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1. Two SumEasyArray · Hash Table
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 exactly one solution exists and you may not use the same element twice.
Example 1
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
// nums[0] + nums[1] = 9
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
- Only one valid answer exists
Test Cases
Case 1
Input: [2,7,11,15], 9
Got: []
Case 2
Input: [3,2,4], 6
Got: []
TSsolution.ts
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// Two Sum — LeetCode #1
// Return indices that sum to target
function twoSum(nums: number[], target: number): number[] {
// TODO: implement
// HashMap approach? Can't remember...
return [];
}
23:47 remaining// Time is running out...
> Brute force O(n²) — too slow
> HashMap approach — but how?
> Interviewer is watching. 😰
✗ FAILING (0/3)TypeScript
1. Two SumEasyArray · Hash Table
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 exactly one solution exists and you may not use the same element twice.
Example 1
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
// nums[0] + nums[1] = 9
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
- Only one valid answer exists
Test Cases
Case 1
Input: [2,7,11,15], 9
Got: [0,1]
Case 2
Input: [3,2,4], 6
Got: [1,2]
TSsolution.ts
Auto-typed by AnswerLabs1
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// O(n) single-pass HashMap solution
function twoSum(nums: number[], target: number): number[] {
const map = new Map<number, number>();
for (let i = 0; i < nums.length; i++) {
const complement = target - nums[i];
if (map.has(complement)) {
return [map.get(complement)!, i];
}
map.set(nums[i], i);
}
return [];
}
✓ SUCCESS (3/3)TypeScript
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