3Sum LeetCode Solution

Problem – 3Sum

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

3Sum LeetCode Solution in Java

public List<List<Integer>> threeSum(int[] num) {
    Arrays.sort(num);
    List<List<Integer>> res = new LinkedList<>(); 
    for (int i = 0; i < num.length-2; i++) {
        if (i == 0 || (i > 0 && num[i] != num[i-1])) {
            int lo = i+1, hi = num.length-1, sum = 0 - num[i];
            while (lo < hi) {
                if (num[lo] + num[hi] == sum) {
                    res.add(Arrays.asList(num[i], num[lo], num[hi]));
                    while (lo < hi && num[lo] == num[lo+1]) lo++;
                    while (lo < hi && num[hi] == num[hi-1]) hi--;
                    lo++; hi--;
                } else if (num[lo] + num[hi] < sum) lo++;
                else hi--;
           }
        }
    }
    return res;
}

3Sum LeetCode Solution in C++

vector<vector<int> > threeSum(vector<int> &num) {
    
    vector<vector<int> > res;

    std::sort(num.begin(), num.end());

    for (int i = 0; i < num.size(); i++) {
        
        int target = -num[i];
        int front = i + 1;
        int back = num.size() - 1;

        while (front < back) {

            int sum = num[front] + num[back];
            
            // Finding answer which start from number num[i]
            if (sum < target)
                front++;

            else if (sum > target)
                back--;

            else {
                vector<int> triplet = {num[i], num[front], num[back]};
                res.push_back(triplet);
                
                // Processing duplicates of Number 2
                // Rolling the front pointer to the next different number forwards
                while (front < back && num[front] == triplet[1]) front++;

                // Processing duplicates of Number 3
                // Rolling the back pointer to the next different number backwards
                while (front < back && num[back] == triplet[2]) back--;
            }
            
        }

        // Processing duplicates of Number 1
        while (i + 1 < num.size() && num[i + 1] == num[i]) 
            i++;

    }
    
    return res;
    
}

3Sum LeetCode Solution in Python

def threeSum(self, nums):
    res = []
    nums.sort()
    for i in xrange(len(nums)-2):
        if i > 0 and nums[i] == nums[i-1]:
            continue
        l, r = i+1, len(nums)-1
        while l < r:
            s = nums[i] + nums[l] + nums[r]
            if s < 0:
                l +=1 
            elif s > 0:
                r -= 1
            else:
                res.append((nums[i], nums[l], nums[r]))
                while l < r and nums[l] == nums[l+1]:
                    l += 1
                while l < r and nums[r] == nums[r-1]:
                    r -= 1
                l += 1; r -= 1
    return res

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