Leetcode - 1. Two Sum

Easy

May 23, 2025

#arrays

#hashing

1. Two Sum

Problem

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?

Approach

We can use a hashmap to solve this problem. The idea is that we want to check if there is another number in the array that can be added to the current number to reach the target. Therefore, we can iterate over the numbers and check if the complement (target - number) exists in the hashmap, if it does, we return the two indices. Otherwise, we add the number to the hashmap with the index as its value. Since it's guranteed that there is a solution, we don't need to worry about what to return if the iteration completes (it never will).

Complexity

Time: O(n)

We only iterate over the input array once

Space: O(n)

We create a hash map that could, at most, contain all the numbers

Solution

func twoSum(nums []int, target int) []int { 
	numMap := make(map[int]int)
	
	for i, num := range nums {
		if idx, ok := numMap[target - num]; ok {
			return []int{idx, i}
		}
		numMap[num] = i
	}
	return []int{}
}