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 −

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

Arnab Chakraborty

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Updated on: 11-Jul-2020

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