package com.leetcode.array;

import com.princeton.stdlib.In;

import java.util.HashMap;

import java.util.Map;

/**

* Created by charles on 4/9/17.

* 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].

*/

public class TwoSum_1 {

/**

* Thought: efficient way to check if complement exists in the array

* if complement exists, we need to lookup its index.

* HashTable is the best way to maintain a mapping of each element in array in its index

*

* use two pass iterations

*/

public int[] twoSum(int[] nums, int target) {

int[] res = new int[0];

Map<Integer, Integer> map = new HashMap<>();

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

map.put(nums[i], i);

}

int complement = 0;

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

complement = target - nums[i];

if (map.containsKey(complement) && map.get(complement) != i) {

// be careful should not use num twice like 3 + 3 = 6

return new int[] {i, map.get(complement)};

}

}

return res;

}

/** one pass iteration, can look current element's complement already exist in table */

public int[] twoSumII(int[] nums, int target) {

Map<Integer, Integer> map = new HashMap<>();

int[] res = new int[0];

int complement = 0;

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

complement = target - nums[i];

if (map.containsKey(complement)) {

return new int[]{map.get(complement), i};

}

map.put(nums[i], i);

}

return res;

}

}