arr){
var toR = new Array(arr.length);
for(var i=0;i<arr.length;i++){
var toRi = new Array(arr[i].length);
for(var j=0;j<arr[i].length;j++){
toRi[j] = arr[i][j];
}
toR[i] = toRi;
}
return toR;
}
function myMinIndex(arr) {
var min = arr[0][0];
var minIndex = 0;
for (var i = 1; i < arr.length; i++) {
if (arr[i][0] < min) {
minIndex = i;
min = arr[i][0];
}
}
return minIndex;
}
function smallest_trailer(grid) {
var n = grid.length;
function st_aux(i,j,grid_aux, acc_mult, nb_z, path) {
if ((i===n-1)&&(j===n-1)) {
var tmp_acc_nbz_f = zero_trailer(grid_aux[i][j],acc_mult,nb_z);
return [tmp_acc_nbz_f[0], path];
}
else if (out_bound(i,j,n)) {
return [MAX_SAFE_INTEGER,[]];
}
else if (grid_aux[i][j]<0) {
return [MAX_SAFE_INTEGER,[]];
}
else {
var tmp_acc_nbz = zero_trailer(grid_aux[i][j],acc_mult,nb_z) ;
grid_aux[i][j]=-1;
var res = [st_aux(i+1,j,deepCopy(grid_aux), tmp_acc_nbz[1], tmp_acc_nbz[0], path+"D"),
st_aux(i-1,j,deepCopy(grid_aux), tmp_acc_nbz[1], tmp_acc_nbz[0], path+"U"),
st_aux(i,j+1,deepCopy(grid_aux), tmp_acc_nbz[1], tmp_acc_nbz[0], path+"R"),
st_aux(i,j-1,deepCopy(grid_aux), tmp_acc_nbz[1], tmp_acc_nbz[0], path+"L")];
return res[myMinIndex(res)];
}
}
return st_aux(0,0,grid, 1, 0, "");
}
myGrid = [[1, 25, 100],[2, 1, 25],[100, 5, 1]];
console.log(smallest_trailer(myGrid)); //[0,"RDDR"]
myGrid = [[1, 2, 100],[25, 1, 5],[100, 25, 1]];
console.log(smallest_trailer(myGrid)); //[0,"DRDR"]
myGrid = [[1, 10, 1, 1, 1],[1, 1, 1, 10, 1],[10, 10, 10, 10, 1],[10, 10, 10, 10, 1],[10, 10, 10, 10, 1]];
console.log(smallest_trailer(myGrid)); //[0,"DRRURRDDDD"]
This is my Dynamic Programming solution.
https://app.codility.com/demo/results/trainingAXFQ5B-SZQ/
For better understanding we can simplify the task and assume that there are no zeros in the matrix (i.e. matrix contains only positive integers), then the Java solution will be the following:
class Solution {
public int solution(int[][] a) {
int minPws[][] = new int[a.length][a[0].length];
int minPws2 = getMinPws(a, minPws, 2);
int minPws5 = getMinPws(a, minPws, 5);
return min(minPws2, minPws5);
}
private int getMinPws(int[][] a, int[][] minPws, int p) {
minPws[0][0] = pws(a[0][0], p);
//Fullfill the first row
for (int j = 1; j < a[0].length; j++) {
minPws[0][j] = minPws[0][j-1] + pws(a[0][j], p);
}
//Fullfill the first column
for (int i = 1; i < a.length; i++) {
minPws[i][0] = minPws[i-1][0] + pws(a[i][0], p);
}
//Fullfill the rest of matrix
for (int i = 1; i < a.length; i++) {
for (int j = 1; j < a[0].length; j++) {
minPws[i][j] = min(minPws[i-1][j], minPws[i][j-1]) + pws(a[i][j], p);
}
}
return minPws[a.length-1][a[0].length-1];
}
private int pws(int n, int p) {
//Only when n > 0
int pws = 0;
while (n % p == 0) {
pws++;
n /= p;
}
return pws;
}
private int min(int a, int b) {
return (a < b) ? a : b;
}
}
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;
}
}
How can I write an algorithm to check if the sum of any two numbers in an array/list matches a given number
with a complexity of nlogn?
