# Farthest distance of a 0 from the center of a 2-D matrix

Given a matrix of odd order mat, the task is to find the farthest distance of a 0 from the center of the matrix. Distance between two elements at locations (i1, j1) and (i2, j2) of the matrix is calculated as |i1- i2| + |j1-j2|. If no 0 occurs in the matrix then print 0 as the result.

Examples:

Input: mat[][] = {{2, 3, 0}, {0, 2, 0}, {0, 1, 1}}
Output: 2

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

Approach:

The center of any matrix with odd order is at index i = j = floor(n/2). Now for finding the farthest distance of any 0 from the center, calculate the distance of each 0 from the center of the matrix as |i-n/2| + |j-n/2| and update the maximum distance as result. Print the result in the end or if the matrix doesn’t contain any 0 then print 0.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the farthest distance` `// of a 0 from the center of the matrix` `#include ` `#define n 3` `using` `namespace` `std;`   `// function to return farthest distance` `// of zero from center of the matrix` `int` `farthestDistance(``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 - n/2) + ``abs``(j - n/2));` `        ``}` `    ``}`   `    ``// return result` `    ``return` `result;` `}`   `// driver program` `int` `main()` `{` `    ``int` `matrix[n][n] = { { 1, 2, 3 }` `                       ``, { 0, 1, 1 }` `                       ``, { 0, 0, 0 } };`   `    ``cout << farthestDistance(matrix);` `    ``return` `0;` `}`

## Java

 `//  Java program to find the farthest distance` `// of a 0 from the center of the matrix`   `import` `java.io.*;`   `class` `GFG {` `   `    `static` `int` `n = ``3``;`     `// function to return farthest distance` `// of zero from center of the matrix` `static` `int` `farthestDistance(``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 - n/``2``) + Math.abs(j - n/``2``));` `        ``}` `    ``}`   `    ``// return result` `    ``return` `result;` `}`   `// driver program`   `    ``public` `static` `void` `main (String[] args) {` `            ``int` `matrix[][] = { { ``1``, ``2``, ``3` `}` `                    ``, { ``0``, ``1``, ``1` `}` `                    ``, { ``0``, ``0``, ``0` `} };`   `    ``System.out.print(farthestDistance(matrix));` `    ``}` `}` `// This code is contributed by anuj_67..`

## Python3

 `# Python3 program to find the farthest distance` `# of a 0 from the center of the matrix`   `n ``=` `3`   `# function to return farthest distance` `# of zero from center of the matrix` `def` `farthestDistance(matrix):` `    ``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 ``-` `n ``/``/` `2``) ``+` `                                     ``abs``(j ``-` `n``/``/``2``))` `        `  `    ``# return result` `    ``return` `result`   `# Driver Code` `matrix ``=` `[[``1``, ``2``, ``3``], ` `          ``[``0``, ``1``, ``1``],` `          ``[``0``, ``0``, ``0``]]`   `print``(farthestDistance(matrix))`   `# This code is contributed by` `# Archana_kumari`

## C#

 `// C# program to find the farthest distance ` `// of a 0 from the center of the matrix ` `using` `System;`   `class` `GFG` `{` `    `  `static` `int` `n = 3; `   `// function to return farthest distance ` `// of zero from center of the matrix ` `static` `int` `farthestDistance(``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 - n / 2) +` `                         ``Math.Abs(j - n / 2)); ` `        ``} ` `    ``} `   `    ``// return result ` `    ``return` `result; ` `} `   `// Driver Code ` `static` `public` `void` `Main ()` `{` `    ``int` `[,]matrix = {{ 1, 2, 3 }, ` `                     ``{ 0, 1, 1 }, ` `                     ``{ 0, 0, 0 }}; `   `    ``Console.WriteLine(farthestDistance(matrix)); ` `} ` `} `   `// This code is contributed by Sachin `

## PHP

 ` 0) ` `                ``\$result` `= max(``\$result``, ` `                          ``abs``(``\$i` `- ``\$n` `/ 2) + ` `                          ``abs``(``\$j` `- ``\$n` `/ 2)); ` `        ``} ` `    ``} `   `    ``// return result ` `    ``return` `\$result``; ` `} `   `// Driver Code` `\$matrix` `= ``array``(``array``( 1, 2, 3 ), ` `                ``array``( 0, 1, 1 ),` `                ``array``( 0, 0, 0 )); `   `echo` `farthestDistance(``\$matrix``); ` `    `  `// This code is contributed by Sach_code` `?>`

## Javascript

 ``

Output

`2`

Complexity Analysis:

• Time Complexity: O(N*M), as we are using nested loops to traverse N*M times.
• Auxiliary Space: O(1), as we are not using any extra space for matrix.

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