Files
leetcode/docs/(01-linear-structures)/hash-table/(problems)/1_two-sum.mdx
T

66 lines
1.5 KiB
Plaintext

---
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"
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 []
```