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

Given two integers n and x, where x <= n. Find the maximum number of one’s in a n * n binary matrix can have such that every sub-matrix of size x * x 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 

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 \left \lfloor (\frac{n}{x})^2 \right \rfloor number of zeroes, in the minimum arrangement. The total number of cells available is n^2 in a NxN matrix.

max(ones) = n^2 - \left \lfloor (\frac{n}{x})^2 \right \rfloor.

Below is the implementation of the above approach:

C++

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// C++ program to get Maximum Number of
// ones in a matrix with given constraints
#include <bits/stdc++.h>
  
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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PHP

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<?php
// PHP program to get Maximum 
// Number of ones in a matrix
// with given constraints
  
// Function that returns 
// the maximum number of ones 
function 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;
$x = 2;
echo getMaxOnes($n, $x);
  
// This code is contributed
// by akt_mit
?>

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Output:

21

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



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I am a problem solving enthusiast and I love competitive programming

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Improved By : jit_t, Mithun Kumar