Three Sum

https://leetcode.com/problems/3sum

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.

Example:
Given array nums = [-1, 0, 1, 2, -1, -4],
A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

Implementation 1 : O(n^3) Time Limit Exceeded 😰

public static List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if(nums == null || nums.length < 3)
            return result;
        int n = nums.length; 
        for(int i = 0; i < n - 2; i++) {
            for(int j = i + 1; j < n - 1; j++) {
            for(int k = j + 1; k < n; k++) {
                 if((nums[i] + nums[j] + nums[k]) == 0 ) {
                  addResult(result, new Integer[] {nums[i], nums[j], nums[k]});
                  }
            }
            }
        }
        return result;
}
private static void addResult(List<List<Integer>> result, Integer[] triplet) {
    Set<Integer> threeSum = new HashSet<Integer>(Arrays.asList(triplet));
    boolean alredayExists = false;
    for(List<Integer> list: result) {
         Set<Integer> res = new HashSet<Integer>(list);
         if(res.equals(threeSum)) {
         alredayExists = true;
         break;
         }
    }
    if(!alredayExists) {
       result.add(Arrays.asList(triplet));
    }    
}

Above implementation have Runtime complexity of O(n^3) and space complexity of O(1)

Runtime Complexity = O(n^3)
Space Complexity   = O(1)

Implementation 2 : O(n^2) Time Limit Exceeded 😰

public static List<Integer[]> threeSum(int[] nums) {
        List<Integer[]> result = new ArrayList<Integer[]>();
        if(nums == null || nums.length < 3)
            return result;
        int n = nums.length; 
        Arrays.sort(nums);
        for(int i = 0; i < n - 2; i++) {
            int left = i + 1;
            int right = n - 1;
            while(left < right) {
                if( (nums[i] + nums[left] + nums[right]) > 0) {
                    right --;
                } else if( (nums[i] + nums[left] + nums[right]) < 0) {
                    left++;
                } else {
                    addResult(result, new Integer[] {nums[i], nums[left], nums[right]});
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
    private static void addResult(List<Integer[]> result, Integer[] triplet) {
        Set<Integer> threeSum = new HashSet<Integer>(Arrays.asList(triplet));
        boolean alredayExists = false;
        for(Integer[] list: result) {
            Set<Integer> res = new HashSet<Integer>(Arrays.asList(list));
            if(res.equals(threeSum)) {
                alredayExists = true;
                break;
            }
        }
        if(!alredayExists) {
            result.add(triplet);
        }
    }

Above implementation have Runtime complexity of O(n^2) and space complexity of O(1)

Runtime Complexity = O(n^2)
Space Complexity   = O(1)

Implementation 3 : O(n^2) Optimization - Handling duplicates in beautiful way 😊

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> response = new ArrayList<>();
        if(nums == null || nums.length < 3)
            return response;
        Arrays.sort(nums);
        int n = nums.length;
        for(int i = 0; i < n-2; i++) {
            if(i > 0 && nums[i] == nums[i-1])
                continue;
            int left = i+1;
            int right = n-1;
            while(left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                if(sum == 0) {
                    response.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    while(left < right && nums[left] == nums[left+1])
                        left++;
                    while(left < right && nums[right] == nums[right-1])
                        right--;
                    left++;
                    right--;
                } else if(sum > 0) {
                    right--;
                } else {
                    left++;
                }
            }
        }
       return response; 
    }
}