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package array;

import java.util.ArrayList;

import java.util.List;

/**

* Created by gouthamvidyapradhan on 11/07/2017. Given an array of integers, return indices of the

* two numbers such that they add up to a specific target.

*

* <p>You may assume that each input would have exactly one solution, and you may not use the same

* element twice.

*

* <p>Example: Given nums = [2, 7, 11, 15], target = 9,

*

* <p>Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].

*

* <p>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.

*/

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;

}

}