LC 1 - Two Sum

Question

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(n^2) time complexity?

Links

Question here and solution here

Solution

concept

We will keep a hash map that stores value: index and we will check if the reminder is in the map. Take note that we will first check whether reminder is in the map or not, and then add the value into the map. This is to handle edge cases such as 2+2=4 since if we add in 2 first we would return True for the reminder since 2 is already in it. This is obviously wrong.

code

class Solution(object):
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        prevMap = {}  # val -> index

        for i, n in enumerate(nums):
            diff = target - n
            if diff in prevMap:
                return [prevMap[diff], i]
            prevMap[n] = i

Complexity

time: $O(n)$
space: $O(n)$