# 001-two-sum

## Question

https://leetcode.com/problems/two-sum/description/

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].
``````

## Thought Process

1. Brute Force
1. Use every element as an anchor and search its right element
2. The time complexity is O(n^2) = (n - 1) + (n - 2) + ... + 1
3. The space complexity is O(1)
2. Optimal (Hash Map)
1. Leverage the power of hashmap and store the number and its index in the map
2. When the map contain target - curNum, we know the search is complete
3. The time complexity is O(n)
4. The space complexity is O(n)

## Solution

``````class Solution {
public int[] twoSum(int[] nums, int target) {
if (nums == null) return null;
int[] result = new int[2];
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
if (map.containsKey(target - nums[i])) {
result[0] = map.get(target - nums[i]);
result[1] = i;
return result;
}
map.put(nums[i], i);
}
return result;
}
}
``````