# Maximum Number of Ones in C++

Suppose we have a matrix M with dimensions w x h, such that every cell has value 0 or 1, and any square sub-matrix of M of size l x l has at most maxOnes number of ones. We have to find the maximum possible number of ones that the matrix M can have.

So, if the input is like w = 3, h = 3, l = 2, maxOnes = 1, then the output will be 4 as in a 3*3 matrix, no 2*2 sub-matrix can have more than 1 one. The best solution that has 4 ones is −

 1 0 1 0 0 0 1 0 1

To solve this, we will follow these steps −

• ret := 0

• make one 2D array sq of size n x n

• for initialize i := 0, when i < height, update (increase i by 1), do −

• for initialize j := 0, when j < width, update (increase j by 1), do −

• increase sq[i mod n, j mod n] by 1

• Define an array v

• for initialize i := 0, when i < n, update (increase i by 1), do −

• for initialize j := 0, when j < n, update (increase j by 1), do −

• insert sq[i, j] at the end of v

• sort the array v in reverse order

• for initialize i := 0, j := 0, when i < size of v and j < maxOnes, update (increase i by 1), (increase j by 1), do −

• return ret

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int maximumNumberOfOnes(int width, int height, int n, int maxOnes) {
int ret = 0;
vector < vector <int> > sq(n, vector <int>(n));
for(int i = 0; i < height; i++){
for(int j = 0; j < width; j++){
sq[i % n][j % n]++;
}
}
vector <int> v;
for(int i = 0; i < n; i++){
for(int j = 0; j < n ; j++){
v.push_back(sq[i][j]);
}
}
sort(v.rbegin(), v.rend());
for(int i = 0, j = 0; i < v.size() && j < maxOnes; i++, j++){
ret += v[i];
}
return ret;
}
};
main(){
Solution ob;
cout << (ob.maximumNumberOfOnes(3,3,2,1));
}

## Input

3, 3, 2, 1

## Output

4