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]
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]
Constraints:
2 <= nums.length <= 104
-109 <= nums[i] <= 109
-109 <= target <= 109
Only one valid answer exists.
Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?
Solutions
Approach 1: Hash Table
Time complexity: \(O(n)\)
Space complexity: \(O(n)\)
Notes
We can use the hash table seen to store the array value and the corresponding index.
Traverse the array nums, when you find target - nums[i] in the hash table, it means that the target value is found, and the index of target - nums[i] and i are returned.
The time complexity is O(n) and the space complexity is O(n). Where nis the length of the array nums.
classSolution:deftwoSum(self,nums:List[int],target:int)->List[int]:seen={}#map elements with their indexs {ele:idx}foridx,eleinenumerate(nums):num=target-eleifnuminseen:return[seen[num],idx]seen[ele]=idx