I'm sure there's a better way, but here's an idea:
Sort array
For every element e in the array, binary search for the complement (sum - e)
Both these operations are O(n log n).
This can be done in O(n) using a hash table. Initialize the table with all numbers in the array, with number as the key, and frequency as the value. Walk through each number in the array, and see if (sum - number) exists in the table. If it does, you have a match. After you've iterated through all numbers in the array, you should have a list of all pairs that sum up to the desired number.
array = initial array
table = hash(array)
S = sum
for each n in array
if table[S-n] exists
print "found numbers" n, S-n
The case where n and table[S-n] refer to the same number twice can be dealt with an extra check, but the complexity remains O(n).
Use a hash table. Insert every number into your hash table, along with its index. Then, let S be your desired sum. For every number array[i] in your initial array, see if S - array[i] exists in your hash table with an index different than i.
Average case is O(n), worst case is O(n^2), so use the binary search solution if you're afraid of the worst case.
Let us say that we want to find two numbers in the array A that when added together equal N.
Sort the array.
Find the largest number in the array that is less than N/2. Set the index of that number as lower.
Initialize upper to be lower + 1.
Set sum = A[lower] + A[upper].
If sum == N, done.
If sum < N, increment upper.
If sum > N, decrement lower.
If either lower or upper is outside the array, done without any matches.
Go back to 4.
The sort can be done in O(n log n). The search is done in linear time.
This is in Java : This even removes the possible duplicates.. Runtime - O(n^2)
private static int[] intArray = {15,5,10,20,25,30};
private static int sum = 35;
private static void algorithm()
{
Map<Integer, Integer> intMap = new Hashtable<Integer, Integer>();
for (int i=0; i<intArray.length; i++)
{
intMap.put(i, intArray[i]);
if(intMap.containsValue(sum - intArray[i]))
System.out.println("Found numbers : "+intArray[i] +" and "+(sum - intArray[i]));
}
System.out.println(intMap);
}
def sum_in(numbers, sum_):
"""whether any two numbers from `numbers` form `sum_`."""
a = set(numbers) # O(n)
return any((sum_ - n) in a for n in a) # O(n)
Example:
>>> sum_in([200, -10, -100], 100)
True
Here's a try in C. This isn't marked homework.
// Assumes a sorted integer array with no duplicates
void printMatching(int array[], int size, int sum)
{
int i = 0, k = size - 1;
int curSum;
while(i < k)
{
curSum = array[i] + array[k];
if(curSum == sum)
{
printf("Found match at indices %d, %d\n", i, k);
i++;k--;
}
else if(curSum < sum)
{
i++;
}
else
{
k--;
}
}
}
Here is some test output using int a[] = { 3, 5, 6, 7, 8, 9, 13, 15, 17 };
Searching for 12..
Found match at indices 0, 5
Found match at indices 1, 3
Searching for 22...
Found match at indices 1, 8
Found match at indices 3, 7
Found match at indices 5, 6
Searching for 4..
Searching for 50..
The search is linear, so O(n). The sort that takes place behind the scenes is going to be O(n*logn) if you use one of the good sorts.
Because of the math behind Big-O, the smaller term in additive terms will effectively drop out of your calculation, and you end up with O(n logn).
