#1631Medium~35 min

Path With Minimum Effort

Time O(m * n * log(m * n)) · Space O(m * n) · Official statement on LeetCode

graph binary search dfs backtracking queue bfs union find dijkstras algorithm

Solutions

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

// Dijkstra algorithm solution
class Solution {
public:
    int minimumEffortPath(vector<vector<int>>& heights) {
        static const vector<pair<int, int>> directions{{0, 1}, {1, 0},
                                                       {0, -1}, {-1, 0}};

        using T = tuple<int, int, int>;
        vector<vector<int>> dist(size(heights), vector<int>(size(heights[0]), numeric_limits<int>::max()));
        dist[0][0] = 0;
        priority_queue<T, vector<T>, greater<T>> min_heap;
        min_heap.emplace(0, 0, 0);
        vector<vector<int>> lookup(size(heights), vector<int>(size(heights[0])));
        while (!empty(min_heap)) {
            const auto [d, r, c] = min_heap.top(); min_heap.pop();
            if (lookup[r][c]) {
                continue;
            }
            lookup[r][c] = true;
            if (r == size(heights) - 1 && c == size(heights[0]) - 1) {
                return d;
            }
            for (const auto& [dr, dc] : directions) {
                int nr = r + dr, nc = c + dc;
                if (!(0 <= nr && nr < size(heights) &&
                      0 <= nc && nc < size(heights[0]) &&
                      !lookup[nr][nc])) {
                    continue;
                }
                int nd = max(d, abs(heights[nr][nc] - heights[r][c]));
                if (nd < dist[nr][nc]) {
                    dist[nr][nc] = nd;
                    min_heap.emplace(nd, nr, nc);
                }
            }
        }
        return -1;
    }
};

// Time:  O(m * n * log(m * n) + m * n * α(m * n)) = O(m * n * log(m * n))
// Space: O(m * n)
// union find solution
class Solution2 {
public:
    int minimumEffortPath(vector<vector<int>>& heights) {
        vector<tuple<int, int, int>> diffs;
        for (int i = 0; i < size(heights); ++i) {
            for (int j = 0; j < size(heights[0]); ++j) {
                if (i > 0) {
                    diffs.emplace_back(abs(heights[i][j] - heights[i - 1][j]), 
                                       index(size(heights[0]), i - 1, j),
                                       index(size(heights[0]), i, j));
                }
                if (j > 0) {
                    diffs.emplace_back(abs(heights[i][j] - heights[i][j - 1]),
                                       index(size(heights[0]), i, j - 1),
                                       index(size(heights[0]), i, j));
                }
            }
        }
        sort(begin(diffs), end(diffs));
        UnionFind union_find(size(heights) * size(heights[0]));
        for (const auto& [d, i, j] : diffs) {
            if (union_find.union_set(i, j)) {
                if (union_find.find_set(index(size(heights[0]), 0, 0)) ==
                    union_find.find_set(index(size(heights[0]), size(heights) - 1, size(heights[0]) - 1))) {
                    return d;
                }
            }
        }
        return 0;
    }

private:
    class UnionFind {
    public:
        UnionFind(const int n)
         : set_(n)
         , rank_(n)
         , count_(n) {
            iota(set_.begin(), set_.end(), 0);
        }

        int find_set(const int x) {
           if (set_[x] != x) {
               set_[x] = find_set(set_[x]);  // Path compression.
           }
           return set_[x];
        }

        bool union_set(const int x, const int y) {
            int x_root = find_set(x), y_root = find_set(y);
            if (x_root == y_root) {
                return false;
            }
            if (rank_[x_root] < rank_[y_root]) {  // Union by rank.
                set_[x_root] = y_root;
            } else if (rank_[x_root] > rank_[y_root]) {
                set_[y_root] = x_root;
            } else {
                set_[y_root] = x_root;
                ++rank_[x_root];
            }
            --count_;
            return true;
        }

        int size() const {
            return count_;
        }

    private:
        vector<int> set_;
        vector<int> rank_;
        int count_;
    };
    
    int index(int n, int i, int j) {
        return i * n + j;
    }
};

// Time:  O(m * n * logh)
// Space: O(m * n)
// bi-bfs solution
class Solution3 {
public:
    int minimumEffortPath(vector<vector<int>>& heights) {
        static const int MAX_H = 1e6;
        int left = 0, right = MAX_H;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (check(heights, mid)) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

private:
    bool check(const vector<vector<int>>& heights, int x) {
        static const vector<pair<int, int>> directions{{0, 1}, {1, 0},
                                                       {0, -1}, {-1, 0}};

