How do you solve Number of Increasing Paths in a Grid on LeetCode?

Number of Increasing Paths in a Grid (LeetCode #2328) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n * log(m * n)) time complexity and O(m * n) space complexity. Key insight: Dynamic programming helps avoid recalculating paths for cells already processed.

What is the time complexity of Number of Increasing Paths in a Grid?

The optimal solution for Number of Increasing Paths in a Grid runs in O(m * n * log(m * n)) time and uses O(m * n) space. The complexity is dominated by the sorting step, which is O(m * n * log(m * n)), followed by a linear pass through the cells to update the DP table.

What is the brute force solution for Number of Increasing Paths in a Grid?

The brute force approach for Number of Increasing Paths in a Grid is Brute Force, with O(m * n * 4^k) time complexity and O(m * n) space complexity. The brute force approach involves exploring all possible paths starting from each cell in the grid. This means recursively checking all adjacent cells to find increasing paths, which can be very slow due to overlapping subproblems. The optimal Optimal Solution approach improves this to O(m * n * log(m * n)).

What are the interview tips for Number of Increasing Paths in a Grid?

Clearly explain your thought process and approach before coding. Consider edge cases, such as single-cell grids or grids with all identical values. Optimize your solution step-by-step, explaining why each improvement is necessary.

What are common mistakes when solving Number of Increasing Paths in a Grid?

Failing to account for all four directions when exploring paths. Not using memoization effectively in the brute force approach.

#2328

Number of Increasing Paths in a Grid

Hard

ArrayDynamic ProgrammingDepth-First SearchBreadth-First SearchGraph TheoryTopological SortMemoizationMatrixDynamic ProgrammingTopological Sorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n * 4^k)	O(m * n * log(m * n))
Space	O(m * n)	O(m * n)

💡

Intuition

Time O(m * n * log(m * n))Space O(m * n)

The optimal solution leverages dynamic programming and topological sorting to efficiently count increasing paths. By processing cells in increasing order, we can build on previously computed results, ensuring we only consider valid increasing paths.

⚙️

Algorithm

3 steps

1Step 1: Create a list of all cells and sort them based on their values.
2Step 2: Initialize a DP array where dp[i][j] represents the number of increasing paths starting from cell (i, j).
3Step 3: Iterate through the sorted list of cells, and for each cell, update its adjacent cells if they have a greater value, adding the current cell's path count to the adjacent cell's count.

solution.py14 lines

1def countPaths(grid):
2    mod = 10**9 + 7
3    m, n = len(grid), len(grid[0])
4    cells = [(grid[i][j], i, j) for i in range(m) for j in range(n)]
5    cells.sort()
6    dp = [[1] * n for _ in range(m)]
7    directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
8    total_paths = 0
9    for value, x, y in cells:
10        for dx, dy in directions:
11            nx, ny = x + dx, y + dy
12            if 0 <= nx < m and 0 <= ny < n and grid[nx][ny] > grid[x][y]:
13                dp[nx][ny] = (dp[nx][ny] + dp[x][y]) % mod
14    return sum(sum(row) for row in dp) % mod

ℹ

Complexity note: The complexity is dominated by the sorting step, which is O(m * n * log(m * n)), followed by a linear pass through the cells to update the DP table.

1Dynamic programming helps avoid recalculating paths for cells already processed.
2Sorting cells by value ensures that we only consider valid increasing paths.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.