This one is O(n)
public static bool doesTargetExistsInList(int Target, int[] inputArray)
{
if (inputArray != null && inputArray.Length > 0 )
{
Hashtable inputHashTable = new Hashtable();
// This hash table will have all the items in the input array and how many times they appeard
Hashtable duplicateItems = new Hashtable();
foreach (int i in inputArray)
{
if (!inputHashTable.ContainsKey(i))
{
inputHashTable.Add(i, Target - i);
duplicateItems.Add(i, 1);
}
else
{
duplicateItems[i] = (int)duplicateItems[i] + 1;
}
}
foreach (DictionaryEntry de in inputHashTable)
{
if ((int)de.Key == (int)de.Value)
{
if ((int)duplicateItems[de.Key] > 1)
return true;
}
else if (inputHashTable.ContainsKey(de.Value))
{
return true;
}
}
}
return false;
}
Here is an algorithm that runs in O(n) if array is already sorted or O(n log n) if it isn't already sorted. Takes cues from lot of other answers here. Code is in Java, but here is a pseudo code as well derived from lot of existing answers, but optimized for duplicates generally
Lucky guess if first and last elements are equal to target
Create a Map with current value and its occurrences
Create a visited Set which contains items we already saw, this optimizes for duplicates such that say with an input of (1,1,1,1,1,1,2) and target 4, we only ever compute for 0 and last element and not all the 1's in the array.
Use these variables to compute existence of target in the array;
set the currentValue to array[ith];
set newTarget to target - currentValue;
set expectedCount to 2 if currentValue equals newTarget or 1 otherwise
AND return true only if
a. we never saw this integer before AND
b. we have some value for the newTarget in the map we created
c. and the count for the newTarget is equal or greater than the expectedCount
OTHERWISE
repeat step 4 till we reach end of array and return false OTHERWISE;
Like I mentioned the best possible use for a visited store is when we have duplicates, it would never help if none of elements are duplicates.
Java Code at https://gist.github.com/eded5dbcee737390acb4
Depends If you want only one sum O(N) or O(N log N) or all sums O(N^2) or O(N^2 log N). In the latter case better uses an FFT>
Step 1 : Sort the array in O(n logn)
Step 2 : Find two indices
0<=i<j<=n in a[0..n] such that a[i]+a[j]==k, where k is given key.
int i=0,j=n;
while(i<j) {
int sum = a[i]+a[j];
if(sum == k)
print(i,j)
else if (sum < k)
i++;
else if (sum > k)
j--;
}
public void sumOfTwoQualToTargetSum()
{
List<int> list= new List<int>();
list.Add(1);
list.Add(3);
list.Add(5);
list.Add(7);
list.Add(9);
int targetsum = 12;
int[] arr = list.ToArray();
for (int i = 0; i < arr.Length; i++)
{
for (int j = 0; j < arr.Length; j++)
{
if ((i != j) && ((arr[i] + arr[j]) == targetsum))
{
Console.Write("i =" + i);
Console.WriteLine("j =" + j);
}
}
}
}
Solved the question in Swift 4.0
Solved in 3 different ways (with 2 different type of return -> Boolean and Indexes)
A) TimeComplexity => 0(n Log n) SpaceComplexity => 0(n).
B) TimeComplexity => 0(n^2) SpaceComplexity => 0(1).
C) TimeComplexity => 0(n) SpaceComplexity => 0(n)
Choose Solution A, B or C depending on TradeOff.
//***********************Solution A*********************//
//This solution returns TRUE if any such two pairs exist in the array
func binarySearch(list: [Int], key: Int, start: Int, end: Int) -> Int? { //Helper Function
if end < start {
return -1
} else {
let midIndex = (start + end) / 2
if list[midIndex] > key {
return binarySearch(list: list, key: key, start: start, end: midIndex - 1)
} else if list[midIndex] < key {
return binarySearch(list: list, key: key, start: midIndex + 1, end: end)
} else {
return midIndex
}
}
}
func twoPairSum(sum : Int, inputArray: [Int]) -> Bool {
//Do this only if array isn't Sorted!