        vector<vector<int>> lookup(size(heights), vector<int>(size(heights[0])));
        unordered_set<pair<int, int>, PairHash<int>> left({{0, 0}});
        unordered_set<pair<int, int>, PairHash<int>> right({{size(heights) - 1, size(heights[0]) - 1}});
        while (!empty(left)) {
            for (const auto& [r, c] : left) {
                lookup[r][c] = true;
            }
            unordered_set<pair<int, int>, PairHash<int>> new_left;
            for (const auto& [r, c] : left) {
                if (right.count(pair(r, c))) {
                    return true;
                }
                for (const auto& [dr, dc] : directions) {
                    int nr = r + dr, nc = c + dc;
                    if (!(0 <= nr && nr < size(heights) &&
                          0 <= nc && nc < size(heights[0]) &&
                          abs(heights[nr][nc] - heights[r][c]) <= x &&
                          !lookup[nr][nc])) {
                        continue;
                    }
                    new_left.emplace(nr, nc);
                }
            }
            left = move(new_left);
            if (size(left) > size(right)) {
                swap(left, right);
            }
        }
        return false;
    }
    
    template <typename T>
    struct PairHash {
        size_t operator()(const pair<T, T>& p) const {
            size_t seed = 0;
            seed ^= std::hash<T>{}(p.first)  + 0x9e3779b9 + (seed<<6) + (seed>>2);
            seed ^= std::hash<T>{}(p.second) + 0x9e3779b9 + (seed<<6) + (seed>>2);
            return seed;
        }
    };
 };

// Time:  O(m * n * logh)
// Space: O(m * n)
// bfs solution
class Solution4 {
public:
    int minimumEffortPath(vector<vector<int>>& heights) {
        static const int MAX_H = 1e6;
        int left = 0, right = MAX_H;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (check(heights, mid)) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

private:
    bool check(const vector<vector<int>>& heights, int x) {
        static const vector<pair<int, int>> directions{{0, 1}, {1, 0},
                                                       {0, -1}, {-1, 0}};

        queue<pair<int, int>> q({{0, 0}});
        vector<vector<int>> lookup(size(heights), vector<int>(size(heights[0])));
        while (!empty(q)) {
            const auto [r, c] = q.front(); q.pop();
            if (r == size(heights) - 1 && c == size(heights[0]) - 1) {
                return true;
            }
            for (const auto& [dr, dc] : directions) {
                int nr = r + dr, nc = c + dc;
                if (!(0 <= nr && nr < size(heights) &&
                      0 <= nc && nc < size(heights[0]) &&
                      abs(heights[nr][nc] - heights[r][c]) <= x &&
                      !lookup[nr][nc])) {
                    continue;
                }
                lookup[nr][nc] = true;
                q.emplace(nr, nc);
            }
        }
        return false;
    }
    
    template <typename T>
    struct PairHash {
        size_t operator()(const pair<T, T>& p) const {
            size_t seed = 0;
            seed ^= std::hash<T>{}(p.first)  + 0x9e3779b9 + (seed<<6) + (seed>>2);
            seed ^= std::hash<T>{}(p.second) + 0x9e3779b9 + (seed<<6) + (seed>>2);
            return seed;
        }
    };
 };
 
// Time:  O(m * n * logh)
// Space: O(m * n)
// dfs solution
class Solution5 {
public:
    int minimumEffortPath(vector<vector<int>>& heights) {
        static const int MAX_H = 1e6;
        int left = 0, right = MAX_H;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (check(heights, mid)) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

private:
    bool check(const vector<vector<int>>& heights, int x) {
        static const vector<pair<int, int>> directions{{0, 1}, {1, 0},
                                                       {0, -1}, {-1, 0}};

        vector<pair<int, int>> stk({{0, 0}});
        vector<vector<int>> lookup(size(heights), vector<int>(size(heights[0])));
        while (!empty(stk)) {
            const auto [r, c] = stk.back(); stk.pop_back();
            if (r == size(heights) - 1 && c == size(heights[0]) - 1) {
                return true;
            }
            for (const auto& [dr, dc] : directions) {
                int nr = r + dr, nc = c + dc;
                if (!(0 <= nr && nr < size(heights) &&
                      0 <= nc && nc < size(heights[0]) &&
                      abs(heights[nr][nc] - heights[r][c]) <= x &&
                      !lookup[nr][nc])) {
                    continue;
                }
                lookup[nr][nc] = true;
                stk.emplace_back(nr, nc);
            }
        }
        return false;
    }
    
    template <typename T>
    struct PairHash {
        size_t operator()(const pair<T, T>& p) const {
            size_t seed = 0;
            seed ^= std::hash<T>{}(p.first)  + 0x9e3779b9 + (seed<<6) + (seed>>2);
            seed ^= std::hash<T>{}(p.second) + 0x9e3779b9 + (seed<<6) + (seed>>2);
            return seed;
        }
    };
 };

Beginner Explanation

What is Path With Minimum Effort?

Path With Minimum Effort (LeetCode #1631) is a Medium problem that primarily trains graph.

How to think about it

Restate the goal in your own words before coding.
Work a tiny example by hand so the invariant becomes obvious.
Identify the pattern — this problem aligns with binary search, dfs backtracking, queue bfs, union find, and dijkstras algorithm.
Only then translate the idea into code.

