# Minimum steps to get 1 at the center of a binary matrix

Given an N * N matrix where N is odd with all 0 values except for a single cell which has the value 1. The task is to find the minimum possible moves to get this 1 to the center of the matrix when in a single move, any two consecutive rows or two consecutive columns can be swapped.

Examples:

Input: mat[][] = {
{0, 0, 1, 0, 0},
{0, 0, 0, 0, 0},
{0, 0, 0, 0, 0},
{0, 0, 0, 0, 0},
{0, 0, 0, 0, 0}}
Output: 2
Operation 1: Swap the first and the second row.
Operation 2: Swap the second and the third row.

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

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

Approach:

• Calculate the position of the center of the matrix as (cI, cJ) = (⌊N / 2⌋, ⌊N / 2⌋).
• Find the position of the 1 in the matrix and store it in (oneI, oneJ).
• Now, the minimum possible moves will be abs(cI – oneI) + abs(cJ – oneJ).

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `const` `int` `N = 5; ` ` `  `// Function to return the ` `// minimum moves required ` `int` `minMoves(``int` `mat[N][N]) ` `{ ` ` `  `    ``// Center of the matrix ` `    ``int` `cI = N / 2, cJ = N / 2; ` ` `  `    ``// Find the position of the 1 ` `    ``int` `oneI = 0, oneJ = 0; ` `    ``for` `(``int` `i = 0; i < N; i++) { ` `        ``for` `(``int` `j = 0; j < N; j++) { ` `            ``if` `(mat[i][j] == 1) { ` `                ``oneI = i; ` `                ``oneJ = j; ` `                ``break``; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``return` `(``abs``(cI - oneI) + ``abs``(cJ - oneJ)); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `mat[N][N] = { { 0, 0, 0, 0, 0 }, ` `                      ``{ 0, 0, 0, 0, 0 }, ` `                      ``{ 0, 0, 0, 0, 0 }, ` `                      ``{ 0, 0, 0, 0, 0 }, ` `                      ``{ 0, 0, 1, 0, 0 } }; ` ` `  `    ``cout << minMoves(mat); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the approach ` `class` `GFG ` `{ ` ` `  `static` `int` `N = ``5``; ` ` `  `// Function to return the ` `// minimum moves required ` `static` `int` `minMoves(``int` `mat[][]) ` `{ ` ` `  `    ``// Center of the matrix ` `    ``int` `cI = N / ``2``, cJ = N / ``2``; ` ` `  `    ``// Find the position of the 1 ` `    ``int` `oneI = ``0``, oneJ = ``0``; ` `    ``for` `(``int` `i = ``0``; i < N; i++) ` `    ``{ ` `        ``for` `(``int` `j = ``0``; j < N; j++)  ` `        ``{ ` `            ``if` `(mat[i][j] == ``1``)  ` `            ``{ ` `                ``oneI = i; ` `                ``oneJ = j; ` `                ``break``; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``return` `(Math.abs(cI - oneI) + Math.abs(cJ - oneJ)); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `mat[][] = { { ``0``, ``0``, ``0``, ``0``, ``0` `}, ` `                    ``{ ``0``, ``0``, ``0``, ``0``, ``0` `}, ` `                    ``{ ``0``, ``0``, ``0``, ``0``, ``0` `}, ` `                    ``{ ``0``, ``0``, ``0``, ``0``, ``0` `}, ` `                    ``{ ``0``, ``0``, ``1``, ``0``, ``0` `} }; ` ` `  `    ``System.out.print(minMoves(mat)); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

 `# Python3 implementation of the approach ` `N ``=` `5` ` `  `# Function to return the ` `# minimum moves required ` `def` `minMoves(mat): ` ` `  `    ``# Center of the matrix ` `    ``cI ``=` `N ``/``/` `2` `    ``cJ ``=` `N ``/``/` `2` ` `  `    ``# Find the position of the 1 ` `    ``oneI ``=` `0` `    ``oneJ ``=` `0` `    ``for` `i ``in` `range``(N): ` `        ``for` `j ``in` `range``(N): ` `            ``if` `(mat[i][j] ``=``=` `1``): ` `                ``oneI ``=` `i ` `                ``oneJ ``=` `j ` `                ``break` ` `  `    ``return` `(``abs``(cI ``-` `oneI) ``+` `abs``(cJ ``-` `oneJ)) ` ` `  `# Driver code ` `mat ``=` `[[``0``, ``0``, ``0``, ``0``, ``0``], ` `       ``[``0``, ``0``, ``0``, ``0``, ``0``], ` `       ``[``0``, ``0``, ``0``, ``0``, ``0``], ` `       ``[``0``, ``0``, ``0``, ``0``, ``0``], ` `       ``[``0``, ``0``, ``1``, ``0``, ``0``]] ` ` `  `print``(minMoves(mat)) ` ` `  `# This code is contributed by Mohit Kumar `

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `static` `int` `N = 5; ` ` `  `// Function to return the ` `// minimum moves required ` `static` `int` `minMoves(``int` `[,]mat) ` `{ ` ` `  `    ``// Center of the matrix ` `    ``int` `cI = N / 2, cJ = N / 2; ` ` `  `    ``// Find the position of the 1 ` `    ``int` `oneI = 0, oneJ = 0; ` `    ``for` `(``int` `i = 0; i < N; i++) ` `    ``{ ` `        ``for` `(``int` `j = 0; j < N; j++)  ` `        ``{ ` `            ``if` `(mat[i, j] == 1)  ` `            ``{ ` `                ``oneI = i; ` `                ``oneJ = j; ` `                ``break``; ` `            ``} ` `        ``} ` `    ``} ` `    ``return` `(Math.Abs(cI - oneI) +  ` `            ``Math.Abs(cJ - oneJ)); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[,]mat = {{ 0, 0, 0, 0, 0 }, ` `                  ``{ 0, 0, 0, 0, 0 }, ` `                  ``{ 0, 0, 0, 0, 0 }, ` `                  ``{ 0, 0, 0, 0, 0 }, ` `                  ``{ 0, 0, 1, 0, 0 }}; ` ` `  `    ``Console.Write(minMoves(mat)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:
```2
```

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