3Sum

Given an arraynumsof n _integers, are there elements _a, b, c _in nums such that _a+b+c= 0? Find all unique triplets in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

Solution

O(n pow 3) solution is as follows, but it does not prevent duplicates.

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();

        for (int i = 0; i < nums.length - 2; i++) {
            for (int j = i + 1; j < nums.length - 1; j++) {
                for (int k = j + 1; k < nums.length; k++) {
                    if (nums[i] + nums[j] + nums[k] == 0) {
                        List<Integer> found = new ArrayList<>();
                        found.add(nums[i]);
                        found.add(nums[j]);
                        found.add(nums[k]);
                        result.add(found);
                    }
                }
            }
        }

        return result;
    }
}

Other option is to use similar approach which is used in 2Sum problem.

Here is the code.

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        Arrays.sort(nums);

        for (int i = 0; i < nums.length - 2; i++) {
            if (i > 0 && nums[i - 1] == nums[i]) { // the same numbers to avoid duplicates
                continue;
            }
            int j = i + 1;
            int k = nums.length - 1;
            while (j < k) {
                int sum = nums[i] + nums[j] + nums[k];
                if (sum == 0) {
                    List<Integer> triplet = new ArrayList<>();
                    triplet.add(nums[i]);
                    triplet.add(nums[j]);
                    triplet.add(nums[k]);
                    result.add(triplet);
                    j++;
                    k--;
                    // avoid duplicates - look at the previous number, if the same, skip it
                    while (j < k && nums[j] == nums[j - 1]) {
                        j++;
                    }
                    while (j < k && nums[k] == nums[k + 1]) {
                        k--;
                    }
                } else if (sum < 0) {
                    j++;
                } else {
                    k--;
                }
            }
        }

        return result;
    }
}