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LeetCode 2503: Maximum Number of Points From Grid Queries Solution

Master LeetCode problem 2503 (Maximum Number of Points From Grid Queries), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation:

To solve this problem, we iterate through each query and traverse the grid starting from the top left cell. We keep track of the maximum points we can obtain for each query based on the given conditions.

  1. Start from the top left cell of the matrix.
  2. For each query:
    • If the current cell value is less than the query value, increment the points and explore adjacent cells.
    • Keep track of visited cells to avoid revisiting them.

Time Complexity: O(mnk) where m and n are the dimensions of the grid and k is the number of queries. Space Complexity: O(m*n) for the visited array. :

Solution Code

class Solution {
    public int[] maxPoints(int[][] grid, int[] queries) {
        int m = grid.length;
        int n = grid[0].length;
        int[] answer = new int[queries.length];

        for (int i = 0; i < queries.length; i++) {
            int points = 0;
            boolean[][] visited = new boolean[m][n];
            dfs(grid, visited, 0, 0, queries[i], m, n, points);
            answer[i] = points;
        }

        return answer;
    }

    private void dfs(int[][] grid, boolean[][] visited, int x, int y, int query, int m, int n, int points) {
        if (x < 0 || x >= m || y < 0 || y >= n || visited[x][y] || grid[x][y] < query) {
            return;
        }

        points++;
        visited[x][y] = true;
        dfs(grid, visited, x + 1, y, query, m, n, points);
        dfs(grid, visited, x - 1, y, query, m, n, points);
        dfs(grid, visited, x, y + 1, query, m, n, points);
        dfs(grid, visited, x, y - 1, query, m, n, points);
    }
}

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