Problem Link

Description


Given an array of integers numsΒ 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.

Β 

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: 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]

Β 

Constraints:

  • 2 <= nums.length <= 104
  • -109 <= nums[i] <= 109
  • -109 <= target <= 109
  • Only one valid answer exists.

Β 

Follow-up:Β Can you come up with an algorithm that is less than O(n2)Β time complexity?

Solution


Python3

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        mp = {}
 
        for i, x in enumerate(nums):
            k = target - x
            if k in mp:
                return [mp[k], i]
            
            mp[x] = i
 
        return None

Python

class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        dict = {}
        for i in range(len(nums)):
            if target-nums[i] not in dict:
                dict[nums[i]]=i
            else:
                return [dict[target-nums[i]],i]