Path With Minimum Effort

📋

Problem

Imagine you are hiking across a mountainous terrain represented as a grid, and you want to find a path from start to finish that minimizes the steepest climb between adjacent steps.

Given a 2D grid heights where heights[i][j] represents the height of cell (i, j), find a path from the top-left corner (0, 0) to the bottom-right corner (n-1, m-1) such that the maximum absolute difference in heights between adjacent cells on the path is minimized. Return this minimum possible maximum difference.

1 ≤ n, m ≤ 1000 ≤ heights[i][j] ≤ 10^6

Edge cases: Single cell grid → output 0Grid with all equal heights → output 0Grid where only one path exists → output is max difference on that path

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IDE

def minimumEffortPath(heights: List[List[int]]) -> int:public int minimumEffortPath(int[][] heights)int minimumEffortPath(vector<vector<int>>& heights)function minimumEffortPath(heights)

def minimumEffortPath(heights):
    # Write your solution here
    pass

class Solution {
    public int minimumEffortPath(int[][] heights) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int minimumEffortPath(vector<vector<int>>& heights) {
    // Write your solution here
    return 0;
}

function minimumEffortPath(heights) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: Sum of all height differences along pathMistakenly summing edge differences instead of minimizing the maximum edge difference.✅ Update effort as max(current effort, edge difference) instead of summing.

Wrong: Greedy local minimum difference path maxUsing greedy approach picking next step with smallest difference without global consideration.✅ Use Dijkstra's algorithm variant with priority queue tracking max edge difference on path.

Wrong: Non-zero output for single cell or uniform gridNot handling base cases where no movement or zero difference exists.✅ Return 0 immediately if grid is 1x1 or all heights equal.

Wrong: Index out of bounds or incorrect boundary checksOff-by-one errors in neighbor exploration causing invalid accesses or missed paths.✅ Ensure neighbor indices are within 0 <= r < rows and 0 <= c < cols before access.

Wrong: Timeout or no output on large inputsUsing brute force DFS/backtracking instead of efficient Dijkstra variant.✅ Implement priority queue based Dijkstra with O(n*m*log(n*m)) complexity.

✓

Test Cases

Focus on handling minimal input sizes and uniform height grids correctly.

Beware of greedy traps and ensure you track max edge difference globally.

Optimize your solution using priority queues and efficient graph traversal.

t1_01basic

Input{"heights":[[1,2,2],[3,8,2],[5,3,5]]}

Expected2

⏱ Performance - must finish in 2000ms

One path is (0,0)->(0,1)->(0,2)->(1,2)->(2,2). The maximum height difference along this path is 2, which is minimal.

💡 Think about how to find a path minimizing the maximum edge weight.

💡 Use a graph traversal with a priority queue to track minimal maximum effort.

💡 Implement Dijkstra's algorithm variant where distance is max edge weight on path.

Why it failed: Incorrect output means the algorithm likely sums differences instead of minimizing the max difference. Fix by tracking max edge difference along paths, not sum.

✓ Correct minimal maximum effort found using proper path evaluation.

t1_02basic

Input{"heights":[[1,10,6,7,9,10,4,9]]}

Expected9

⏱ Performance - must finish in 2000ms

Single row path from left to right; max difference is max adjacent difference which is 9 (between 1 and 10).

💡 Consider that in a single row, path is linear and max difference is max adjacent difference.

💡 Check all adjacent pairs and find the maximum absolute difference.

💡 The minimal maximum effort equals the largest height difference between consecutive cells.

Why it failed: If output is less than 9, code likely ignores some edges or uses sum instead of max difference. Fix by using max of adjacent differences along path.

✓ Correctly identified max adjacent difference as minimal maximum effort.

t2_01edge

Input{"heights":[[42]]}

Expected0

⏱ Performance - must finish in 2000ms

Single cell grid; no movement needed, so effort is zero.

💡 Check base case when grid size is 1x1.

💡 No edges exist, so minimal effort must be zero.

💡 Return 0 immediately if start equals end.

Why it failed: If output is not zero, code likely does not handle single cell base case. Fix by returning 0 when grid has one cell.

✓ Correctly handles single cell grid with zero effort.

t2_02edge

Input{"heights":[[5,5],[5,5]]}

Expected0

⏱ Performance - must finish in 2000ms

All heights equal; any path has zero height difference.

💡 Consider that all adjacent cells have zero difference.

💡 Minimal maximum effort must be zero for uniform height grids.

💡 Check that max difference calculation correctly handles zero differences.

Why it failed: If output is not zero, code likely does not handle zero differences properly. Fix by ensuring max difference is computed correctly.

