Find Triplets with Given Sum in an Unsorted Array

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] equals to a target sum

Notice that the solution set must not contain duplicate triplets.

Example 1

Input: nums = [-1, 0, 1, 2, -1, -4], target = 0
Output: [[-1,-1,2],[-1,0,1]]
Explanation: The values in each solution set sum up to 0

Example 2

Input: nums = [0, 1, 1, 2, -1, 3], target = 4
Output: [[-1,2,3],[0,1,3],[1,1,2]]
Explanation: The values in each solution set sum up to 4

Approach 1: Brute Force

Idea

The idea is to simply run three loops and check for the target sum. To avoid duplicates, we will insert the values into a set.

Implementation

vector<vector<int>> threeSum(vector<int>& nums, const int target) {
    int n = nums.size();
    sort(nums.begin(), nums.end());
    set<vector<int>> tripletValues;

    for (int i = 0; i < n; ++i) {
        for (int j = i + 1; j < n; ++j) {
            for (int k = j + 1; k < n; ++k) {
                if (nums[i] + nums[j] + nums[k] == target) {
                    tripletValues.insert({nums[i], nums[j], nums[k]});
                }
            }
        }
    }

    return vector<vector<int>>(tripletValues.begin(), tripletValues.end());
}

Complexity Analysis

Time Complexity: $O(N^3 + N^3log(N^3)) = O(N^3logN)$ since we are running three nested loops. Inserting the results into the set takes logarithmic time.
Space Complexity: $O(N^3)$ extra for the set.

Approach 2: Sort and Two Pointers

This problem can be reduced to two sum. First sort the array and then let us fix one number to nums[i] and find for pairs of sum target - nums[i] for the subarray nums[i + 1...n - 1]. The last part can be solved using the two pointers approach of two sum (approach 3 of the problem two sum).

Implementation

vector<vector<int>> threeSum(vector<int>& nums, const int target) {
    sort(nums.begin(), nums.end());
    int n = nums.size();
    vector<vector<int>> triplets;
    for (int i = 0; i < n; ++i) {
        // nums[i] != nums[i - 1] is used to ignore the duplicate solutions
        if (i == 0 or nums[i] != nums[i - 1]) {
            int cur = nums[i];
            int required = target - cur;

            int left = i + 1, right = n - 1;

            // this part is equivalent to two sum
            while (left < right) {
                int curSum = nums[left] + nums[right];
                if (curSum == required) {
                    triplets.push_back({nums[i], nums[left], nums[right]});
                    while(left < right and nums[right] == nums[right - 1]) --right;
                    while(left < right and nums[left] == nums[left + 1]) ++left;
                    --right;
                    ++left;
                } else if (curSum > required) {
                    --right;
                } else {
                    ++left;
                }
            }
        }
    }
    return triplets;
}

Complexity Analysis

Time Complexity: $O(N^2)$ $ for the outer for and inner while loop.
Space Complexity: $O(1)$ extra. No extra space is required.

Published 31 Jul 2021

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