Find Triplets Sum equal to given Target in Java with Explanation | 3 Sum Problem

Return all unique triplets sum to given target | Leetcode Problem

Problem Description :

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.

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].
The order of the output and the order of the triplets does not matter.

Example 2 :

Input: nums = [0,1,1]
Output: []
Explanation: Any triplet does not sum up to 0.

Here we can solve this problem using two pointer approach. 

Solution 1 : Java solution for finding triplets sum to given target

import java.util.ArrayList;

import java.util.Arrays;

import java.util.List;

import java.util.Scanner;

public class TripletsSum {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        

        System.out.println("Enter array size : ");

        int size = sc.nextInt();

        int[] array = new int[size];

        

        // Taking user inputs

        System.out.println("Enter array elements :");

        for (int i = 0; i < array.length; i++) {

            array[i] = sc.nextInt();

        }

        

        System.out.println(threeSum(array));

    }

    

    public static List<List<Integer>> threeSum(int[] nums) {

            

        List<List<Integer>> ans = new ArrayList<>();

            

        Arrays.sort(nums);

            

        for (int i = 0; i < nums.length-2; i++) {

            

            if (i == 0 || (i > 0 && nums[i] != nums[i-1])) {

                

                int start = i+1;

                int end = nums.length-1;

                int sum = 0 - nums[i];

                

                while (start < end) {

                    if (nums[start]+nums[end] == sum) {

                        ans.add(Arrays.asList(nums[i], nums[start], nums[end]));

                        

                        while (start < end && nums[start] == nums[start+1]) {

                            start++;

                        }

                        

                        while (start < end && nums[end] == nums[end-1]) {

                            end--;

                        }

                        

                        start++;

                        end--;

                    } else if (nums[start] + nums[end] < sum) {

                        start++;

                    } else {

                        end--;

                    }

                    

                }

                

            }

        }

        return ans;

    }

} 

Solution Explanation :

• We have to return all possible unique triplets sum. so declare List of List where we can store all triplets.
• Sort the given nums array using Arrays.sort() method.
• We traverse through entire given nums array.
• In if loop, skip the repeated number so it does not loop through duplicate numbers again.
• Here we are using two pointer approach so take start as current + 1, end as array length index and sum as current ith number. (Subtract with 0 so we will get positive number).
• Start While loop until start is less than end.
• Check if addition of start and end become to sum. If so add start, end and ith index into ans list whichever we are returning at last as all triplets sum.
• We have to find out all triplets sum that are present in given nums array, so if we find one triplets, move start and end pointers. Skip duplicates numbers from given nums array.
• At last, return ans list which have all unique triplets to given target.

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