What is the time complexity of Path With Minimum Effort?

The optimal solution has a time complexity of O(m*n*log(m*n)).

What is the space complexity of Path With Minimum Effort?

The space complexity is O(m*n).

What data structures are used to solve Path With Minimum Effort?

This problem uses Array, Binary Search, Depth-First Search, Breadth-First Search, Union Find, Heap (Priority Queue), Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Path With Minimum Effort?

Insight 1: The problem can be modeled as a shortest path problem where the 'distance' between nodes (cells) is the 'effort' (maximum absolute height difference between adjacent cells). Insight 2: Since we're looking for the *minimum* effort, algorithms like Dijkstra or a variation of BFS are applicable. A binary search on the possible range of effort values can also be combined with a graph traversal to find if a path exists with a given maximum effort.

Which companies ask LeetCode 1631. Path With Minimum Effort in interviews?

This problem is frequently asked at Cohesity, Snowflake.

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Path With Minimum Effort - LeetCode 1631 Solution

Path With Minimum Effort - Complete Solution Guide

Path With Minimum Effort is LeetCode problem 1631, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 consec

Detailed Explanation

The problem asks us to find the path with the minimum 'effort' from the top-left cell (0, 0) to the bottom-right cell (rows-1, cols-1) in a 2D grid called 'heights'. We can move up, down, left, or right. The 'effort' of a path is defined as the maximum absolute difference in heights between any two consecutive cells in that path. The goal is to return the minimum possible effort among all possible paths from the starting cell to the destination cell.

Solution Approach

The solution employs Dijkstra's algorithm using a min-heap to find the path with the minimum effort. We maintain a 'dist' array to store the minimum effort required to reach each cell from the starting cell. The min-heap stores tuples of (effort, row, col), allowing us to always explore the cell with the least effort. The algorithm iteratively extracts the cell with the minimum effort from the heap, updates the 'dist' values of its neighbors if a path with lower effort is found, and adds the neighbors to the heap. The algorithm terminates when the bottom-right cell is reached, and the effort value associated with it represents the minimum effort required.

Step-by-Step Algorithm

Step 1: Initialize a 'dist' array of the same size as the 'heights' array, filling it with infinity. 'dist[0][0]' is initialized to 0 because the effort to reach the starting cell from itself is 0.
Step 2: Create a min-heap to store (effort, row, col) tuples. Insert the starting cell (0, 0) with an effort of 0 into the heap.
Step 3: While the heap is not empty:
Step 4: Extract the (effort, row, col) tuple with the minimum effort from the heap.
Step 5: If the current effort is greater than the current minimum effort to reach (row, col) as stored in 'dist[row][col]', continue to the next iteration (this means we have already found a better path to this cell).
Step 6: If the current cell is the destination cell (bottom-right), return the current effort (this is the minimum effort path).
Step 7: Iterate through the four possible directions (up, down, left, right). For each neighbor:
Step 8: Calculate the 'new_effort' as the absolute difference in height between the current cell and the neighbor cell.
Step 9: Calculate the 'max_effort' as the maximum between the current effort and the 'new_effort' (this is the effort of the path up to the neighbor).
Step 10: If the 'max_effort' is less than the current minimum effort to reach the neighbor as stored in 'dist[neighbor_row][neighbor_col]', update 'dist[neighbor_row][neighbor_col]' with 'max_effort', and insert (max_effort, neighbor_row, neighbor_col) into the min-heap.

Key Insights

Insight 1: The problem can be modeled as a shortest path problem where the 'distance' between nodes (cells) is the 'effort' (maximum absolute height difference between adjacent cells).
Insight 2: Since we're looking for the *minimum* effort, algorithms like Dijkstra or a variation of BFS are applicable. A binary search on the possible range of effort values can also be combined with a graph traversal to find if a path exists with a given maximum effort.
Insight 3: Using a priority queue (min-heap) helps to explore the paths with the least effort encountered so far, guaranteeing we find the minimum effort path first.

Complexity Analysis

Time Complexity: O(m*n*log(m*n))

Space Complexity: O(m*n)

Topics

This problem involves: Array, Binary Search, Depth-First Search, Breadth-First Search, Union Find, Heap (Priority Queue), Matrix.

Companies

Asked at: Cohesity, Snowflake.

Path With Minimum Effort | LeetCode 1631

Medium

Array Binary Search Depth-First Search Breadth-First Search Union Find Heap (Priority Queue)Matrix

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

Solution Code

Available in:

import java.util.PriorityQueue;

class Solution {
    public int minimumEffortPath(int[][] heights) {
        int rows = heights.length;
        int cols = heights[0].length;
        
        int[][] dist = new int[rows][cols];
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                dist[i][j] = Integer.MAX_VALUE;
            }
        }
        dist[0][0] = 0;
        
        PriorityQueue<int[]> min_heap = new PriorityQueue<>((a, b) -> a[0] - b[0]);  // (effort, row, col)
        min_heap.offer(new int[]{0, 0, 0});
        
        while (!min_heap.isEmpty()) {
            int[] curr = min_heap.poll();
            int effort = curr[0];
            int r = curr[1];
            int c = curr[2];
            
            if (effort > dist[r][c]) {
                continue;
            }
            
            if (r == rows - 1 && c == cols - 1) {
                return effort;
            }
            
            int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
            for (int[] dir : directions) {
                int nr = r + dir[0];
                int nc = c + dir[1];
                
                if (0 <= nr && nr < rows && 0 <= nc && nc < cols) {
                    int new_effort = Math.abs(heights[r][c] - heights[nr][nc]);
                    int max_effort = Math.max(effort, new_effort);
                    
                    if (max_effort < dist[nr][nc]) {
                        dist[nr][nc] = max_effort;
                        min_heap.offer(new int[]{max_effort, nr, nc});
                    }
                }
            }
        }
        
        return 0;
    }
}

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Problem Info

Description

Problem 1631: Path With Minimum Effort - Detailed explanation and solution approaches.

