How do you solve Path With Minimum Effort on LeetCode?

Path With Minimum Effort (LeetCode #1631) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log(max_diff)) time complexity and O(n) space complexity. Key insight: The problem can be visualized as a graph where each cell is a node and edges represent height differences.

What is the brute force solution for Path With Minimum Effort?

The brute force approach for Path With Minimum Effort is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves exploring all possible paths from the top-left to the bottom-right cell, calculating the maximum height difference for each path. This is straightforward but inefficient as it doesn't prune any paths. The optimal Optimal Solution approach improves this to O(n log(max_diff)).

What are the interview tips for Path With Minimum Effort?

Always clarify the problem statement and constraints before diving into coding. Think about edge cases and how they might affect your solution. Practice explaining your thought process as you code, as communication is key in interviews.

What are common mistakes when solving Path With Minimum Effort?

Failing to correctly implement the BFS or DFS traversal logic. Not considering edge cases like a single cell or all cells having the same height.

#1631

Path With Minimum Effort

Medium

ArrayBinary SearchDepth-First SearchBreadth-First SearchUnion-FindHeap (Priority Queue)MatrixGraph Traversal (BFS/DFS)Binary Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log(max_diff))
Space	O(1)	O(n)

💡

Intuition

Time O(n log(max_diff))Space O(n)

The optimal approach uses a binary search combined with a graph traversal (like BFS or Dijkstra's algorithm) to efficiently find the minimum effort required. We treat the problem as finding the smallest maximum edge weight in a path from the start to the end.

⚙️

Algorithm

3 steps

1Step 1: Define the binary search range between 0 and the maximum height difference in the grid.
2Step 2: For each mid value in the binary search, check if there exists a path from (0, 0) to (rows-1, columns-1) using only edges with a difference less than or equal to mid.
3Step 3: If a path exists, try for a smaller mid value; otherwise, increase mid.

solution.py27 lines

1from collections import deque
2
3def minEffort(heights):
4    def canReach(mid):
5        directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
6        queue = deque([(0, 0)])
7        visited = set((0, 0))
8        while queue:
9            x, y = queue.popleft()
10            if (x, y) == (len(heights) - 1, len(heights[0]) - 1):
11                return True
12            for dx, dy in directions:
13                nx, ny = x + dx, y + dy
14                if 0 <= nx < len(heights) and 0 <= ny < len(heights[0]) and (nx, ny) not in visited:
15                    if abs(heights[nx][ny] - heights[x][y]) <= mid:
16                        visited.add((nx, ny))
17                        queue.append((nx, ny))
18        return False
19
20    left, right = 0, max(max(row) for row in heights) - min(min(row) for row in heights)
21    while left < right:
22        mid = (left + right) // 2
23        if canReach(mid):
24            right = mid
25        else:
26            left = mid + 1
27    return left

ℹ

Complexity note: The time complexity is O(n log(max_diff)) where n is the number of cells and max_diff is the maximum height difference in the grid. The space complexity is O(n) due to the queue used for BFS.

1The problem can be visualized as a graph where each cell is a node and edges represent height differences.
2Binary search helps efficiently narrow down the minimum effort required.

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