Minimum steps to reach any of the boundary edges of a matrix | Set-2

Given an N X M matrix, where ai, j = 1 denotes the cell is not empty, ai, j = 0 denotes the cell is empty and ai, j = 2, denotes that you are standing at that cell. You can move vertically up or down and horizontally left or right to any cell which is empty. The task is to find the minimum number of steps to reach any boundary edge of the matrix. Print -1 if not possible to reach any of the boundary edges.

Note: There will be only one cell with value 2 in the entire matrix.

Examples:



Input: matrix[] = {1, 1, 1, 0, 1}
                  {1, 0, 2, 0, 1} 
                  {0, 0, 1, 0, 1}
                  {1, 0, 1, 1, 0} 
Output: 2
Move to the right and then move 
upwards to reach the nearest boundary
edge. 

Input: matrix[] = {1, 1, 1, 1, 1}
                  {1, 0, 2, 0, 1} 
                  {1, 0, 1, 0, 1}
                  {1, 1, 1, 1, 1}
Output: -1 

Approach: The problem has been solved in Set-1 using Dynamic Programming. In this article, we will be discussing a method using BFS. Given below are the steps to solve the above problem.

  • Find the index of the ‘2’ in the matrix.
  • Check if this index is a boundary edge or not, if it is, then no moves are required.
  • Insert the index x and index y of ‘2’ in the queue with moves as 0.
  • Use a 2-D vis array to mark the visiting positions in the matrix.
  • Iterate till the queue is empty or we reach any boundary edge.
  • Get the front element(x, y, val = moves) in the queue and mark vis[x][y] as visited. Do all the possible moves(right, left, up and down) possible.
  • Re-insert val+1 and their indexes of all the valid moves to the queue.
  • If the x and y become the boundary edges any time return val.
  • If all the moves are made, and the queue is empty, then it is not possible, hence return -1.

Below is the implementation of the above approach:

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// C++ program to find Minimum steps
// to reach any of the boundary
// edges of a matrix
#include <bits/stdc++.h>
using namespace std;
#define r 4
#define c 5
  
// Function to check validity
bool check(int i, int j, int n, int m, int mat[r])
{
    if (i >= 0 && i < n && j >= 0 && j < m) {
        if (mat[i][j] == 0)
            return true;
    }
    return false;
}
  
// Function to find out minimum steps
int findMinSteps(int mat[r], int n, int m)
{
  
    int indx, indy;
    indx = indy = -1;
  
    // Find index of only 2 in matrix
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (mat[i][j] == 2) {
                indx = i, indy = j;
                break;
            }
        }
        if (indx != -1)
            break;
    }
  
    // Push elements in the queue
    queue<pair<int, pair<int, int> > > q;
  
    // Push the position 2 with moves as 0
    q.push(make_pair(0, make_pair(indx, indy)));
  
    // If already at boundary edge
    if (check(indx, indy, n, m, mat))
        return 0;
  
    // Marks the visit
    bool vis[r];
    memset(vis, 0, sizeof vis);
  
    // Iterate in the queue
    while (!q.empty()) {
        // Get the front of the queue
        auto it = q.front();
  
        // Pop the first element from the queue
        q.pop();
  
        // Get the position
        int x = it.second.first;
        int y = it.second.second;
  
        // Moves
        int val = it.first;
  
        // If a boundary edge
        if (x == 0 || x == (n - 1) || y == 0 || y == (m - 1)) {
            return val;
        }
  
        // Marks the visited array
        vis[x][y] = 1;
  
        // If a move is possible
        if (check(x - 1, y, n, m, mat)) {
  
            // If not visited previously
            if (!vis[x - 1][y])
                q.push(make_pair(val + 1, make_pair(x - 1, y)));
        }
  
        // If a move is possible
        if (check(x + 1, y, n, m, mat)) {
  
            // If not visited previously
            if (!vis[x + 1][y])
                q.push(make_pair(val + 1, make_pair(x + 1, y)));
        }
  
        // If a move is possible
        if (check(x, y + 1, n, m, mat)) {
  
            // If not visited previously
            if (!vis[x][y + 1])
                q.push(make_pair(val + 1, make_pair(x, y + 1)));
        }
  
        // If a move is possible
        if (check(x, y - 1, n, m, mat)) {
  
            // If not visited previously
            if (!vis[x][y - 1])
                q.push(make_pair(val + 1, make_pair(x, y - 1)));
        }
    }
  
    return -1;
}
  
// Driver Code
int main()
{
    int mat[r] = { { 1, 1, 1, 0, 1 },
                      { 1, 0, 2, 0, 1 },
                      { 0, 0, 1, 0, 1 },
                      { 1, 0, 1, 1, 0 } };
  
    cout << findMinSteps(mat, r, c);
}

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Output:

2

Time Complexity: O(N^2)
Auxiliary Space: O(N^2)



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