# 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
```

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

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:

 `// CPP for maximum |i-j| such that Aij = 0 ` `#include ` `#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; ` `} `

 `// 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)); ` `} ` ` `  `} `

 `# 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 `

 `// 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 `

 `

Output:
```2
```

Time complexity: O(n^2)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

Article Tags :
Practice Tags :