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 

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;
}

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

<script>

// JavaScript implementation of the approach

const N = 5;

// Function to return the
// minimum moves required
function minMoves(mat)
{

    // Center of the matrix
    let cI = parseInt(N / 2), cJ = parseInt(N / 2);

    // Find the position of the 1
    let oneI = 0, oneJ = 0;
    for (let i = 0; i < N; i++) {
        for (let 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
    let 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 ] ];

    document.write(minMoves(mat));

</script>

Output:

2

Time Complexity: O(N^2)

Auxiliary Space Complexity: O(1)