Solving Problem: Two Sum

(Python)

Problem Statement:

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?

Solution:

1. [Naive Approach] Generating all Possible Pairs – O(n²) time and O(1) space

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        for i in range(0, len(nums)):
            for j in range(i+1, len(nums)):
                if nums[i] + nums[j] == target:
                    two_sum = [i,j]
                    break
        return two_sum

2. [Expected Approach] Using Hash Set – O(n) time and O(n) space

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        s = set()
        for index, num in enumerate(nums):
            complement = target - num
            if complement in s:
                return [next(index for index,num in enumerate(nums) if num == complement), index]
            s.add(num)
        return []

Reference: Two Sum – LeetCode