LeetCode #1631 — MEDIUM

Path With Minimum Effort

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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). You are situated in the top-left cell, (0, 0), and you hope to travel to the bottom-right cell, (rows-1, columns-1) (i.e., 0-indexed). You can move up, down, left, or right, and you wish to find a route that requires the minimum effort.

A route's effort is the maximum absolute difference in heights between two consecutive cells of the route.

Return the minimum effort required to travel from the top-left cell to the bottom-right cell.

Example 1:

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.

Example 2:

Input: heights = [[1,2,3],[3,8,4],[5,3,5]]
Output: 1
Explanation: The route of [1,2,3,4,5] has a maximum absolute difference of 1 in consecutive cells, which is better than route [1,3,5,3,5].

Example 3:

Input: heights = [[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]]
Output: 0
Explanation: This route does not require any effort.

Constraints:

rows == heights.length
columns == heights[i].length
1 <= rows, columns <= 100
1 <= heights[i][j] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: 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). You are situated in the top-left cell, (0, 0), and you hope to travel to the bottom-right cell, (rows-1, columns-1) (i.e., 0-indexed). You can move up, down, left, or right, and you wish to find a route that requires the minimum effort. A route's effort is the maximum absolute difference in heights between two consecutive cells of the route. Return the minimum effort required to travel from the top-left cell to the bottom-right cell.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Union-Find

Example 1

[[1,2,2],[3,8,2],[5,3,5]]

Example 2

[[1,2,3],[3,8,4],[5,3,5]]

Example 3

[[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]]

Core Insight

What unlocks the optimal approach

Consider the grid as a graph, where adjacent cells have an edge with cost of the difference between the cells.
If you are given threshold k, check if it is possible to go from (0, 0) to (n-1, m-1) using only edges of ≤ k cost.
Binary search the k value.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1631: Path With Minimum Effort
class UnionFind {
    private final int[] p;
    private final int[] size;

    public UnionFind(int n) {
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            size[i] = 1;
        }
    }

    public int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

    public boolean union(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) {
            return false;
        }
        if (size[pa] > size[pb]) {
            p[pb] = pa;
            size[pa] += size[pb];
        } else {
            p[pa] = pb;
            size[pb] += size[pa];
        }
        return true;
    }

    public boolean connected(int a, int b) {
        return find(a) == find(b);
    }
}

class Solution {
    public int minimumEffortPath(int[][] heights) {
        int m = heights.length, n = heights[0].length;
        UnionFind uf = new UnionFind(m * n);
        List<int[]> edges = new ArrayList<>();
        int[] dirs = {1, 0, 1};
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                for (int k = 0; k < 2; ++k) {
                    int x = i + dirs[k], y = j + dirs[k + 1];
                    if (x >= 0 && x < m && y >= 0 && y < n) {
                        int d = Math.abs(heights[i][j] - heights[x][y]);
                        edges.add(new int[] {d, i * n + j, x * n + y});
                    }
                }
            }
        }
        Collections.sort(edges, (a, b) -> a[0] - b[0]);
        for (int[] e : edges) {
            uf.union(e[1], e[2]);
            if (uf.connected(0, m * n - 1)) {
                return e[0];
            }
        }
        return 0;
    }
}

// Accepted solution for LeetCode #1631: Path With Minimum Effort
type unionFind struct {
	p, size []int
}

func newUnionFind(n int) *unionFind {
	p := make([]int, n)
	size := make([]int, n)
	for i := range p {
		p[i] = i
		size[i] = 1
	}
	return &unionFind{p, size}
}

func (uf *unionFind) find(x int) int {
	if uf.p[x] != x {
		uf.p[x] = uf.find(uf.p[x])
	}
	return uf.p[x]
}

func (uf *unionFind) union(a, b int) bool {
	pa, pb := uf.find(a), uf.find(b)
	if pa == pb {
		return false
	}
	if uf.size[pa] > uf.size[pb] {
		uf.p[pb] = pa
		uf.size[pa] += uf.size[pb]
	} else {
		uf.p[pa] = pb
		uf.size[pb] += uf.size[pa]
	}
	return true
}

func (uf *unionFind) connected(a, b int) bool {
	return uf.find(a) == uf.find(b)
}

func minimumEffortPath(heights [][]int) int {
	m, n := len(heights), len(heights[0])
	edges := make([][3]int, 0, m*n*2)
	dirs := [3]int{0, 1, 0}
	for i, row := range heights {
		for j, h := range row {
			for k := 0; k < 2; k++ {
				x, y := i+dirs[k], j+dirs[k+1]
				if x >= 0 && x < m && y >= 0 && y < n {
					edges = append(edges, [3]int{abs(h - heights[x][y]), i*n + j, x*n + y})
				}
			}
		}
	}
	sort.Slice(edges, func(i, j int) bool { return edges[i][0] < edges[j][0] })
	uf := newUnionFind(m * n)
	for _, e := range edges {
		uf.union(e[1], e[2])
		if uf.connected(0, m*n-1) {
			return e[0]
		}
	}
	return 0
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #1631: Path With Minimum Effort
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

