What is the time complexity of Maximum Gap?

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

What data structures are used to solve Maximum Gap?

This problem uses Array, Sorting, Bucket Sort, Radix Sort. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Gap?

Insight 1: Directly sorting the array using comparison-based sorting algorithms (e.g., merge sort, quicksort) would result in O(n log n) time complexity, violating the linear time constraint. Insight 2: The problem can be solved in linear time using bucket sort, or a variation of it, since the range of input numbers is known and the linear space is allowed.

Which companies ask LeetCode 164. Maximum Gap in interviews?

This problem is frequently asked at DoorDash.

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Maximum Gap - LeetCode 164 Solution

Maximum Gap - Complete Solution Guide

Maximum Gap is LeetCode problem 164, 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

Given an integer array nums , return the maximum difference between two successive elements in its sorted form . If the array contains less than two elements, return 0 . You must write an algorithm that runs in linear time and uses linear extra space. Example 1: Input: nums = [3,6,9,1] Output: 3 Explanation: The sorted form of the array is [1,3,6,9], either (3,6) or (6,9) has the maximum difference 3. Example 2: Input: nums = [10] Output: 0 Explanation: The array contains less than 2 elements, t

Detailed Explanation

The problem requires finding the largest difference between adjacent elements in a sorted version of a given array of integers. If the array has fewer than two elements, the maximum gap is 0. Crucially, the problem insists on a linear time and linear space solution, effectively ruling out comparison sorts (like merge sort or quicksort) directly.

Solution Approach

The solution utilizes a bucket sort approach. The core idea is to divide the range of numbers into buckets of equal size. Then, for each number in the input array, we determine its corresponding bucket and update the minimum and maximum values within that bucket. Finally, we iterate through the buckets and find the maximum gap between the maximum value of one bucket and the minimum value of the next non-empty bucket.

Step-by-Step Algorithm

Step 1: Handle the base case: If the array has fewer than 2 elements, return 0.
Step 2: Find the minimum and maximum values in the array.
Step 3: Calculate the bucket size. To ensure linear time, the bucket size is typically calculated as `max(1, (max_val - min_val) // (n - 1))`. Using `n-1` aims to have at least one empty bucket.
Step 4: Determine the number of buckets required: `(max_val - min_val) // bucket_size + 1`.
Step 5: Initialize the buckets. Each bucket stores the minimum and maximum values encountered within that bucket. Initialize the min value as infinity and max value as negative infinity.
Step 6: Iterate through the input array. For each number, calculate its bucket index: `(num - min_val) // bucket_size`. Update the minimum and maximum values for the corresponding bucket.
Step 7: Iterate through the buckets, maintaining the maximum gap found so far. The gap between two successive values is the difference between the current bucket's minimum and the previous bucket's maximum. Skip empty buckets.
Step 8: Return the maximum gap found.

Key Insights

Insight 1: Directly sorting the array using comparison-based sorting algorithms (e.g., merge sort, quicksort) would result in O(n log n) time complexity, violating the linear time constraint.
Insight 2: The problem can be solved in linear time using bucket sort, or a variation of it, since the range of input numbers is known and the linear space is allowed.
Insight 3: The Pigeonhole Principle guarantees that if we divide the range of values into n-1 buckets, at least one bucket must be empty if there are n elements in the input array. The maximum gap must therefore occur between elements in different buckets.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Sorting, Bucket Sort, Radix Sort.

Companies

Asked at: DoorDash.

Maximum Gap | LeetCode 164

Medium

Array Sorting Bucket Sort Radix Sort

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int maximumGap(int[] nums) {
        int n = nums.length;
        if (n < 2) {
            return 0;
        }

        int min_val = nums[0];
        int max_val = nums[0];
        for (int num : nums) {
            min_val = Math.min(min_val, num);
            max_val = Math.max(max_val, num);
        }

        if (min_val == max_val) {
            return 0;
        }

        int bucket_size = Math.max(1, (max_val - min_val) / (n - 1));
        int num_buckets = (max_val - min_val) / bucket_size + 1;
        
        int[][] buckets = new int[num_buckets][2];
        for (int i = 0; i < num_buckets; i++) {
            buckets[i][0] = Integer.MAX_VALUE;
            buckets[i][1] = Integer.MIN_VALUE;
        }
        
        for (int num : nums) {
            int idx = (num - min_val) / bucket_size;
            buckets[idx][0] = Math.min(buckets[idx][0], num);
            buckets[idx][1] = Math.max(buckets[idx][1], num);
        }
        
        int max_gap = 0;
        int prev_max = buckets[0][1];
        
        for (int i = 1; i < num_buckets; i++) {
            if (buckets[i][0] == Integer.MAX_VALUE) {
                continue;
            }
            
            max_gap = Math.max(max_gap, buckets[i][0] - prev_max);
            prev_max = buckets[i][1];
            
        }
        
        return max_gap;
    }
}

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Problem Info

Description

Problem 164: Maximum Gap - Detailed explanation and solution approaches.