Topics

Array Binary Search Depth-First Search Breadth-First Search Union Find Heap (Priority Queue)Matrix

Companies

CohesitySnowflake

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1631

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Initialize a 'dist' array of the same size as the 'heights' array, filling it with infinity. 'dist[0][0]' is initialized to 0 because the effort to reach the starting cell from itself is 0.
2
Step 2: Create a min-heap to store (effort, row, col) tuples. Insert the starting cell (0, 0) with an effort of 0 into the heap.
3
Step 3: While the heap is not empty:
4
Step 4: Extract the (effort, row, col) tuple with the minimum effort from the heap.
5
Step 5: If the current effort is greater than the current minimum effort to reach (row, col) as stored in 'dist[row][col]', continue to the next iteration (this means we have already found a better path to this cell).
6
Step 6: If the current cell is the destination cell (bottom-right), return the current effort (this is the minimum effort path).
7
Step 7: Iterate through the four possible directions (up, down, left, right). For each neighbor:
8
Step 8: Calculate the 'new_effort' as the absolute difference in height between the current cell and the neighbor cell.
9
Step 9: Calculate the 'max_effort' as the maximum between the current effort and the 'new_effort' (this is the effort of the path up to the neighbor).
10
Step 10: If the 'max_effort' is less than the current minimum effort to reach the neighbor as stored in 'dist[neighbor_row][neighbor_col]', update 'dist[neighbor_row][neighbor_col]' with 'max_effort', and insert (max_effort, neighbor_row, neighbor_col) into the min-heap.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n*log(m*n))

💾

Space Complexity

O(m*n)

📝

Why?

Each cell is visited at most once. The min-heap can contain at most m*n elements. The heap operations (insertion and extraction) take O(log(m*n)) time. Therefore, the overall time complexity is O(m*n*log(m*n)). The space complexity is dominated by the 'dist' array and the min-heap, both of which require O(m*n) space.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Binary Search + DFS/BFS

Instead of using Dijkstra's algorithm, we can perform a binary search on the possible effort range (0 to maximum height difference). For each 'effort' value during binary search, we can use DFS or BFS to check if a path exists from the top-left to the bottom-right cell, where the absolute height difference between adjacent cells is at most 'effort'.

Time: O(m*n*log(H)), where H is the maximum height difference and m, n are dimensions of the grid.Space: O(m*n)

Trade-offs: Binary search + DFS/BFS may be faster if the range of height values (H) is significantly smaller than the number of cells (m*n). However, it can be slower if the DFS/BFS exploration becomes expensive for each 'effort' value.

Union Find

Sort all possible edges (connections between adjacent cells) based on their effort (absolute height difference). Then, iteratively add edges from smallest effort to largest effort. After adding each edge, check if the top-left and bottom-right cells are connected using Union Find. The first effort value at which they become connected is the minimum effort required.

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

Trade-offs: Union Find's performance depends on the efficiency of the Union Find implementation. It might be slower in practice due to the overhead of sorting and repeated Union Find operations.

✨

Key Insights

1Insight 1: The problem can be modeled as a shortest path problem where the 'distance' between nodes (cells) is the 'effort' (maximum absolute height difference between adjacent cells).
2Insight 2: Since we're looking for the *minimum* effort, algorithms like Dijkstra or a variation of BFS are applicable. A binary search on the possible range of effort values can also be combined with a graph traversal to find if a path exists with a given maximum effort.
3Insight 3: Using a priority queue (min-heap) helps to explore the paths with the least effort encountered so far, guaranteeing we find the minimum effort path first.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: The input 'heights' array contains only one row or one column. The algorithm correctly computes the effort as it iterates through the neighbors.
Edge case 2: The starting cell and the destination cell have the same height and all adjacent cells have the same height. The algorithm will correctly return 0 as the effort.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not initializing the 'dist' array with infinity, which can lead to incorrect results because the algorithm won't be able to determine the minimum effort to reach each cell.

✗Common mistake 2: Not using a priority queue (min-heap) to prioritize cells with lower effort, which can cause the algorithm to explore paths with higher effort unnecessarily, resulting in a higher time complexity or incorrect answer.

✗Common mistake 3: Not checking if the current effort is greater than the current best effort to reach a cell (dist[r][c]) before exploring its neighbors, leading to unnecessary computations and potential TLE.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Pathfinding in a terrain map where the 'heights' represent elevation and the 'effort' represents the steepness of the terrain. This could be used for autonomous vehicle navigation or route planning for hikers.

🏦Real-world scenario 2: Network routing where the 'heights' represent network congestion or latency and the 'effort' represents the maximum latency encountered along a path. The algorithm can find the path with the minimum maximum latency.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 12, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Path With Minimum Effort"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?