3Sum

LeetCode 15 | Difficulty: Medium

Medium

Problem Description

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation: 
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.

Example 2:

Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.

Example 3:

Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.

Constraints:

- `3 <= nums.length <= 3000`

- `-10^5 <= nums[i] <= 10^5`

Topics: Array, Two Pointers, Sorting

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Approach

Two Pointers

Use two pointers to traverse the array, reducing the search space at each step. This avoids the need for nested loops, bringing complexity from O(n²) to O(n) or O(n log n) if sorting is involved.

When to use

Array is sorted or can be sorted, and you need to find pairs/triplets that satisfy a condition.

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Solutions

Solution 1: C# (Best: 156 ms)

Metric	Value
Runtime	156 ms
Memory	48.2 MB
Date	2021-11-09

Solution

public class Solution {
    public IList<IList<int>> ThreeSum(int[] nums) {
       IList<IList<int>> result = new List<IList<int>>();
    if (nums.Length < 3)
    {
        return result;
    }
    Array.Sort(nums);
    for (int i = 0; i < nums.Length - 2; i++)
    {
        if(i>0 && nums[i] == nums[i-1]) continue;
        if(nums[i] > 0) break;
        int left = i + 1;
        int right = nums.Length - 1;
        while (left < right)
        {
            int sum = nums[i] + nums[left] + nums[right];
            if (sum == 0)
            {
                result.Add(new List<int> { nums[i], nums[left], nums[right] });
                left++;
                right--;
                while (left < right && nums[left] == nums[left - 1]) left++;
                while (left < right && nums[right] == nums[right + 1]) right--;
            }
            else if (sum < 0)
            {
                left++;
            }
            else
            {
                right--;
            }
        }
    }

    return result;
    }
}

📜 2 more C# submission(s)

Submission (2021-11-09) — 164 ms, 47.9 MB

public class Solution {
    public IList<IList<int>> ThreeSum(int[] nums) {
       IList<IList<int>> result = new List<IList<int>>();
    if (nums.Length < 3)
    {
        return result;
    }
    Array.Sort(nums);
    for (int i = 0; i < nums.Length - 2; i++)
    {
        if(i>0 && nums[i] == nums[i-1]) continue;
        int left = i + 1;
        int right = nums.Length - 1;
        while (left < right)
        {
            int sum = nums[i] + nums[left] + nums[right];
            if (sum == 0)
            {
                result.Add(new List<int> { nums[i], nums[left], nums[right] });
                left++;
                right--;
                while (left < right && nums[left] == nums[left - 1]) left++;
                while (left < right && nums[right] == nums[right + 1]) right--;
            }
            else if (sum < 0)
            {
                left++;
            }
            else
            {
                right--;
            }
        }
    }

    return result;
    }
}

Submission (2017-07-24) — 742 ms, N/A

public class Solution {
    public IList<IList<int>> ThreeSum(int[] nums) {
        IList<IList<int>> result = new List<IList<int>>();
    Array.Sort(nums);
    for (int i = 0; i+2 < nums.Length; i++)
    {
        if(i>0 && nums[i]==nums[i-1]) continue;
        int j = i + 1, k = nums.Length - 1;
        int target = -nums[i];
        while (j < k)
        {
            if (nums[j] + nums[k] == target)
            {
                result.Add(new List<int> { nums[i], nums[j], nums[k]});
                j++;
                k--;
                while(j<k && nums[j] == nums[j-1]) j++;
                while(j<k && nums[k] == nums[k+1]) k--;
            }
            else if (nums[j] + nums[k] > target)
            {
                k--;
            }
            else j++;               
        }       
    }
    return result;
    }
}

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Complexity Analysis

Approach	Time	Space
Two Pointers	$O(n)$	$O(1)$
Sort + Process	$O(n log n)$	$O(1) to O(n)$

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Interview Tips

Key Points

• Discuss the brute force approach first, then optimize. Explain your thought process.
• Ask: "Can I sort the array?" — Sorting often enables two-pointer techniques.
• LeetCode provides 3 hint(s) for this problem — try solving without them first.

💡 Hints

Hint 1: So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand!

Hint 2: For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next number y, which is value - x where value is the input parameter. Can we change our array somehow so that this search becomes faster?

Hint 3: The second train of thought for two-sum is, without changing the array, can we use additional space somehow? Like maybe a hash map to speed up the search?