Easy
Two Sum
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.
If your fine with O(n^2) time complexity, go with a nested loops. If you want the quickest possible solution, try using a hash table.
*hover to un-blur*
class BruteForceSolution: def twoSum(self, nums: List[int], target: int) -> List[int]: n = len(nums) for i in range(n - 1): for j in range(i + 1, n): if nums[i] + nums[j] == target: return [i, j] return [] # No solution foundclass Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: n = len(nums) for i in range(n - 1): for j in range(i + 1, n): if nums[i] + nums[j] == target: return [i, j] return [] # No solution found class HashTableSolution: def twoSum(self, nums: List[int], target: int) -> List[int]: numMap = {} n = len(nums) for i in range(n): complement = target - nums[i] if complement in numMap: return [numMap[complement], i] numMap[nums[i]] = i return [] # No solution found