# Minimize steps required to move all 1’s in a matrix to a given index

Given a binary matrix mat[][] of size NxM and two integers X and Y, the task is to find the minimum number steps required to move all 1’s in the given matrix to the cell (X, Y), where, one step involves moving a cell left, right, up or down.
Examples:

Input: mat[][] = { {1, 0, 1}, {0, 1, 0}, {1, 0, 1} }, X = 1, Y = 1
Output:
Explanation:
Cells (0, 0), (0, 2), (1, 1), (2, 0) and (2, 2) consists of 1.
Moving 1 at index (0, 0) to (1, 1) requires 2 steps
(0, 0) -> (0, 1) ->(1, 0)
Moving 1 at index (0, 2) to (1, 1) requires 2 steps
Moving 1 at index (2, 0) to (1, 1) requires 2 steps
Moving 1 at index (2, 2) to (1, 1) requires 2 steps
Therefore, 8 steps are required.

Input: mat[][] = { {1, 0, 0, 0}, {0, 1, 0, 1}, {1, 0, 1, 1} }, X = 0, Y = 2
Output: 15

Approach:
The idea is to traverse the given matrix and find the cells consisting of 1. For any cell (i, j) consisting of 1, minimum steps required to reach (X, Y) based on the given directions is given by:

Minimum steps = abs(X - i) + abs(Y - j)

Calculate the total number of steps required using the above formula for each cell that contains 1 in the given matrix mat[][].

Below is the implementation of the above approach:

 // C++ program to calculate  // the minimum steps  // required to reach  // a given cell from all  // cells consisting of 1's  #include   using namespace std;     // Function to calculate and  // return the minimum  // number of steps required  // to move all 1s to (X, Y)  int findAns(vector > mat,              int x, int y,              int n, int m)  {      int ans = 0;         // Iterate the given matrix      for (int i = 0; i < n; i++) {             for (int j = 0; j < m; j++) {                 // Update the answer with              // minimum moves required              // for the given element              // to reach the given index              if (mat[i][j] == 1) {                     ans += abs(x - i)                      + abs(y - j);              }          }      }         // Return the number      // of steps      return ans;  }     // Driver Code  int main()  {      // Given matrix      vector > mat          = { { 1, 0, 0, 0 },              { 0, 1, 0, 1 },              { 1, 0, 1, 1 } };         // Given position      int x = 0, y = 2;         // Function Call      cout << findAns(mat, x, y,                      mat.size(),                      mat[0].size())          << endl;      return 0;  }

 // Java program to calculate the  // minimum steps required to reach // a given cell from all cells  // consisting of 1's  import java.util.*;    class GFG{    // Function to calculate and // return the minimum number // of steps required to move // all 1s to (X, Y) static int findAns(int [][]mat,                    int x, int y,                    int n, int m) {     int ans = 0;        // Iterate the given matrix     for(int i = 0; i < n; i++)     {         for(int j = 0; j < m; j++)          {                            // Update the answer with             // minimum moves required             // for the given element             // to reach the given index             if (mat[i][j] == 1)             {                 ans += Math.abs(x - i) +                        Math.abs(y - j);             }         }     }        // Return the number     // of steps     return ans; }    // Driver code public static void main(String[] args) {            // Given matrix     int [][]mat = { { 1, 0, 0, 0 },                     { 0, 1, 0, 1 },                     { 1, 0, 1, 1 } };        // Given position     int x = 0, y = 2;        // Function Call     System.out.print(findAns(mat, x, y,                              mat.length,                              mat[0].length) + "\n"); } }    // This code is contributed by amal kumar choubey

 # Python3 program to calculate  # the minimum steps required to  # reach a given cell from all  # cells consisting of 1's     # Function to calculate and  # return the minimum number  # of steps required to move  # all 1s to (X, Y)  def findAns(mat, x, y, n, m):             ans = 0            # Iterate the given matrix      for i in range(n):          for j in range(m):                             # Update the answer with              # minimum moves required              # for the given element              # to reach the given index              if (mat[i][j] == 1):                  ans += (abs(x - i) +                         abs(y - j))             # Return the number      # of steps      return ans         # Driver Code     # Given matrix  mat = [ [ 1, 0, 0, 0 ],          [ 0, 1, 0, 1 ],          [ 1, 0, 1, 1 ] ]     # Given position  x = 0 y = 2    # Function call  print(findAns(mat, x, y, len(mat),                          len(mat[0])))     # This code is contributed by shubhamsingh10

 // C# program to calculate the  // minimum steps required to reach // a given cell from all cells  // consisting of 1's  using System;    class GFG{    // Function to calculate and // return the minimum number // of steps required to move // all 1s to (X, Y) static int findAns(int [,]mat,                    int x, int y,                    int n, int m) {     int ans = 0;        // Iterate the given matrix     for(int i = 0; i < n; i++)     {         for(int j = 0; j < m; j++)          {                            // Update the answer with             // minimum moves required             // for the given element             // to reach the given index             if (mat[i, j] == 1)             {                 ans += Math.Abs(x - i) +                        Math.Abs(y - j);             }         }     }        // Return the number     // of steps     return ans; }    // Driver code public static void Main(String[] args) {            // Given matrix     int [,]mat = { { 1, 0, 0, 0 },                    { 0, 1, 0, 1 },                    { 1, 0, 1, 1 } };        // Given position     int x = 0, y = 2;        // Function call     Console.Write(findAns(mat, x, y,                           mat.GetLength(0),                           mat.GetLength(1)) + "\n"); } }    // This code is contributed by Rajput-Ji

Output:
15

Time Complexity: O(N*M)
Auxiliary Space: O(1)

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