// Accepted solution for LeetCode #1631: Path With Minimum Effort
struct Solution;

use std::cmp::Reverse;
use std::collections::BinaryHeap;

impl Solution {
    fn minimum_effort_path(heights: Vec<Vec<i32>>) -> i32 {
        let n = heights.len();
        let m = heights[0].len();
        let mut visited: Vec<Vec<bool>> = vec![vec![false; m]; n];
        let mut queue: BinaryHeap<(Reverse<i32>, usize, usize)> = BinaryHeap::new();
        let mut res = 0;
        queue.push((Reverse(0), 0, 0));
        while let Some((Reverse(effort), i, j)) = queue.pop() {
            res = res.max(effort);
            if i == n - 1 && j == m - 1 {
                break;
            }
            visited[i][j] = true;
            if i > 0 && !visited[i - 1][j] {
                let diff = heights[i][j] - heights[i - 1][j];
                queue.push((Reverse(diff.abs()), i - 1, j));
            }
            if j > 0 && !visited[i][j - 1] {
                let diff = heights[i][j] - heights[i][j - 1];
                queue.push((Reverse(diff.abs()), i, j - 1));
            }
            if i + 1 < n && !visited[i + 1][j] {
                let diff = heights[i][j] - heights[i + 1][j];
                queue.push((Reverse(diff.abs()), i + 1, j));
            }
            if j + 1 < m && !visited[i][j + 1] {
                let diff = heights[i][j] - heights[i][j + 1];
                queue.push((Reverse(diff.abs()), i, j + 1));
            }
        }
        res
    }
}

#[test]
fn test() {
    let heights = vec_vec_i32![[1, 2, 2], [3, 8, 2], [5, 3, 5]];
    let res = 2;
    assert_eq!(Solution::minimum_effort_path(heights), res);
    let heights = vec_vec_i32![[1, 2, 3], [3, 8, 4], [5, 3, 5]];
    let res = 1;
    assert_eq!(Solution::minimum_effort_path(heights), res);
    let heights = vec_vec_i32![
        [1, 2, 1, 1, 1],
        [1, 2, 1, 2, 1],
        [1, 2, 1, 2, 1],
        [1, 2, 1, 2, 1],
        [1, 1, 1, 2, 1]
    ];
    let res = 0;
    assert_eq!(Solution::minimum_effort_path(heights), res);
}

// Accepted solution for LeetCode #1631: Path With Minimum Effort
class UnionFind {
    private p: number[];
    private size: number[];

    constructor(n: number) {
        this.p = Array.from({ length: n }, (_, i) => i);
        this.size = Array(n).fill(1);
    }

    find(x: number): number {
        if (this.p[x] !== x) {
            this.p[x] = this.find(this.p[x]);
        }
        return this.p[x];
    }

    union(a: number, b: number): boolean {
        const pa = this.find(a);
        const pb = this.find(b);
        if (pa === pb) {
            return false;
        }
        if (this.size[pa] > this.size[pb]) {
            this.p[pb] = pa;
            this.size[pa] += this.size[pb];
        } else {
            this.p[pa] = pb;
            this.size[pb] += this.size[pa];
        }
        return true;
    }

    connected(a: number, b: number): boolean {
        return this.find(a) === this.find(b);
    }
}

function minimumEffortPath(heights: number[][]): number {
    const m = heights.length;
    const n = heights[0].length;
    const uf = new UnionFind(m * n);
    const edges: number[][] = [];
    const dirs = [1, 0, 1];

    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            for (let k = 0; k < 2; ++k) {
                const x = i + dirs[k];
                const y = j + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n) {
                    const d = Math.abs(heights[i][j] - heights[x][y]);
                    edges.push([d, i * n + j, x * n + y]);
                }
            }
        }
    }

    edges.sort((a, b) => a[0] - b[0]);

    for (const [h, a, b] of edges) {
        uf.union(a, b);
        if (uf.connected(0, m * n - 1)) {
            return h;
        }
    }

    return 0;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n × log(m × n)

Space

O(m × n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.