let sortedArray = inputArray.sorted()
for (currentIndex, value) in sortedArray.enumerated() {
if let indexReturned = binarySearch(list: sortedArray, key: sum - value, start: 0, end: sortedArray.count-1) {
if indexReturned != -1 && (indexReturned != currentIndex) {
return true
}
}
}
return false
}
//***********************Solution B*********************//
//This solution returns the indexes of the two pair elements if any such two pairs exists in the array
func twoPairSum(_ nums: [Int], _ target: Int) -> [Int] {
for currentIndex in 0..<nums.count {
for nextIndex in currentIndex+1..<nums.count {
if calculateSum(firstElement: nums[currentIndex], secondElement: nums[nextIndex], target: target) {
return [currentIndex, nextIndex]
}
}
}
return []
}
func calculateSum (firstElement: Int, secondElement: Int, target: Int) -> Bool {//Helper Function
return (firstElement + secondElement) == target
}
//*******************Solution C*********************//
//This solution returns the indexes of the two pair elements if any such two pairs exists in the array
func twoPairSum(_ nums: [Int], _ target: Int) -> [Int] {
var dict = [Int: Int]()
for (index, value) in nums.enumerated() {
dict[value] = index
}
for (index, value) in nums.enumerated() {
let otherIndex = dict[(target - value)]
if otherIndex != nil && otherIndex != index {
return [index, otherIndex!]
}
}
return []
}
This question is missing some more details into it. Like what is the return value, the limitation on the input.
I have seen some questions related to that, which can be this question with extra requirement, to return the actual elements that result in the input
Here is my version of the solution, it should be O(n).
import java.util.*;
public class IntegerSum {
private static final int[] NUMBERS = {1,2,3,4,5,6,7,8,9,10};
public static void main(String[] args) {
int[] result = IntegerSum.isSumExist(7);
System.out.println(Arrays.toString(result));
}
/**
* n = x + y
* 7 = 1 + 6
* 7 - 1 = 6
* 7 - 6 = 1
* The subtraction of one element in the array should result into the value of the other element if it exist;
*/
public static int[] isSumExist(int n) {
// validate the input, based on the question
// This to return the values that actually result in the sum. which is even more tricky
int[] output = new int[2];
Map resultsMap = new HashMap<Integer, Integer>();
// O(n)
for (int number : NUMBERS) {
if ( number > n )
throw new IllegalStateException("The number is not in the array.");
if ( resultsMap.containsKey(number) ) {
output[0] = number;
output[1] = (Integer) resultsMap.get(number);
return output;
}
resultsMap.put(n - number, number);
}
throw new IllegalStateException("The number is not in the array.");
}
}
Design an algorithm to find all pairs of integers within an array which sum to a specified value.
I have tried this problem using a hash table to store entries for the sum of array elements, but it is not an efficient solution.
What algorithm can I use to solve this efficiently?
I don't see why the hash table approach is inefficient, at least in algorithm analysis terms - in memory locality terms admittedly, it can be quite bad. Anyway, scan the array twice...
First scan - put all the array elements in the hash table - O(n) total. Individual inserts are only amortized O(1), but a neat thing about how amortized analysis works means the O(n) is absolute - not amortized.
Second scan - check for (sum - current) in the hash table - O(n) total.
This beats the O(n log n) sort-and-search methods, at least in theory.
Then, note that you can combine the two scans into one. You can spot a pair as soon as you encounter the second of that pair during the first scan. In pseudocode...
for i in array.range
hashset.insert (array [i])
diff = sum - array [i]
if hashset.includes (diff)
output diff, array [i]
If you need positions of the items, use a hashmap and store item positions in it. If you need to cope with duplicates, you might need to store counts in a hashmap. For positions and duplicates, you might need a hashmap of start pointers for linked lists of positions.
This makes assumptions about the hash table implementation, but fairly safe ones given the usual implementations in most current languages and libraries.
BTW - combining the scans shouldn't be seen as an optimisation. The iteration overhead should be insignificant. Memory locality issues could make a single pass slightly more efficient for very large arrays, but the real memory locality issues will be in the hashtable lookups anyway.
