# Maximum number of ones in a N*N matrix with given constraints

Given two integers and , where . Find the maximum number of one’s in a binary matrix can have such that every sub-matrix of size has atleast one cell as zero.

Examples:

Input:5 3
Output: Maximum number of ones = 24
The matrix will be:
1 1 1 1 1
1 1 1 1 1
1 1 0 1 1
1 1 1 1 1
1 1 1 1 1

Input:5 2
Output: Maximum number of ones = 21
The matrix will be:
1 1 1 1 1
1 0 1 0 1
1 1 1 1 1
1 0 1 0 1
1 1 1 1 1 

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach The problem can be solved using a greedy approach. Place a zero at the right-bottom corner of the first square sub-matrix, i.e. the sub-matrix with coordinates (1, 1) and (x, x), and create the rest of the matrix symmetrically, we can get the minimum number of zeros, or, the maximum number of ones. Thus by observing, a common conclusion can be drawn that there are number of zeroes, in the minimum arrangement. The total number of cells available is in a NxN matrix. .

Below is the implementation of the above approach:

## C++

 // C++ program to get Maximum Number of  // ones in a matrix with given constraints  #include     using namespace std;     // Function that returns the maximum   // number  of ones   int getMaxOnes(int n, int x)  {      // Minimum number of zeroes      int zeroes = (n / x);      zeroes = zeroes * zeroes;         // Totol cells = square of the size of the matrices      int total = n * n;         // Intialising the answer      int ans =  total - zeroes;         return ans;   }     // Driver code  int main()  {      // Intitalising the variables      int n = 5;      int x = 2;                cout << getMaxOnes(n, x);         return 0;  }

## Java

 // Java program to get Maximum  // Number of ones in a matrix   // with given constraints  import java.io.*;     class GFG  {         // Function that returns   // the maximum number of ones   static int getMaxOnes(int n,                         int x)  {      // Minimum number of zeroes      int zeroes = (n / x);      zeroes = zeroes * zeroes;         // Totol cells = square of      // the size of the matrices      int total = n * n;         // Intialising the answer      int ans = total - zeroes;         return ans;   }     // Driver code  public static void main (String[] args)   {     // Intitalising the variables  int n = 5;  int x = 2;  System.out.println(getMaxOnes(n, x));  }  }     // This code is contributed  // by akt_mit

## Python3

 # Python3 program to get   # Maximum Number of ones   # in a matrix with given  # constraints     # Function that returns   # the maximum number of ones   def getMaxOnes(n, x):             # Minimum number      # of zeroes      zeroes = (int)(n / x);      zeroes = zeroes * zeroes;         # Totol cells = square of      # the size of the matrices      total = n * n;         # Intialising       # the answer      ans = total - zeroes;         return ans;      # Driver code     # Intitalising the variables  n = 5;  x = 2;  print(getMaxOnes(n, x));     # This code is contributed  # by mits

## C#

 // C# program to get Maximum  // Number of ones in a matrix   // with given constraints  using System;     class GFG  {         // Function that returns   // the maximum number of ones   static int getMaxOnes(int n,                         int x)  {      // Minimum number of zeroes      int zeroes = (n / x);      zeroes = zeroes * zeroes;         // Totol cells = square of      // the size of the matrices      int total = n * n;         // Intialising the answer      int ans = total - zeroes;         return ans;   }     // Driver code  static public void Main ()  {                 // Intitalising the      // variables      int n = 5;      int x = 2;      Console.WriteLine(getMaxOnes(n, x));  }  }     // This code is contributed  // by ajit

## PHP

 

Output:

21


Time Complexity: O(1)
Auxiliary Space: O(1)

My Personal Notes arrow_drop_up I am a problem solving enthusiast and I love competitive programming

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Improved By : jit_t, Mithun Kumar