LeetCode 2503: Maximum Number of Points From Grid Queries
LeetCode 2503 Solution 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.
- Start from the top left cell of the matrix.
- 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. :
LeetCode 2503 Solutions in Java, C++, Python
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);
}
}Interactive Code Editor for LeetCode 2503
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