# 1. Two Sum

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

Approach 1: Brute force.

Approach 2: Hash table.

Approach 3: Sort then use two pointers.

# Code (Python)

Approach 3:

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        sorted_nums = sorted((num, i) for i, num in enumerate(nums))
        l, r = 0, len(sorted_nums) - 1
        while l < r:
            if sorted_nums[l][0] + sorted_nums[r][0] == target:
                return [sorted_nums[l][1], sorted_nums[r][1]]
            elif sorted_nums[l][0] + sorted_nums[r][0] < target:
                l += 1
            else:
                r -= 1

# Code (C++)

Approach 1:

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> res;
        int n = nums.size();
        if (n <= 1)
        {
            return res;
        }
        
        for (int i = 0; i < n - 1; ++i)
        {
            for (int j = i + 1; j < n; ++j)
            {
                if (nums[i] + nums[j] == target)
                {
                    res.push_back(i);
                    res.push_back(j);
                }                
            }
        }
        return res;
    }
};

Approach 2:

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target)
    {
        vector<int> res;
        int n = nums.size();
        if (n <= 1)
        {
            return res;
        }
            
        unordered_map<int,int> umap;
        for (int i = 0; i < n; ++i)
        {
            int tmp = target - nums[i];
            if (umap.find(tmp) != umap.end())
            {
                res.push_back(i);
                res.push_back(umap[tmp]);
                break;
            }
            umap[nums[i]] = i;
        }
        return res;
    }
};