Given an array nums of 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.

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Example:

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

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

Solution:

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class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        
        if (nums.length < 3) {
            return Collections.emptyList();
        }
        
        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<>();
        Set<String> uniq = new HashSet<>();
        Map<Integer, Integer> valueToIndex = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            valueToIndex.put(nums[i], i);
        }
        
        int i = 0;
        
        while (i < nums.length - 1) {
            for (int j = i + 1; j < nums.length; j++) {
                Integer sum = -(nums[i] + nums[j]);
                Integer sumIndex = valueToIndex.get(sum);
                if (sumIndex != null 
                    && sumIndex != i 
                    && sumIndex != j
                    && sumIndex > j
                    && uniq.contains("" + nums[i] + nums[j] + sumIndex) == false) {
                    result.add(List.of(nums[i], nums[j], sum));
                    uniq.add("" + nums[i] + nums[j] + sumIndex);
                }
            }  
            i++;
        }
        return result;
    }
}

Solution Without Space

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class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        int n = nums.length;
        if (n < 3) return Collections.emptyList();
        Arrays.sort(nums);
        List<List<Integer>> res = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) { // add only first occurence
                continue;
            } 
            int lo = i + 1;
            int hi = n - 1;
            int sum = - nums[i];  
            while (lo < hi) {
                if (nums[lo] + nums[hi] == sum) {
                    res.add(List.of(nums[i], nums[lo], nums[hi]));
                    while (lo < hi && nums[lo] == nums[lo + 1]) lo++; // skip duplicates
                    while (lo < hi && nums[hi] == nums[hi - 1]) hi--;
                    lo++;
                    hi--;
                } else if (nums[lo] + nums[hi] < sum) {
                    lo++;
                } else {
                    hi--;
                }
            }
        }
        return res;
    }
}

Solution 2021-11-23

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class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        int n = nums.length;
        Arrays.sort(nums);
        List<List<Integer>> res = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            if (i > 0 && nums[i] == nums[i-1]) continue;
            int lo = i + 1;
            int hi = n - 1;
            int target = -nums[i];
            while (lo < hi) {
                if (nums[lo] + nums[hi] > target) {
                    hi--;
                } else if (nums[lo] + nums[hi] < target) {
                    lo++;
                } else {
                    res.add(List.of(nums[i], nums[lo], nums[hi]));
                    while (lo < hi && nums[lo] == nums[lo + 1]) lo++;
                    while (lo < hi && nums[hi - 1] == nums[hi]) hi--;
                    lo++;
                    hi--;
                }
            }
        }
        return res;
    }
}