LeetCode 1: Two Sum

Date: January 15, 2026
Category: Arrays | HashMap
Time Taken: 15 minutes
Difficulty: Easy

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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.

Link: Two Sum

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My Approach

Initial thought:

• The most brute-force solution is to use 2 for loops to compare each pair of values and check if they add up to the target. Its time complexity would be O(N²) and space complexity would be O(1).

• The second solution I considered was to sort the array and use the two-pointer method to compare the sum of the maximum and minimum values to the target. However, this won't work because we need to return their original indices.

• The third solution is to map the values to their indices and use a for loop to check if the difference between the target and the current value exists in the hashmap. This solution will have a space and time complexity of O(N) for both.

Final solution:

• First, create a HashMap to store the values and their indices.

• Use a for loop with enumerate to get the indices and values of the array.

• Check if diff = target - currValue is in the HashMap. If it is, return the current index and the index of the value in the HashMap.

• If it's not there, check if the target is less than the sum; if so, decrement the right pointer. If the target is greater than the sum, increment the left pointer.

Why this works:
This works because we only need to compare the target with the difference and store all the values in the hashmap that currently don't match our difference.

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Solution Code

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        hashmap = {}

        for currIndex, currValue in enumerate(nums):
            diff = target - currValue
            if diff in hashmap:
                return [currIndex, hashmap[diff]]
            else:
                hashmap[currValue] = currIndex

Complexity:

Time: O(N) Space: O(N)

Key Takeaway: Using enumerate makes it easy to get both the index and value from an array.

Pattern: Hashing.

Mistakes I Made: I did not read the complete problem and rushed to solve it, which resulted in time being lost.

Series: 90 Days of Data Engineering Progress: 6/90 problems completed

Tags: #DEQuest #LeetCode #SQL #DataEngineering #BuildInPublic