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

• Last Updated : 30 Aug, 2022

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++

 `// 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

 `// 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

 `# 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#

 `// 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`

## Javascript

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Output:

`15`

Time Complexity: O(N*M), since two nested loops are used the time taken by the algorithm to complete all operations is N*M.
Auxiliary Space: O(1) since no extra array is used so the space taken by the algorithm is constant

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