Minimum Distance from Boundary to Target on a 2D Matrix

Last Updated : 02 Oct, 2023

Given a matrix of size (n*m) and a point 'p' (x, y). The matrix is given in the form of 0's and 1's. In one move you can go in any of the four possible directions (up, down, left, or right) inside the boundary of the matrix whose value is 1. The task is to find the minimum number of moves required to reach 'p', when started from the boundary of the matrix. Print -1 if 'p' can not be reached. Note: All indexing used is zero-based.

Examples:

Input: n = 5, m = 5
0 1 0 0 0
0 1 1 0 0
0 1 1 1 0
0 0 0 1 1
0 0 0 1 0

(x, y) = (1, 2)

Output: 2
Explanation: Some possible paths are:

(4, 3) -> (3, 3) -> (2, 3) -> (2, 2) -> (1, 2). Steps = 4.
(3, 4) -> (3, 3) -> (2, 3) -> (2, 2) -> (1, 2). Steps = 4.
(0, 1) -> (1, 1) -> (1, 2). Steps = 2.
And many more. Out of which 2 is the minimum possible steps.

Input: n = 5, m = 5
0 1 0 0 0
0 1 0 0 0
0 1 1 1 0
0 0 0 1 1
0 0 0 1 0

(x, y) = (1, 2)

Output: -1
Explanation: Index (1, 2) is already zero so 'q' can never reach to this index hence the output is -1.

Approach: This can be solved with the following idea:

We can do BFS for this problem. Start BFS from the index (x, y) and traverse in all four possible directions. Because we are doing BFS here so the shortest path will which leads out of the bounds of the matrix will appear first hence as soon as it meets stop the BFS and return the answer.

Below are the steps involved in this approach-

Start the BFS from the index (x, y) with moves as zero.
Go in all four unvisited directions whose value is 1.
Increase the moves by one for each BFS call.
When any index reaches out of the bounds the number of moves till that point is our answer.

See the below code implementation to understand better:

C++

// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;

class Solution {

    int bfs(vector<vector<int> >& matrix, int& x, int& y)
    {
        if (matrix[x][y] == 0)

            return -1;

        // Queue for bfs traversal
        queue<vector<int> > q;

        int n = matrix.size(), m = matrix[0].size();
        q.push({ x, y, 0 });

        int ans = -1;
        vector<vector<int> > visited(n, vector<int>(m, 0));

        while (q.size()) {
            vector<int> v = q.front();
            int i = v[0], j = v[1], moves = v[2];
            q.pop();

            if (i >= n || i < 0 || j >= m || j < 0) {

                // As soon as p reaches
                // out of the bounds stop
                // bfs calls
                ans = moves;

                break;
            }
            else {
                if (visited[i][j])
                    continue;
                visited[i][j] = 1;

                // Bfs calls in all
                // possible directions
                if (i + 1 >= n || matrix[i + 1][j])
                    q.push({ i + 1, j, moves + 1 });

                if (i - 1 < 0 || matrix[i - 1][j])
                    q.push({ i - 1, j, moves + 1 });

                if (j + 1 >= m || matrix[i][j + 1])
                    q.push({ i, j + 1, moves + 1 });

                if (j - 1 < 0 || matrix[i][j - 1])
                    q.push({ i, j - 1, moves + 1 });
            }
        }

        return ans;
    }

public:
    int minMoves(vector<vector<int> >& matrix, int x, int y)
    {
        int ans = bfs(matrix, x, y);

        return (ans >= 1e7) ? -1 : ans - 1;
    }
};

// Driver code
int main()
{

    vector<vector<int> > matrix = { { 0, 1, 0, 0, 0 },
                                    { 0, 1, 1, 0, 0 },
                                    { 0, 1, 1, 1, 0 },
                                    { 0, 0, 0, 1, 1 },
                                    { 0, 0, 0, 1, 0 } };

    int x = 1, y = 2;

    Solution obj;

    // Function call
    cout << obj.minMoves(matrix, x, y) << endl;

    return 0;
}

Java

// Java code for the above approach:
import java.util.ArrayDeque;
import java.util.Queue;

class Solution {
    public int bfs(int[][] matrix, int x, int y) {
        if (matrix[x][y] == 0) {
            return -1;
        }
      
