Leetcode #2328: Number of Increasing Paths in a Grid
In this guide, we solve Leetcode #2328 Number of Increasing Paths in a Grid in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an m x n integer matrix grid, where you can move from a cell to any adjacent cell in all 4 directions. Return the number of strictly increasing paths in the grid such that you can start from any cell and end at any cell.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Depth-First Search, Breadth-First Search, Graph, Topological Sort, Memoization, Array, Dynamic Programming, Matrix
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: grid = [[1,1],[3,4]]
Output: 8
Explanation: The strictly increasing paths are:
- Paths with length 1: [1], [1], [3], [4].
- Paths with length 2: [1 -> 3], [1 -> 4], [3 -> 4].
- Paths with length 3: [1 -> 3 -> 4].
The total number of paths is 4 + 3 + 1 = 8.
Python Solution
class Solution:
def countPaths(self, grid: List[List[int]]) -> int:
def dfs(i: int, j: int) -> int:
ans = 1
for a, b in pairwise((-1, 0, 1, 0, -1)):
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n and grid[i][j] < grid[x][y]:
ans = (ans + dfs(x, y)) % mod
return ans
mod = 10**9 + 7
m, n = len(grid), len(grid[0])
return sum(dfs(i, j) for i in range(m) for j in range(n)) % mod
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.