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

• Last Updated : 07 Jun, 2021

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:
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:

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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 ``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

 `// 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

 `# 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#

 `// 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`

## Javascript

 ``
Output:
`2`

My Personal Notes arrow_drop_up