✓ Correctly returns zero for uniform height grid.

t2_03edge

Input{"heights":[[1,1000],[1000,1]]}

Expected999

⏱ Performance - must finish in 2000ms

Only two possible paths; minimal max difference is 999 along any path.

💡 Enumerate all possible paths and find max difference on each.

💡 Minimal max difference is the smallest among these maximums.

💡 Check that algorithm explores all paths or uses correct graph traversal.

Why it failed: If output is less than 999, code likely misses some paths or uses incorrect max difference logic. Fix by exploring all paths or using correct priority queue logic.

✓ Correctly finds minimal maximum effort as 999.

t3_01corner

Input{"heights":[[1,100,1],[1,100,1],[1,1,1]]}

Expected0

⏱ Performance - must finish in 2000ms

Greedy approach picking smallest immediate difference fails; correct path goes around high differences with max difference 0.

💡 Beware of greedy approach that picks next step with smallest difference.

💡 Use Dijkstra variant to consider max edge difference along entire path.

💡 Track max difference on path, not just local differences.

Why it failed: If output is 99 or 100, code uses greedy local min difference instead of global min max difference. Fix by using priority queue with max edge tracking.

✓ Correctly avoids greedy trap and finds minimal max difference 0.

t3_02corner

Input{"heights":[[1,2],[2,3]]}

Expected1

⏱ Performance - must finish in 2000ms

Tests confusion between 0/1 knapsack style sum vs max edge difference; minimal max difference is 1, not sum 3.

💡 Do not sum differences; minimal effort is max difference on path.

💡 Check that path cost is max edge difference, not sum.

💡 Use max() function to update effort along path.

Why it failed: If output is 3, code sums differences instead of taking max. Fix by updating effort as max(current effort, edge difference).

✓ Correctly computes minimal max difference as 1.

t3_03corner

Input{"heights":[[1,2,3],[3,8,4],[5,3,5]]}

Expected2

⏱ Performance - must finish in 2000ms

Off-by-one errors in indexing or boundary checks can cause wrong results; correct minimal max difference is 2.

💡 Verify boundary checks when exploring neighbors.

💡 Ensure indices are within grid before accessing heights.

💡 Use correct row and column limits in loops.

Why it failed: If output differs, code likely has off-by-one errors causing invalid accesses or missed paths. Fix boundary checks to be 0 <= r < rows and 0 <= c < cols.

✓ Correctly handles boundaries and finds minimal max difference 2.

t4_01performance

Input{"_description":"n=100 at constraint boundary - executor generates this input"}

Expectednull

⏱ Performance - must finish in 2000ms

Grid size n=100 by m=100; O(n*m*log(n*m)) Dijkstra variant must complete within 2 seconds.

💡 Use priority queue to implement Dijkstra efficiently.

💡 Avoid brute force DFS which is exponential.

💡 Optimize neighbor checks and data structures.

Why it failed: TLE indicates brute force or inefficient approach used. Fix by implementing Dijkstra's algorithm with priority queue and proper pruning.

✓ Algorithm runs within time limit using O(n*m*log(n*m)) complexity.

Practice

1. You are given a list of courses and their prerequisite pairs. You need to determine if it is possible to finish all courses without running into a circular dependency. Which algorithmic approach guarantees an optimal solution to detect such cycles in the course dependency graph?

easy

A. Perform a topological sort using Kahn's Algorithm (BFS) to detect cycles by checking if all nodes can be processed.

B. Use a brute force approach that tries all possible course orders to find a valid sequence.

C. Apply dynamic programming to count the number of ways to complete courses without cycles.

D. Use a greedy algorithm that always picks the course with the fewest prerequisites first.

Solution

1. Step 1: Understand the problem as cycle detection in a directed graph

   The problem requires detecting if a cycle exists in the course prerequisite graph, which invalidates any course order.

2. Step 2: Identify the algorithm that detects cycles efficiently

   Kahn's Algorithm performs a topological sort using BFS and in-degree counts. If all nodes are processed, no cycle exists; otherwise, a cycle is detected.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Topological sort with BFS is the standard cycle detection method for this problem [OK]

Hint: Topological sort detects cycles by processing all nodes [OK]

Common Mistakes:

• Greedy approach fails on complex dependencies
• Dynamic programming is not suited for cycle detection
• Brute force is inefficient and impractical

2. What is the time complexity of the BFS-based algorithm to find the maximum area of an island in an m x n grid, where each cell is visited at most once?

medium

A. O((m * n)^2) -- BFS could revisit cells multiple times in worst case

B. O(m + n) -- because BFS visits each row and column once

C. O(m * n) -- each cell is visited once and neighbors checked constant times

D. O(m * n) plus extra O(m * n) space for the visited set

Solution

1. Step 1: Identify BFS visits each cell at most once

   Each cell is enqueued and dequeued once, so total operations proportional to number of cells.

