Leetcode: Two Sum

Aug 2, 2025

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Inputs

Problem 1: Two Sum

Description

Given an array of integers nums and an integer target, return the indices of the two numbers such that they add up to target.

Each input is guaranteed to have exactly one solution.
You may not use the same element twice.
The answer can be returned in any order.

Examples

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.

Follow Up

Can you come up with an algorithm that is less than (O(n^2)) time complexity?

Steps

As you iterate over the array, keep adding elements to a list.
While considering a particular element, calculate the difference from target that is required.
If this difference is in the list, we have found a pair

Implementation

target_hashmap = {}

for i, num in enumerate(nums):
    difference = target - num
    if difference in target_hashmap:
        print("found")
    else:
        target_hashmap[num] = i

found

Submitted Code

from typing import List

def twoSum(nums: List[int], target: int) -> List[int]:
    target_hashmap = {}

    for i, num in enumerate(nums):
        difference = target - num
        if difference in target_hashmap:
            # print("found")
            return (i, target_hashmap[difference])
        else:
            target_hashmap[num] = i

nums = [2,7,11,15]
target = 9

twoSum(nums, target)

(1, 0)