Shortest Path by Removing K walls

Given a 2-D binary matrix of size n*m, where 0 represents an empty space while 1 represents a wall you cannot walk through. You are also given an integer k.
You can walk up, down, left, or right. Given that you can remove up to k walls, return the minimum number of steps to walk from the top left corner (0, 0) to the bottom right corner (n-1, m-1).

Note: If there is no way to walk from the top left corner to the bottom right corner, return -1.

• Time : O(nmk)
• Space : O(nmk)

class Solution {
    bool vis[51][51][51*51];

public:
    int shotestPath(vector<vector<int>> mat, int n, int m, int k) {
        if(n == 1 and m == 1) return 0;
        memset(vis, false, sizeof vis);
        queue<array<int,3>> q;
        int dy[4]{-1,0,1,0}, dx[4]{0,1,0,-1};
        q.push({0,0,k});
        vis[0][0][k] = true;
        int res = 1;
        
        while(!q.empty()) {
            int sz = q.size();
            while(sz--) {
                auto A = q.front(); q.pop();
                auto y = A[0], x = A[1], b = A[2];
                for(int i = 0; i < 4; i++) {
                    int ny = y + dy[i], nx = x + dx[i];
                    if(0 <= ny and ny < n and 0 <= nx and nx < m) {
                        if(mat[ny][nx] == 0 and !vis[ny][nx][b]) {
                            if(ny == n - 1 and nx == m - 1) return res;
                            vis[ny][nx][b] = true;
                            q.push({ny,nx,b});
                        } else if(mat[ny][nx] == 1 and b and !vis[ny][nx][b-1]) {
                            if(ny == n - 1 and nx == m - 1) return res;
                            vis[ny][nx][b-1] = true;
                            q.push({ny,nx,b-1});
                        }
                    }
                }
            }
            res++;
        }

        return -1;
    }
};