1. Two Sum

EasyLeetCode

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 <= 10^4
• -10^9 <= nums[i] <= 10^9
• -10^9 <= target <= 10^9
• Only one valid answer exists.

How to solve the problem

• Brute Force

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

• HashMap

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        saveDict = {}
        for i, num in enumerate(nums):
            remainder = target - num
            if remainder in saveDict:
                return [saveDict[remainder], i]
            saveDict[num] = i

Complexity

• Brute Force: Time complexity: O(n^2), Space complexity: O(1)
• HashMap: Time complexity: O(n), Space complexity: O(n)