          // Queue for bfs traversal
        Queue<int[]> q = new ArrayDeque<>();
        int n = matrix.length;
        int m = matrix[0].length;
        q.offer(new int[]{x, y, 0});
        int ans = -1;
        int[][] visited = new int[n][m];
        while (!q.isEmpty()) {
            int[] curr = q.poll();
            int i = curr[0];
            int j = curr[1];
            int moves = curr[2];
              
          
              if (i >= n || i < 0 || j >= m || j < 0) {
              // As soon as p reaches
            // out of the bounds stop
            // bfs calls
                ans = moves;
                break;
            } else {
                if (visited[i][j] == 1) {
                    continue;
                }
                visited[i][j] = 1;
              
                  // Bfs calls in all
                // possible directions
                if (i + 1 >= n || matrix[i + 1][j] == 1) {
                    q.offer(new int[]{i + 1, j, moves + 1});
                }
                if (i - 1 < 0 || matrix[i - 1][j] == 1) {
                    q.offer(new int[]{i - 1, j, moves + 1});
                }
                if (j + 1 >= m || matrix[i][j + 1] == 1) {
                    q.offer(new int[]{i, j + 1, moves + 1});
                }
                if (j - 1 < 0 || matrix[i][j - 1] == 1) {
                    q.offer(new int[]{i, j - 1, moves + 1});
                }
            }
        }
        return ans;
    }
    
    public int minMoves(int[][] matrix, int x, int y) {
        Solution obj = new Solution();
        int ans = obj.bfs(matrix, x, y);
        return ans >= 1e7 ? -1 : ans - 1;
    }
}

// Driver code
public class Main {
    public static void main(String[] args) {
        int[][] matrix = {{0, 1, 0, 0, 0},
                          {0, 1, 1, 0, 0},
                          {0, 1, 1, 1, 0},
                          {0, 0, 0, 1, 1},
                          {0, 0, 0, 1, 0}};
        int x = 1;
        int y = 2;
        Solution obj = new Solution();
        System.out.println(obj.minMoves(matrix, x, y));
    }
}

Python3

# python code for the above approach:

from collections import deque

class Solution:

    def bfs(self, matrix, x, y):
        if matrix[x][y] == 0:
            return -1

        # Queue for BFS traversal
        q = deque()
        n = len(matrix)
        m = len(matrix[0])
        q.append([x, y, 0])

        ans = -1
        visited = [[0] * m for _ in range(n)]

        while q:
            i, j, moves = q.popleft()

            if i >= n or i < 0 or j >= m or j < 0:
                # As soon as the cell reaches out of bounds, stop BFS
                ans = moves
                break
            else:
                if visited[i][j]:
                    continue
                visited[i][j] = 1

                # BFS calls in all possible directions
                if i + 1 >= n or matrix[i + 1][j]:
                    q.append([i + 1, j, moves + 1])

                if i - 1 < 0 or matrix[i - 1][j]:
                    q.append([i - 1, j, moves + 1])

                if j + 1 >= m or matrix[i][j + 1]:
                    q.append([i, j + 1, moves + 1])

                if j - 1 < 0 or matrix[i][j - 1]:
                    q.append([i, j - 1, moves + 1])

        return ans

    def minMoves(self, matrix, x, y):
        obj = Solution()
        ans = obj.bfs(matrix, x, y)

        return -1 if ans >= 1e7 else ans - 1


# Driver code
def main():
    matrix = [[0, 1, 0, 0, 0],
              [0, 1, 1, 0, 0],
              [0, 1, 1, 1, 0],
              [0, 0, 0, 1, 1],
              [0, 0, 0, 1, 0]]

    x = 1
    y = 2

    obj = Solution()

    # Function call
    print(obj.minMoves(matrix, x, y))


main()

using System;
using System.Collections.Generic;

class Solution
{
    private int BFS(int[][] matrix, ref int x, ref int y)
    {
        if (matrix[x][y] == 0)
            return -1;

