Maximum difference of indices (i, j) such that A[i][j] = 0 in the given matrix

Given a matrix of order n*n, the task is to find the maximum value of |i-j| such that Aij = 0. Given matrix must contain at least one 0.

Examples:

Input: matrix[][] = 
{{2, 3, 0},
{0, 2, 0},
{0, 1, 1}}
Output: 2
mat(0, 2) has a value 0 and difference of index is maximum i.e. 2.


Input: matrix[][] = 
{{2, 3, 4},
{0, 2, 0},
{6, 1, 1}}
Output: 1

Approach: For finding the maximum value of |i-j| such that Aij = 0, traverse the whole matrix and for each occurrence of zero calculate the mod of (i-j) and store it corresponding to same position in an auxiliary matrix. At last, find the maximum value from the auxiliary matrix.
Apart from using an auxiliary matrix, the maximum value of |i-j| can be stored in a variable and can be updated while its calculation. this will save the extra use of space.



Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP for maximum |i-j| such that Aij = 0
#include <bits/stdc++.h>
#define n 4
using namespace std;
  
// function to return maximum |i-j| such that Aij = 0
int calculateDiff(int matrix[][n])
{
  
    int result = 0;
  
    // traverse the matrix
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (matrix[i][j] == 0)
                result = max(result, abs(i - j));
        }
    }
  
    // return result
    return result;
}
  
// driver program
int main()
{
    int matrix[n][n] = { { 2, 3, 0, 1 },
                         { 0, 2, 0, 1 },
                         { 0, 1, 1, 3 },
                         { 1, 2, 3, 3 } };
  
    cout << calculateDiff(matrix);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for maximum |i-j| such that Aij = 0
import java.math.*;
class GFG {
      
static int n = 4;
  
// function to return maximum |i-j| such that Aij = 0
static int calculateDiff(int matrix[][])
{
  
    int result = 0;
  
    // traverse the matrix
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (matrix[i][j] == 0)
                result = Math.max(result, Math.abs(i - j));
        }
    }
  
    // return result
    return result;
}
  
// driver program
public static void main(String args[])
{
    int matrix[][] = new int[][] {{ 2, 3, 0, 1 },
                        { 0, 2, 0, 1 },
                        { 0, 1, 1, 3 },
                        { 1, 2, 3, 3 } };
  
    System.out.println(calculateDiff(matrix));
}
  
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program for maximum 
# |i-j| such that Aij = 0
  
# function to return maximum 
# |i-j| such that Aij = 0
def calculateDiff(matrix, n):
      
    result = 0
      
    # traverse the matrix
    for i in range(0, n):
        for j in range(0, n):
            if(matrix[i][j] == 0):
                result = max(result, abs(i - j))
                  
    return result
      
# Driver code
if __name__=='__main__':
    matrix = [[2, 3, 0, 1],
              [0, 2, 0, 1],
              [0, 1, 1, 3],
              [1, 2, 3, 3]]
    n = len(matrix)
    print(calculateDiff(matrix, n))
      
# This code is contributed by
# Kirti_Mangal

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# for maximum |i-j| such that Aij = 0
using System;
  
class GFG 
{
static int n = 4;
  
// function to return maximum |i-j|
// such that Aij = 0
static int calculateDiff(int [,]matrix)
{
    int result = 0;
  
    // traverse the matrix
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++) 
        {
            if (matrix[i, j] == 0)
                result = Math.Max(result,
                         Math.Abs(i - j));
        }
    }
  
    // return result
    return result;
}
  
// Driver code
static void Main()
{
    int [,]matrix = new int[,] 
    {
        { 2, 3, 0, 1 },
        { 0, 2, 0, 1 },
        { 0, 1, 1, 3 },
        { 1, 2, 3, 3 }
    };
  
    Console.WriteLine(calculateDiff(matrix));;
}
}
  
// This code is contributed by ANKITRAI1

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP for maximum |i-j| such that Aij = 0
  
// function to return maximum |i-j| 
// such that Aij = 0
function calculateDiff($matrix)
{
    $n = 4;
    $result = 0;
  
    // traverse the matrix
    for ($i = 0; $i < $n; $i++) 
    {
        for ($j = 0; $j < $n; $j++) 
        {
            if ($matrix[$i][$j] == 0)
                $result = max($result
                          abs($i - $j));
        }
    }
  
    // return result
    return $result;
}
  
// Driver Code
$matrix = array(array( 2, 3, 0, 1 ),
                array( 0, 2, 0, 1 ),
                array( 0, 1, 1, 3 ),
                array( 1, 2, 3, 3 ));
  
echo calculateDiff($matrix);
  
// This code is contributed
// by Akanksha Rai

chevron_right


Output:

2

Time complexity: O(n^2)



My Personal Notes arrow_drop_up

Discovering ways to develop a plane for soaring career goals

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.