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
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.
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Below is the implementation of the above approach:
C++
// 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; // Total cells = square of the size of the matrices int total = n * n; // Initialising the answer int ans = total - zeroes; return ans; } // Driver code int main() { // Initialising 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; // Total cells = square of // the size of the matrices int total = n * n; // Initialising the answer int ans = total - zeroes; return ans; } // Driver code public static void main (String[] args) { // Initialising 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; # Total cells = square of # the size of the matrices total = n * n; # Initialising # the answer ans = total - zeroes; return ans; # Driver code # Initialising 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; // Total cells = square of // the size of the matrices int total = n * n; // Initialising the answer int ans = total - zeroes; return ans; } // Driver code static public void Main () { // Initialising the // variables int n = 5; int x = 2; Console.WriteLine(getMaxOnes(n, x)); } } // This code is contributed // by ajit |
PHP
<?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 ; // Total cells = square of // the size of the matrices $total = $n * $n ; // Initialising // the answer $ans = $total - $zeroes ; return $ans ; } // Driver code // Initialising // the variables $n = 5; $x = 2; echo getMaxOnes( $n , $x ); // This code is contributed // by akt_mit ?> |
Javascript
<script> // Javascript 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 let zeroes = parseInt(n / x, 10); zeroes = zeroes * zeroes; // Total cells = square of the // size of the matrices let total = n * n; // Initialising the answer let ans = total - zeroes; return ans; } // Driver Code // Initialising the variables let n = 5; let x = 2; document.write(getMaxOnes(n, x)); // This code is contributed by mukesh07 </script> |
Output
21
Complexity Analysis:
- Time Complexity: O(1)
- Auxiliary Space: O(1)