As Far from Land as Possible LeetCode solution in C++

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class Solution {
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  vector<int> dr = {0, 0, 1, -1};
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  vector<int> dc = {1, -1, 0, 0};
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 public:
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  bool isPos(int r, int c, int n) {
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    return (r >= 0 && c >= 0 && r < n && c < n);
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  }
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  int maxDistance(vector<vector<int>> &grid) {
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    int n = grid.size();
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    vector<vector<int>> dist(n, vector<int>(n, -1));
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    queue<vector<int>> pq;
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    for (int i = 0; i < n; i++) {
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      for (int j = 0; j < n; j++) {
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        if (1 == grid[i][j]) {
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          dist[i][j] = 0;
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          pq.push({i, j});
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        }
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      }
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    }
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    int ans = 0;
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    while (!pq.empty()) {
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      vector<int> vect = pq.front();
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      pq.pop();
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      int r = vect[0];
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      int c = vect[1];
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      if (0 == grid[r][c]) {
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        ans = max(ans, dist[r][c]);
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      }
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      for (int i = 0; i < 4; i++) {
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        int nr = r + dr[i];
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        int nc = c + dc[i];
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        if (isPos(nr, nc, n) && -1 == dist[nr][nc]) {
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          dist[nr][nc] = dist[r][c] + 1;
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          pq.push({nr, nc});
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        }
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      }
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    }
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    return (ans == 0 ? -1 : ans);
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  }
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};