# Find two elements from the given which sums to the target number – Java

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 1:

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

## Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

## Example 3:

Input: nums = [3,3], target = 6
Output: [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 hash set, if yes return their
index, else add this to hash-set and continue.

### Java Code :

```class Solution {
class NumIndex {
int i, e;

NumIndex(int i, int e) {
this.i = i;
this.e = e;
}
}
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.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;
}
}```