What is the time complexity of 3Sum Closest?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve 3Sum Closest?

This problem uses Array, Two Pointers, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving 3Sum Closest?

Insight 1: Sorting the array allows us to use a two-pointer approach, significantly improving time complexity. Insight 2: The two-pointer technique efficiently explores all possible pairs of numbers for each fixed number in the sorted array.

Which companies ask LeetCode 16. 3Sum Closest in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Google and 6 more companies.

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3Sum Closest - LeetCode 16 Solution

3Sum Closest - Complete Solution Guide

3Sum Closest is LeetCode problem 16, 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 of length n and an integer target , find three integers in nums such that the sum is closest to target . Return the sum of the three integers . You may assume that each input would have exactly one solution. Example 1: Input: nums = [-1,2,1,-4], target = 1 Output: 2 Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2). Example 2: Input: nums = [0,0,0], target = 1 Output: 0 Explanation: The sum that is closest to the target is 0. (0 + 0 + 0 = 0). C

Detailed Explanation

The problem requires finding three numbers within a given array whose sum is closest to a specified target value. The input is an array of integers `nums` and a target integer `target`. The output is the sum of three integers from `nums` that has the minimum absolute difference with the `target`. We are guaranteed that there will be one unique answer.

Solution Approach

The solution uses a combination of sorting and the two-pointer technique. First, the input array is sorted. Then, the algorithm iterates through the array, fixing one number at a time. For each fixed number, two pointers are used to find a pair of numbers in the remaining part of the array such that their sum, combined with the fixed number, is closest to the target. The algorithm keeps track of the closest sum found so far and updates it whenever a closer sum is encountered.

Step-by-Step Algorithm

Step 1: Sort the input array `nums` in ascending order.
Step 2: Initialize `closest_sum` to the sum of the first three elements of the sorted array. This serves as the initial closest sum.
Step 3: Iterate through the array from index `i = 0` to `n - 2` (where `n` is the length of the array).
Step 4: For each index `i`, set `left = i + 1` and `right = n - 1`. These are the two pointers.
Step 5: While `left < right`, calculate the `current_sum` as `nums[i] + nums[left] + nums[right]`.
Step 6: If `current_sum` is equal to the `target`, return the `target` because it's the perfect match.
Step 7: If the absolute difference between `current_sum` and `target` is less than the absolute difference between `closest_sum` and `target`, update `closest_sum` to `current_sum`.
Step 8: If `current_sum` is less than `target`, increment `left` to consider a larger number.
Step 9: If `current_sum` is greater than `target`, decrement `right` to consider a smaller number.
Step 10: Repeat steps 5-9 until `left` and `right` meet.
Step 11: After the outer loop completes, return the `closest_sum`.

Key Insights

Insight 1: Sorting the array allows us to use a two-pointer approach, significantly improving time complexity.
Insight 2: The two-pointer technique efficiently explores all possible pairs of numbers for each fixed number in the sorted array.
Insight 3: Keeping track of the `closest_sum` and updating it whenever a sum closer to the target is found is crucial.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, Sorting.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Google, Meta, Microsoft, Uber, Yahoo, Zoho, tcs.

3Sum Closest | LeetCode 16

Medium

Array Two Pointers Sorting

Time: O(n^2)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);
        int n = nums.length;
        int closest_sum = nums[0] + nums[1] + nums[2];

        for (int i = 0; i < n - 2; i++) {
            if (i > 0 && nums[i] == nums[i-1]) {
                continue;
            }

            int left = i + 1;
            int right = n - 1;
            while (left < right) {
                int current_sum = nums[i] + nums[left] + nums[right];

                if (current_sum == target) {
                    return target;
                }

                if (Math.abs(current_sum - target) < Math.abs(closest_sum - target)) {
                    closest_sum = current_sum;
                }

                if (current_sum < target) {
                    left += 1;
                } else {
                    right -= 1;
                }
            }
        }

        return closest_sum;
    }
}

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Problem Info

Description

Problem 16: 3Sum Closest - Detailed explanation and solution approaches.

Topics

Array Two Pointers Sorting

Companies

AdobeAmazonAppleBloombergGoogleMetaMicrosoftUber

Stats

Difficulty:Medium

Languages:4

Companies:11

Problem #:16

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

11 steps

1
Step 1: Sort the input array `nums` in ascending order.
2
Step 2: Initialize `closest_sum` to the sum of the first three elements of the sorted array. This serves as the initial closest sum.
3
Step 3: Iterate through the array from index `i = 0` to `n - 2` (where `n` is the length of the array).
4
Step 4: For each index `i`, set `left = i + 1` and `right = n - 1`. These are the two pointers.
5
Step 5: While `left < right`, calculate the `current_sum` as `nums[i] + nums[left] + nums[right]`.
6
Step 6: If `current_sum` is equal to the `target`, return the `target` because it's the perfect match.
7
Step 7: If the absolute difference between `current_sum` and `target` is less than the absolute difference between `closest_sum` and `target`, update `closest_sum` to `current_sum`.
8
Step 8: If `current_sum` is less than `target`, increment `left` to consider a larger number.
9
Step 9: If `current_sum` is greater than `target`, decrement `right` to consider a smaller number.
10
Step 10: Repeat steps 5-9 until `left` and `right` meet.
11
Step 11: After the outer loop completes, return the `closest_sum`.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the nested loops. The outer loop iterates `n-2` times, and the inner `while` loop iterates at most `n` times in total in its lifetime. Hence the overall time complexity becomes O(n*n) = O(n^2). The space complexity is constant because we are only using a few variables to store indices and the closest sum, irrespective of the input size.

📖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

Brute-Force Approach

Iterate through all possible combinations of three numbers in the array, calculate their sum, and find the sum closest to the target.

Time: O(n^3)Space: O(1)

Trade-offs: Much simpler to implement but significantly less efficient (O(n^3) vs O(n^2)). Not suitable for larger input arrays. The two-pointer approach is preferred due to better performance.

✨

Key Insights

1Insight 1: Sorting the array allows us to use a two-pointer approach, significantly improving time complexity.
2Insight 2: The two-pointer technique efficiently explores all possible pairs of numbers for each fixed number in the sorted array.
3Insight 3: Keeping track of the `closest_sum` and updating it whenever a sum closer to the target is found is crucial.

📚

Key Terms Explained

For Students

Array

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Sorting

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

Edge Cases

Edge case 1: Input array with duplicate numbers. The `if i > 0 and nums[i] == nums[i-1]: continue` line prevents redundant calculations by skipping duplicate numbers in the outer loop.
Edge case 2: Small input arrays (e.g., size 3). The algorithm works correctly even for the smallest possible input satisfying the constraints.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the array. Sorting is crucial for the two-pointer approach to work correctly and efficiently.

✗Common mistake 2: Not handling duplicate numbers correctly. Failing to skip duplicate numbers can lead to redundant calculations and potentially incorrect results in some cases.

✗Common mistake 3: Not initializing `closest_sum` properly. The initial value should be the sum of the first three elements after sorting. Some programmers might initialize it with a maximum integer value, which can cause issues if all possible sums are negative.

✅

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: Recommendation systems: Finding the combination of products or services that best match a user's preferences and budget.

🏦Real-world scenario 2: Financial modeling: Identifying the portfolio allocation that yields a return closest to a desired target, given various asset options.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 17, 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

❓

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.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

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 "3Sum Closest"?

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

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