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

**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++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `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; ` `} ` |

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## Java

`// 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 ` |

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## Python3

`# 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 ` |

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## C#

`// 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 ` |

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**Output:**

2

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