Leetcode #1631: Path With Minimum Effort
In this guide, we solve Leetcode #1631 Path With Minimum Effort 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 a hiker preparing for an upcoming hike. You are given heights, a 2D array of size rows x columns, where heights[row][col] represents the height of cell (row, col).
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Breadth-First Search, Union Find, Array, Binary Search, Matrix, Heap (Priority Queue)
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: heights = [[1,2,2],[3,8,2],[5,3,5]]
Output: 2
Explanation: The route of [1,3,5,3,5] has a maximum absolute difference of 2 in consecutive cells.
This is better than the route of [1,2,2,2,5], where the maximum absolute difference is 3.
Python Solution
class UnionFind:
def __init__(self, n):
self.p = list(range(n))
self.size = [1] * n
def find(self, x):
if self.p[x] != x:
self.p[x] = self.find(self.p[x])
return self.p[x]
def union(self, a, b):
pa, pb = self.find(a), self.find(b)
if pa == pb:
return False
if self.size[pa] > self.size[pb]:
self.p[pb] = pa
self.size[pa] += self.size[pb]
else:
self.p[pa] = pb
self.size[pb] += self.size[pa]
return True
def connected(self, a, b):
return self.find(a) == self.find(b)
class Solution:
def minimumEffortPath(self, heights: List[List[int]]) -> int:
m, n = len(heights), len(heights[0])
uf = UnionFind(m * n)
e = []
dirs = (0, 1, 0)
for i in range(m):
for j in range(n):
for a, b in pairwise(dirs):
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n:
e.append(
(abs(heights[i][j] - heights[x][y]), i * n + j, x * n + y)
)
e.sort()
for h, a, b in e:
uf.union(a, b)
if uf.connected(0, m * n - 1):
return h
return 0
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.