IMO the only real reason to combine the scans is because you only want each pair reported once - handling that in a two-scan approach would be a bit more hassle.
If the array is sorted:
Let i = 0, j = end of array, sum = the value you are looking for,
then do:
If i+j = sum, then output (i,j).
If i+j < sum, then move i to the right one position.
If i+j > sum, then move j to the left one position.
Time complexity: O(n). Space complexity: O(1).
If the array is not sorted, there are a few ways to approach this problem:
Sort the array and then use the above approach.
HashMap:
Store all elements in a HashMap.
a+b=sum, so b=sum-a. For each element a of the array, look up b from the HashMap.
HashMap lookup takes amortized O(1).
Time complexity: O(n). Space complexity: O(n).
BitMap:
Iterate through the input to create a bitmap where each bit corresponds to an element value. Say the input is {2,5,8}, then we toggle the bitmap array's indices 2, 5 and 8 from binary 0 to 1. This takes O(1) per element, thus O(n) in total.
Go through the input again. We know b=sum-a, so for every element a in the input, look up its b, which can be done in O(1) since it's a bitmap index. This also takes O(n) in total.
Time complexity: O(n) + O(n) = O(n). Space complexity: bitmap space = O(n).
You don't even need to store all the elements in hashmap, and then scan. You can scan during the first iteration itself.
void foo(int[] A, int sum) {
HashSet<Integer> set = new HashSet<Integer>();
for (int e : A) {
if (set.contains(sum-e)) {
System.out.println(e + "," + (sum-e));
// deal with the duplicated case
set.remove(sum-e);
} else {
set.add(e);
}
}
}
How about sorting the array, then marching in from both ends?
Assume required sum = R
sort the array
for each number in the array A(n), do a binary search to find the number A(x) such that A(n) + A(x) = R
If you don't mind spending O(M) in space, where M is the sum you are seeking, you can do this in O(N + M) time. Set sums[i] = 1 when i <= M on a single pass over N, then check (sums[i] && sums[M-i]) on a single pass over M/2.
#include <iostream>
using namespace std;
#define MAX 15
int main()
{
int array[MAX] = {-12,-6,-4,-2,0,1,2,4,6,7,8,12,13,20,24};
const int find_sum = 0;
int max_index = MAX - 1;
int min_index = 0;
while(min_index < max_index)
{
if(array[min_index] + array[max_index-min_index] == find_sum)
{
cout << array[min_index] << " & " << array[max_index-min_index] << " Matched" << endl;
return 0;
}
if(array[min_index]+array[max_index-min_index] < find_sum)
{
min_index++;
//max_index++;
}
if(array[min_index]+array[max_index-min_index] > find_sum)
{
max_index--;
}
}
cout << "NO MATCH" << endl;
return 0;
}
//-12 & 12 matched
Implemented in Python 2.7:
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
We can use C++ STL map to solve this
void subsetSum(int arr[], int n, int sum)
{
map<int, int>Map;
for(int i=0; i<n; i++)
{
Map[arr[i]]++;
if(Map.count(sum-arr[i]))
{
cout<<arr[i]<<" "<<sum-arr[i]<<"\n";
}
}
}
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;
}
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 and don't forget to bump it if its the best answer!!
A solution that takes into account duplicates and uses every number only one time:
void printPairs(int[] numbers, int S) {
// toMap(numbers) converts the numbers array to a map, where
// Key is a number from the original array
// Value is a count of occurrences of this number in the array
Map<Integer, Integer> numbersMap = toMap(numbers);
for (Entry<Integer, Integer> entry : numbersMap.entrySet()) {
if (entry.getValue().equals(0)) {
continue;
}
int number = entry.getKey();
int complement = S - number;
if (numbersMap.containsKey(complement) && numbersMap.get(complement) > 0) {
for (int j = 0; j < min(numbersMap.get(number),
numbersMap.get(complement)); j++) {
if (number.equals(complement) && numbersMap.get(number) < 2) {
break;
}
System.out.println(number, complement);
numbersMap.put(number, numbersMap.get(number) - 1);
numbersMap.put(complement, numbersMap.get(complement) - 1);
}
}
}
}
Hashtable solution, in Ruby (quite straightforward to understand):
value = 100
pairs = [1,99,5,95]
hash_of_pairs = {}
pairs.map! do |pair|
# Adds to hashtable the pair
hash_of_pairs[pair] = pair
# Finds the value the pair needs
new_pair = hash_of_pairs[value - pair]
# Returns the pair whenever the pair exists, or nil
[new_pair, pair] if !new_pair.nil?
