[LeetCode] [easy] Two Sum (brute-force, map)

1. Problem

Given: numbers array and target.

 

Select two element and sum, if output is target return these index.

 

2. brute-force

time-complexity of brute-force is O(N^2)

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

 

 

3. Hash Map

time-complexity use of Hash Map is O(N).

if use 'map' then time is O(NlgN), because map is BST. 

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> m;
 
        for(int i = 0; i < nums.size(); i++)
        {
            if(m.count(nums[i]))
                 return {m[nums[i]], i};
        
            m[target-nums[i]] = i;
            
        }
        return {};
    }
};

(4) Two Sum - LeetCode