Why this problem matters

It sits in the sweet spot of interview difficulty: multiple valid approaches, clear trade-offs. Official solution notes mention: Binary Search, DFS, BFS, Bi-BFS.

AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.

Interview Walkthrough

Interview approach for Path With Minimum Effort

Opening (30–60 seconds)

Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
State a brute force so the interviewer knows you can solve it naively.
Propose the optimal direction tied to binary search, dfs backtracking, queue bfs, union find, and dijkstras algorithm.

Core solution narrative

Define the state you track (pointers, DP cell, set membership, stack top, etc.).
Explain the transition when you process the next element.
Call out time (O(m * n * log(m * n))) and space (O(m * n)) before coding.
Code cleanly; narrate variable names.

What interviewers listen for

Correctness on edge cases
Complexity honesty
Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)

Follow-up questions they may ask

Can you solve it with less memory?
What if the input stream is infinite / doesn't fit in RAM?
How would tests look for adversarial inputs?

Optimized Approach

Optimized solution notes

The reference solutions on AlgoForge target O(m * n * log(m * n)) time and O(m * n) space.

Pattern focus: binary search, dfs backtracking, queue bfs, union find, and dijkstras algorithm

Use the pattern as a checklist:

binary search — confirm the invariant holds after each step
dfs backtracking — confirm the invariant holds after each step
queue bfs — confirm the invariant holds after each step
union find — confirm the invariant holds after each step
dijkstras algorithm — confirm the invariant holds after each step

Multiple methods appear in the source solutions — compare them and explain when each is preferable.

Implementation tips

Prefer readable names over micro-optimizations in interviews.
Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
After AC-level logic, re-scan for off-by-one and null checks.

Complexity Analysis

Complexity

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

How to justify this in an interview

Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
Space: include hash maps, recursion stack, and output allocation when the problem asks for it.

If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.

Common Mistakes

Common mistakes on Path With Minimum Effort

Skipping edge cases — empty collections, single-element inputs, max constraints.
Wrong invariant for binary search, dfs backtracking, queue bfs, union find, and dijkstras algorithm — updating state too early or too late.
Mutating input unexpectedly when the problem forbids it.
Off-by-one in windows, ranges, or binary search bounds.
Ignoring overflow / precision for integer arithmetic problems.
Overengineering — jumping to an advanced structure when a simpler approach works.

Alternative Approaches

Alternatives

The source file includes more than one method. Compare:

Primary optimized path — best complexity for typical interviews.
Secondary approach — often brute force, sorting-based, or space-optimized variant.

Practice articulating when you would pick each (constraints, readability, follow-ups).

Edge Cases

Edge cases checklist

Minimum input size
Maximum input size / time limits
Duplicates and already-sorted input
Negative numbers / zeros (if applicable)
Disconnected structures (graphs/trees)
Single path vs branching recursion depth

Pattern Recognition

Spotting this pattern

Signal phrases that point to binary search, dfs backtracking, queue bfs, union find, and dijkstras algorithm:

Sorted input or ability to sort without changing the answer class
Need for contiguous subarray / substring → consider sliding window
Need for O(1) membership → hash set/map
Optimal substructure + overlapping subproblems → DP
Connectivity / components → graph DFS/BFS or Union-Find

Primary topics: graph.

Follow-up Interview Questions

Follow-ups

How does the solution change if the input is a stream?
Can you solve it in-place?
What if duplicates must be handled differently?
How would you parallelize the approach?
Design tests that would break a buggy implementation.

Practice Recommendations

What to practice next

Re-solve Path With Minimum Effort in a second language (cpp, python).
Drill 3–5 more problems tagged graph.
Teach the solution out loud in under 5 minutes.
Add this problem to your revision calendar in 3 days and 14 days.

Visualization

Conceptual diagram for Path With Minimum Effort: show input structure (graph), highlight the moving parts of the binary search, dfs backtracking, queue bfs, union find, and dijkstras algorithm approach, and annotate each step with the maintained invariant and complexity.

Study checklist

Read the official problem statement on LeetCode
Solve on paper / whiteboard first
Implement the binary search, dfs backtracking, queue bfs, union find, and dijkstras algorithm approach
Verify edge cases from the checklist
State time and space complexity aloud
Compare with the AlgoForge reference solution
Schedule a revision session

Revision notes

Path With Minimum Effort (#1631) — Medium. Pattern: binary search, dfs backtracking, queue bfs, union find, and dijkstras algorithm. Complexity: O(m * n * log(m * n)) time / O(m * n) space. Re-derive the invariant before coding.

FAQs

What is the time complexity of Path With Minimum Effort?+

The reference solutions aim for O(m * n * log(m * n)) time and O(m * n) space. Always re-derive complexity from the code you write in the interview.

What pattern does Path With Minimum Effort use?+

It primarily maps to binary search, dfs backtracking, queue bfs, union find, and dijkstras algorithm, within the broader topic of graph.

Is Path With Minimum Effort good for interviews?+

Yes — as a Medium problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.

Where can I read the official statement?+

Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/path-with-minimum-effort/

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