        // Queue for BFS traversal
        Queue<int[]> q = new Queue<int[]>();

        int n = matrix.Length;
        int m = matrix[0].Length;
        q.Enqueue(new int[] { x, y, 0 });

        int ans = -1;
        int[][] visited = new int[n][];
        for (int i = 0; i < n; i++)
        {
            visited[i] = new int[m];
            for (int j = 0; j < m; j++)
            {
                visited[i][j] = 0;
            }
        }

        while (q.Count > 0)
        {
            int[] v = q.Dequeue();
            int i = v[0], j = v[1], moves = v[2];

            if (i >= n || i < 0 || j >= m || j < 0)
            {
                // As soon as (x, y) reaches out of bounds, stop BFS
                ans = moves;
                break;
            }
            else
            {
                if (visited[i][j] == 1)
                    continue;
                visited[i][j] = 1;

                // BFS in all possible directions
                if (i + 1 >= n || matrix[i + 1][j] == 1)
                    q.Enqueue(new int[] { i + 1, j, moves + 1 });

                if (i - 1 < 0 || matrix[i - 1][j] == 1)
                    q.Enqueue(new int[] { i - 1, j, moves + 1 });

                if (j + 1 >= m || matrix[i][j + 1] == 1)
                    q.Enqueue(new int[] { i, j + 1, moves + 1 });

                if (j - 1 < 0 || matrix[i][j - 1] == 1)
                    q.Enqueue(new int[] { i, j - 1, moves + 1 });
            }
        }

        return ans;
    }

    public int MinMoves(int[][] matrix, int x, int y)
    {
        int ans = BFS(matrix, ref x, ref y);
        return (ans >= 1e7) ? -1 : ans - 1;
    }
}

class Program
{
    static void Main()
    {
        int[][] matrix = {
            new int[] { 0, 1, 0, 0, 0 },
            new int[] { 0, 1, 1, 0, 0 },
            new int[] { 0, 1, 1, 1, 0 },
            new int[] { 0, 0, 0, 1, 1 },
            new int[] { 0, 0, 0, 1, 0 }
        };

        int x = 1, y = 2;

        Solution obj = new Solution();

        // Function call
        Console.WriteLine(obj.MinMoves(matrix, x, y));
    }
}

JavaScript

class Solution {
  
  // Function for breadth first search
  bfs(matrix, x, y) {
    if (matrix[x][y] === 0) {
      return -1;
    }
    
    // Queue for bfs traversal
    const q = [];
    const n = matrix.length;
    const m = matrix[0].length;
    q.push([x, y, 0]);
    let ans = -1;
    const visited = Array.from({ length: n }, () => Array(m).fill(0));
    while (q.length > 0) {
      const [i, j, moves] = q.shift();
      if (i >= n || i < 0 || j >= m || j < 0) {
        
        // As soon as p reaches
        // out of the bounds stop
        // bfs calls
        ans = moves;
        break;
      } else {
        if (visited[i][j]) {
          continue;
        }
        visited[i][j] = 1;
        
        // Bfs calls in all
        // possible directions
        if (i + 1 >= n || matrix[i + 1][j]) {
          q.push([i + 1, j, moves + 1]);
        }
        if (i - 1 < 0 || matrix[i - 1][j]) {
          q.push([i - 1, j, moves + 1]);
        }
        if (j + 1 >= m || matrix[i][j + 1]) {
          q.push([i, j + 1, moves + 1]);
        }
        if (j - 1 < 0 || matrix[i][j - 1]) {
          q.push([i, j - 1, moves + 1]);
        }
      }
    }
    return ans;
  }
  
  // Function to find minimum moves
  minMoves(matrix, x, y) {
    const obj = new Solution();
    const ans = obj.bfs(matrix, x, y);
    return ans >= 1e7 ? -1 : ans - 1;
  }
}


  // Test case
  const matrix = [[0, 1, 0, 0, 0],
                  [0, 1, 1, 0, 0],
                  [0, 1, 1, 1, 0],
                  [0, 0, 0, 1, 1],
                  [0, 0, 0, 1, 0]];
  const x = 1;
  const y = 2;
  const obj = new Solution();
  console.log(obj.minMoves(matrix, x, y));

Output

Time Complexity: O(n*m)
Auxiliary Space: O(n*m)

Minimum Distance from a given Cell to all other Cells of a Matrix

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Minimum Distance from Boundary to Target on a 2D Matrix

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