LeetCode 2328: Number of Increasing Paths in a Grid Solution

Master LeetCode problem 2328 (Number of Increasing Paths in a Grid), 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.

2328. Number of Increasing Paths in a Grid

Hard

Array Dynamic Programming Depth-First Search Breadth-First Search Graph Topological Sort Memoization Matrix

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We will create a 2D DP array to keep track of the number of strictly increasing paths ending at each cell. We will then iterate through the grid and update the DP array accordingly. Finally, we will sum up all the values in the DP array to get the total number of strictly increasing paths in the grid.

Algorithm:
1. Initialize a 2D DP array of size m x n with all values set to 0.
2. Iterate through the grid:
  - For each cell (i, j), update the DP array with the sum of paths ending at adjacent cells with values less than grid[i][j].
3. Sum up all values in the DP array to get the total number of strictly increasing paths.
4. Return the result modulo 10^9 + 7.
Time Complexity: O(m*n) where m is the number of rows and n is the number of columns in the grid.
Space Complexity: O(m*n) for the DP array.

Solution Code

class Solution {
    public int numIncreasingPaths(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int[][] dp = new int[m][n];

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                dp[i][j] = 1;
            }
        }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i > 0 && grid[i][j] > grid[i - 1][j]) {
                    dp[i][j] += dp[i - 1][j];
                }
                if (j > 0 && grid[i][j] > grid[i][j - 1]) {
                    dp[i][j] += dp[i][j - 1];
                }
                dp[i][j] %= 1000000007;
            }
        }

        int result = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                result += dp[i][j];
                result %= 1000000007;
            }
        }

        return result;
    }
}

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