# LeetCode challenge – Two Sum – Java Solution

This is a solution for the LeetCode challenge – Two Sum written in Java ( Source )

Given an array of integers, return indices of the two numbers such that they add up to a specific target.You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].

Solution: O(n log n). Wrap index and element in a class and sort in increasing order. Do a two pointer sum and compare. An alternative solution is to use hashing which is a O(n) solution – For each element e check if element (target – e) is already found in hashset, if yes return their index, else add this to hash-set and continue.

import java.util.ArrayList; import java.util.List; public class TwoSum { class NumIndex { int i, e; NumIndex(int i, int e) { this.i = i; this.e = e; } } public static void main(String[] args) { int[] nums = {3, 2, 4}; int[] ans = new TwoSum().twoSum(nums, 6); for (int i : ans) System.out.println(i); } public int[] twoSum(int[] nums, int target) { List<NumIndex> list = new ArrayList<>(); for (int i = 0; i < nums.length; i++) { NumIndex n = new NumIndex(i, nums[i]); list.add(n); } list.sort((o1, o2) -> Integer.compare(o1.e, o2.e)); int[] ans = new int[2]; for (int i = 0, j = nums.length - 1; i < j; ) { NumIndex numi = list.get(i); NumIndex numj = list.get(j); int sum = numi.e + numj.e; if (sum == target) { ans[0] = numi.i; ans[1] = numj.i; return ans; } else if (sum > target) { j--; } else i++; } return ans; } }