Replace O’s with X’s

Given a matrix mat of size N x M where every element is either ‘O’ or ‘X’.
Replace all ‘O’ with ‘X’ that are surrounded by ‘X’.
A ‘O’ (or a set of ‘O’) is considered to be by surrounded by ‘X’ if there are ‘X’ at locations just below, just above, just left and just right of it.

• Time : O(nm)
• Space : O(n + m)

class Solution {
    void bfs(vector<vector<char>>& mat, int sy, int sx, int n, int m) {
        int dy[4]{-1,0,1,0}, dx[4]{0,1,0,-1};
        queue<pair<int,int>> q;
        q.push({sy,sx});
        mat[sy][sx] = '#';
        while(!q.empty()) {
            auto p = q.front(); q.pop();
            auto y = p.first, x = p.second;
            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 and mat[ny][nx] == 'O') {
                    mat[ny][nx] = '#';
                    q.push({ny,nx});
                }
            }
        }
    }
public:
    vector<vector<char>> fill(int n, int m, vector<vector<char>> mat) {
        for(int i = 0; i < n; i++) {
            if(mat[i][0] == 'O') bfs(mat,i,0,n,m);
            if(mat[i][m-1] == 'O') bfs(mat,i,m-1,n,m);
        }
        for(int j = 0; j < m; j++) {
            if(mat[0][j] == 'O') bfs(mat,0,j,n,m);
            if(mat[n-1][j] == 'O') bfs(mat,n-1,j,n,m);
        }
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(mat[i][j] == 'O') mat[i][j] = 'X';
                else if(mat[i][j] == '#') mat[i][j] = 'O';
            }
        }
        return mat;
    }
};