2. Step 2: Each cell checks up to 4 neighbors in constant time

   Neighbor checks are constant per cell, so total is O(m*n).

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Linear in grid size, no repeated visits [OK]

Hint: Each cell visited once -> O(m*n) time [OK]

Common Mistakes:

• Assuming BFS revisits cells multiple times
• Confusing rows+cols with total cells
• Ignoring constant neighbor checks

3. What is the time complexity of the BFS-based algorithm to count connected components in an undirected graph with n nodes and e edges?

medium

A. O(e log n) due to sorting adjacency lists before BFS

B. O(n²) because BFS visits all nodes and edges repeatedly

C. O(n + e) because each node and edge is visited once during BFS

D. O(n + e²) because edges are checked multiple times

Solution

1. Step 1: Identify BFS visits each node once

   Each node is enqueued and dequeued exactly once, so node visits are O(n).

2. Step 2: Identify BFS visits each edge twice

   Each undirected edge is checked from both endpoints, so edge visits are O(e).

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Sum of node and edge visits -> O(n + e) [OK]

Hint: BFS visits nodes and edges once each [OK]

Common Mistakes:

• Assuming BFS repeats visits causing O(n²)
• Confusing edge visits with sorting cost

4. The following code attempts to solve the Surrounded Regions problem using BFS. Identify the line that contains a subtle bug that can cause incorrect flipping of boundary 'O's.

from collections import deque

def solve(board):
    if not board or not board[0]:
        return
    m, n = len(board), len(board[0])
    queue = deque()

    for i in range(m):
        for j in [0, n-1]:
            if board[i][j] == 'O':
                queue.append((i, j))
                board[i][j] = 'X'  # Bug here
    for j in range(n):
        for i in [0, m-1]:
            if board[i][j] == 'O':
                queue.append((i, j))
                board[i][j] = 'X'  # Bug here

    while queue:
        x, y = queue.popleft()
        for dx, dy in [(-1,0),(1,0),(0,-1),(0,1)]:
            nx, ny = x + dx, y + dy
            if 0 <= nx < m and 0 <= ny < n and board[nx][ny] == 'O':
                board[nx][ny] = 'S'
                queue.append((nx, ny))

    for i in range(m):
        for j in range(n):
            if board[i][j] == 'O':
                board[i][j] = 'X'
            elif board[i][j] == 'S':
                board[i][j] = 'O'

medium

A. Line that flips remaining 'O's to 'X' after BFS.

B. Line that appends boundary 'O's to the queue.

C. Line that checks neighbors inside the BFS loop.

D. Line that sets boundary 'O's to 'X' instead of marking safe with 'S'.

Solution

1. Step 1: Identify the purpose of marking boundary 'O's

   Boundary 'O's must be marked as safe (e.g., 'S'), not flipped to 'X' immediately.

2. Step 2: Locate the bug

   The lines setting boundary 'O's to 'X' instead of 'S' cause these safe cells to be flipped incorrectly.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Marking safe cells incorrectly flips boundary 'O's [OK]

Hint: Boundary 'O's must be marked safe, not flipped immediately [OK]

Common Mistakes:

• Flipping boundary 'O's before BFS causes incorrect final board.
• Not marking visited cells leads to infinite loops.
• Only checking immediate neighbors without full BFS traversal.

5. Suppose the problem is modified so that you can only move in 4 directions (up, down, left, right) instead of 8. Which of the following statements about the shortest path algorithm is true?

hard

A. Greedy Best-First Search guarantees shortest path with 4-direction moves

B. Bidirectional BFS still finds the shortest path, but the path length may be longer due to restricted moves

C. DFS with backtracking becomes optimal since fewer directions reduce complexity

D. Dynamic Programming is required to handle the reduced movement directions

Solution

1. Step 1: Understand movement restriction impact

   Reducing to 4 directions limits possible paths, potentially increasing shortest path length.

2. Step 2: Evaluate algorithm suitability

   Bidirectional BFS still finds shortest path efficiently; DFS and greedy do not guarantee shortest path.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Bidirectional BFS adapts to movement constraints, still optimal [OK]

Hint: Bidirectional BFS adapts to movement constraints [OK]

Common Mistakes:

• Assuming DFS or greedy become optimal
• Thinking DP is needed for movement changes