PROBLEM DESCRIPTION
Given a 2D binary matrix A(0-based index) of dimensions NxM. Find the minimum number of steps required to reach from (0,0) to (X, Y). Note: You can only move left, right, up and down, and only through cells that contain 1.
SOLUTION
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class Solution {
int shortestDistance(int n, int m, int A[][], int X, int Y) {
int count = 0;
// Edge case: Check if the starting point or the destination is blocked
if(A[0][0] == 0 || A[X][Y] == 0)
return -1;
// Create a queue to store the coordinates for BFS
Queue<int[]> q = new LinkedList<>();
// Start BFS from the top-left corner (0,0)
q.add(new int[]{0, 0});
// Continue BFS until there are no more cells to explore
while(!q.isEmpty()){
// Get the number of elements in the current level of BFS
int qSize = q.size();
// Process each cell in the current level
for(int i = 0; i < qSize; i++){
int currentX = q.peek()[0];
int currentY = q.peek()[1];
q.poll(); // Remove the current cell from the queue
// Mark the current cell as visited by setting it to 0
A[currentX][currentY] = 0;
// Return the number of steps if the destination is reached
if(currentX == X && currentY == Y)
return count;
// Arrays to move in four possible directions: right, left, down, up
int[] x = {0, 0, 1, -1};
int[] y = {1, -1, 0, 0};
// Explore all four possible directions
for(int k = 0; k < 4; k++){
int nextX = currentX + x[k];
int nextY = currentY + y[k];
// Check if the new position is within bounds and not yet visited
if(isValid(n, m, nextX, nextY) && A[nextX][nextY] == 1){
// Add the new position to the queue
q.add(new int[]{nextX, nextY});
// Mark the new position as visited
A[nextX][nextY] = 0;
}
}
}
// We did not reach the destination, so need another step
count++;
}
// Return -1 if the destination cannot be reached
return -1;
}
// Helper function to check if a given cell (i, j) is within the bounds of the matrix
boolean isValid(int n, int m, int i, int j){
if(i < 0 || j < 0 || i >= n || j >= m)
return false;
return true;
}
};