3 Sum Problem LeetCode

 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.


Good Approach


class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
         List<List<Integer>> result = new ArrayList<>();
        //Arrays.sort(nums); // Step 1: Sort the array
        int n = nums.length;

       Set<List<Integer>> st = new HashSet<>();

        for (int i = 0; i < n; i++) {
            Set<Integer> hashset = new HashSet<>();
            for (int j = i + 1; j < n; j++) {
                //Calculate the 3rd element:
                int third = -(nums[i] + nums[j]);

                //Find the element in the set:
                if (hashset.contains(third)) {
                    List<Integer> temp = Arrays.asList(nums[i], nums[j], third);
                    Collections.sort(temp);
                    st.add(temp);
                }
                hashset.add(nums[j]);
            }
        }

        // store the set elements in the answer:
        List<List<Integer>> ans = new ArrayList<>(st);
        return ans;
    }
}



Optimal Approach

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
         List<List<Integer>> result = new ArrayList<>();
        Arrays.sort(nums); // Step 1: Sort the array
        int n = nums.length;

        for (int i = 0; i < n - 2; i++) { // Step 2: Fix one number (nums[i])
            if (i > 0 && nums[i] == nums[i - 1]) continue; // Skip duplicates

            int left = i + 1, right = n - 1; // Two pointers
            int target = -nums[i]; // We need two numbers that sum to this value

            while (left < right) { // Step 3: Find two numbers that sum to `-nums[i]`
                int sum = nums[left] + nums[right];

                if (sum == target) { // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));

                    // Move left and right pointers to skip duplicates
                    while (left < right && nums[left] == nums[left + 1]) left++;
                    while (left < right && nums[right] == nums[right - 1]) right--;

                    left++;
                    right--;
                } else if (sum < target) {
                    left++; // Increase sum
                } else {
                    right--; // Decrease sum
                }
            }
        }
        return result;
    }
}



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