end.compact! # Cleans up the array, removing all nil values
print pairs # [[1,99], [5,95]]
#Test
public void hasPairWithSum() {
assertFalse(hasPairWithSum_Ordered_Logarithmic(new int[] { 1, 2, 3, 9 }, 8));
assertTrue(hasPairWithSum_Ordered_Logarithmic(new int[] { 1, 2, 4, 4 }, 8));
assertFalse(hasPairWithSum_Ordered_Linear(new int[] { 1, 2, 3, 9 }, 8));
assertTrue(hasPairWithSum_Ordered_Linear(new int[] { 1, 2, 4, 4 }, 8));
assertFalse(hasPairWithSum_Unsorted_Linear(new int[] { 9, 1, 3, 2 }, 8));
assertTrue(hasPairWithSum_Unsorted_Linear(new int[] { 4, 2, 1, 4 }, 8));
assertFalse(hasPairWithSum_Unsorted_Quadratic(new int[] { 9, 1, 3, 2 }, 8));
assertTrue(hasPairWithSum_Unsorted_Quadratic(new int[] { 4, 2, 1, 4 }, 8));
}
private boolean hasPairWithSum_Ordered_Logarithmic(int[] data, int sum) {
for (int i = 0; i < data.length; i++) {
int current = data[i];
int complement = sum - current;
int foundIndex = Arrays.binarySearch(data, complement);
if (foundIndex >= 0 && foundIndex != i) {
return true;
}
}
return false;
}
private boolean hasPairWithSum_Ordered_Linear(int[] data, int sum) {
int low = 0;
int high = data.length - 1;
while (low < high) {
int total = data[low] + data[high];
if (total == sum) {
return true;
} else if (total < sum) {
low++;
} else {
high--;
}
}
return false;
}
private boolean hasPairWithSum_Unsorted_Linear(int[] data, int sum) {
Set<Integer> complements = Sets.newHashSet();
for (int current : data) {
if (complements.contains(current)) {
return true;
}
complements.add(sum - current);
}
return false;
}
private boolean hasPairWithSum_Unsorted_Quadratic(int[] data, int sum) {
for (int i = 0; i < data.length; i++) {
int current = data[i];
int complement = sum - current;
for (int j = 0; j < data.length; j++) {
if (data[j] == complement && i != j) {
return true;
}
}
}
return false;
}
Creating a hash table and then looking for value in it.
function sum_exist(num : number, arr : any[]) {
var number_seen = {};
for(let item of arr){
if(num - item in number_seen){
return true
}
number_seen[item] = 0;
}
return false;
}
Test case (using Jest)
test('Given a list of numbers, return whether any two sums equal to the set number.', () => {
expect(sum_exist(17 , [10, 15, 3, 7])).toEqual(true);
});
test('Given a list of numbers, return whether any two sums equal to the set number.', () => {
expect(sum_exist(16 , [10, 15, 3, 7])).toEqual(false);
});
#python 3.x
def sum_pairs(list_data, number):
list_data.sort()
left = 0
right = len(list_data)-1
pairs = []
while left < right:
if list_data[left]+list_data[right] == number:
find_pairs = [list_data[left], list_data[right]]
pairs.append(find_pairs)
right = right-1
elif list_data[left]+list_data[right] < number:
left = left+1
else:
right = right-1
return pairs