---
title: '1. Two Sum'
description: You are given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target
icon: dot
topics:
  - { name: "Array", slug: "array" }
  - { name: "Hash Table", slug: "hash-table" }
video: "KLlXCFG5TnA"
issue: "https://github.com/prdlk/leetcode/issues/29"
companies:
  - { name: "Amazon", period: "Thirty Days", frequency: 100 }
  - { name: "Google", period: "Thirty Days", frequency: 100 }
  - { name: "Meta", period: "Thirty Days", frequency: 100 }
  - { name: "Microsoft", period: "Thirty Days", frequency: 100 }
  - { name: "tcs", period: "Thirty Days", frequency: 79.6 }
  - { name: "Ola Cabs", period: "Thirty Days", frequency: 76.7 }
  - { name: "Bloomberg", period: "Thirty Days", frequency: 69.8 }
  - { name: "Infosys", period: "Thirty Days", frequency: 68.6 }
  - { name: "Altimetrik", period: "Three Months", frequency: 100 }
  - { name: "Capgemini", period: "Three Months", frequency: 100 }
  - { name: "Microstrategy", period: "Three Months", frequency: 100 }
  - { name: "MongoDB", period: "Three Months", frequency: 100 }
  - { name: "Cognizant", period: "Three Months", frequency: 88.7 }
  - { name: "Apple", period: "Three Months", frequency: 80.4 }
  - { name: "Barclays", period: "Six Months", frequency: 100 }
  - { name: "Cisco", period: "Six Months", frequency: 100 }
  - { name: "Deloitte", period: "Six Months", frequency: 100 }
  - { name: "Expedia", period: "Six Months", frequency: 100 }
  - { name: "NetApp", period: "Six Months", frequency: 100 }
  - { name: "Samsung", period: "Six Months", frequency: 100 }
  - { name: "Sony", period: "Six Months", frequency: 100 }
  - { name: "Oracle", period: "Six Months", frequency: 91.2 }
  - { name: "Qualcomm", period: "Six Months", frequency: 80.7 }
  - { name: "IBM", period: "Six Months", frequency: 77.2 }
  - { name: "eBay", period: "Six Months", frequency: 75.2 }
  - { name: "Walmart Labs", period: "Six Months", frequency: 70.5 }
  - { name: "Flipkart", period: "Six Months", frequency: 57.1 }
  - { name: "Goldman Sachs", period: "Six Months", frequency: 44.8 }
sidebar:
  label: 'Two Sum'
  badge: 'Easy'
---

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

### 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.

## Approach

```mermaid
flowchart TD
  S(["twoSum(nums, target)"]) --> I["seen = {} — value to index"]
  I --> L{"more (i, num) in nums?"}
  L -- no --> E(["return []"])
  L -- yes --> C["complement = target - num"]
  C --> H{"complement in seen?"}
  H -- yes --> R(["return [seen[complement], i]"])
  H -- no --> W["seen[num] = i"]
  W --> L
```

## Solution

```py
class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        seen = {}
        for i, num in enumerate(nums):
            complement = target - num
            if complement in seen:
                return [seen[complement], i]
            seen[num] = i
        return []
```