📌Problem

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Constraints:

• 2 <= nums.length <= $10^3$
• $-10^9$ <= nums[i] <= $10^9$
• $-10^9$ <= target <= $10^9$
• Only one valid answer exists.

leetcode 에서 풀기

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📝Solution

✏️ Brute Force

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        for i in range(len(nums)):
            for j in range(i+1,len(nums)):
                if nums[i]+nums[j] == target:
                    return [i,j]

• Time complexity : $O(N^2)$

✏️ Hash-table

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        
        hashMap={nums[idx]:idx for idx in range(len(nums))} # {num : index}
        
        for idx in range(len(nums)):
            complement=target-nums[idx]
            if hashMap.get(complement,False) and hashMap.get(complement,False) != idx:
                return [idx,hashMap.get(complement)]           

• Time complexity : $O(N)$