Topics

Array Sorting Bucket Sort Radix Sort

Companies

DoorDash

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:164

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

8 steps

1
Step 1: Handle the base case: If the array has fewer than 2 elements, return 0.
2
Step 2: Find the minimum and maximum values in the array.
3
Step 3: Calculate the bucket size. To ensure linear time, the bucket size is typically calculated as `max(1, (max_val - min_val) // (n - 1))`. Using `n-1` aims to have at least one empty bucket.
4
Step 4: Determine the number of buckets required: `(max_val - min_val) // bucket_size + 1`.
5
Step 5: Initialize the buckets. Each bucket stores the minimum and maximum values encountered within that bucket. Initialize the min value as infinity and max value as negative infinity.
6
Step 6: Iterate through the input array. For each number, calculate its bucket index: `(num - min_val) // bucket_size`. Update the minimum and maximum values for the corresponding bucket.
7
Step 7: Iterate through the buckets, maintaining the maximum gap found so far. The gap between two successive values is the difference between the current bucket's minimum and the previous bucket's maximum. Skip empty buckets.
8
Step 8: Return the maximum gap found.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The linear time complexity is achieved by using a non-comparison based sorting strategy. The bucket operations are done in constant time per element. The buckets use linear space since we have at most n buckets.

📖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

Radix Sort

Radix sort is another non-comparison based sorting algorithm that can sort integers in linear time under certain conditions. It sorts elements by processing individual digits or bits.

Time: O(nk), where n is the number of elements and k is the number of digits in the largest number. If k is constant or proportional to log(n), this approaches O(n).Space: O(n + r), where r is the range of digits (e.g., 10 for decimal digits).

Trade-offs: Radix sort can be faster than bucket sort in some cases, especially when the numbers are uniformly distributed. However, it can be more complex to implement. Bucket sort is generally simpler and suitable when the data range is well-defined.

✨

Key Insights

1Insight 1: Directly sorting the array using comparison-based sorting algorithms (e.g., merge sort, quicksort) would result in O(n log n) time complexity, violating the linear time constraint.
2Insight 2: The problem can be solved in linear time using bucket sort, or a variation of it, since the range of input numbers is known and the linear space is allowed.
3Insight 3: The Pigeonhole Principle guarantees that if we divide the range of values into n-1 buckets, at least one bucket must be empty if there are n elements in the input array. The maximum gap must therefore occur between elements in different buckets.

📚

Key Terms Explained

For Students

Array

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Array with less than two elements: The code explicitly handles this case by returning 0.
Edge case 2: All elements in the array are the same: The code correctly handles this by returning 0 after identifying min_val == max_val.
Edge case 3: Array with very large values: The code handles large integer values by using `int` or `long` data types that can accommodate numbers up to 10^9 as per the constraints. The bucket index calculation `(num - min_val) // bucket_size` should not cause overflow issues if performed correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using comparison-based sorting algorithms (like quicksort or mergesort) which have O(n log n) time complexity, thus failing to meet the linear time requirement.

✗Common mistake 2: Not considering edge cases such as arrays with fewer than two elements or arrays where all elements are the same. Failing to handle these cases can lead to incorrect results.

✗Common mistake 3: Integer overflow in bucket index calculation: When calculating the bucket index `(num - min_val) // bucket_size`, ensure that intermediate results do not cause integer overflow, particularly when dealing with large input numbers. Using proper casting and data types can help.

✗Common mistake 4: Incorrect bucket size calculation: An incorrect bucket size might lead to either too many empty buckets or all the elements falling into the same bucket, defeating the purpose of bucket sort and resulting in incorrect maximum gap calculation.

✅

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: Analyzing data ranges in sensor readings: Identifying the largest difference between consecutive sensor readings within a specific time frame. For instance, finding the largest temperature difference recorded by a weather sensor within a day.

🏦Real-world scenario 2: Financial data analysis: Determining the largest price gap between consecutive stock prices during a trading day. This information can be valuable for identifying volatility and potential trading opportunities.

💡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: November 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Gap"?

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 sorting techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?