3Sum Leetcode Solution

In this post, we are going to solve the 3Sum Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Table of Contents

Problem

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
-10⁵ <= nums[i] <= 10⁵

Now, let’s see the leetcode solution of 3Sum Leetcode Solution.

3Sum Leetcode Solution in Python

class Solution:
  def threeSum(self, nums: List[int]) -> List[List[int]]:
    if len(nums) < 3:
      return []

    ans = []

    nums.sort()

    for i in range(len(nums) - 2):
      if i > 0 and nums[i] == nums[i - 1]:
        continue
      l = i + 1
      r = len(nums) - 1
      while l < r:
        summ = nums[i] + nums[l] + nums[r]
        if summ == 0:
          ans.append((nums[i], nums[l], nums[r]))
          l += 1
          r -= 1
          while nums[l] == nums[l - 1] and l < r:
            l += 1
          while nums[r] == nums[r + 1] and l < r:
            r -= 1
        elif summ < 0:
          l += 1
        else:
          r -= 1

    return ans

3Sum Leetcode Solution in CPP

class Solution {
 public:
  vector<vector<int>> threeSum(vector<int>& nums) {
    if (nums.size() < 3)
      return {};

    vector<vector<int>> ans;

    sort(begin(nums), end(nums));

    for (int i = 0; i + 2 < nums.size(); ++i) {
      if (i > 0 && nums[i] == nums[i - 1])
        continue;
      // Choose nums[i] as the first num in the triplet,
      // and search the remaining nums in [i + 1, n - 1]
      int l = i + 1;
      int r = nums.size() - 1;
      while (l < r) {
        const int sum = nums[i] + nums[l] + nums[r];
        if (sum == 0) {
          ans.push_back({nums[i], nums[l++], nums[r--]});
          while (l < r && nums[l] == nums[l - 1])
            ++l;
          while (l < r && nums[r] == nums[r + 1])
            --r;
        } else if (sum < 0) {
          ++l;
        } else {
          --r;
        }
      }
    }

    return ans;
  }
};

3Sum Leetcode Solution in Java

class Solution {
  public List<List<Integer>> threeSum(int[] nums) {
    if (nums.length < 3)
      return new ArrayList<>();

    List<List<Integer>> ans = new ArrayList<>();

    Arrays.sort(nums);

    for (int i = 0; i + 2 < nums.length; ++i) {
      if (i > 0 && nums[i] == nums[i - 1])
        continue;
      // Choose nums[i] as the first num in the triplet,
      // and search the remaining nums in [i + 1, n - 1]
      int l = i + 1;
      int r = nums.length - 1;
      while (l < r) {
        final int sum = nums[i] + nums[l] + nums[r];
        if (sum == 0) {
          ans.add(Arrays.asList(nums[i], nums[l++], nums[r--]));
          while (l < r && nums[l] == nums[l - 1])
            ++l;
          while (l < r && nums[r] == nums[r + 1])
            --r;
        } else if (sum < 0) {
          ++l;
        } else {
          --r;
        }
      }
    }

    return ans;
  }
}

Note: This problem 3Sum is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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