Vasily Guzov 1 year ago
parent 8dcb175459
commit a54b5b8996

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import { twoSum } from "./twoSum";
export default twoSum;

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# [twoSum](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted)
## описание задачи
Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length.
Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2.
The tests are generated such that there is exactly one solution. You may not use the same element twice.
Your solution must use only constant extra space.
Example 1:
- Input: numbers = [2,7,11,15], target = 9
- Output: [1,2]
- Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2].
Example 2:
- Input: numbers = [2,3,4], target = 6
- Output: [1,3]
- Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].
Example 3:
- Input: numbers = [-1,0], target = -1
- Output: [1,2]
- Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].
Constraints:
2 <= numbers.length <= 3 * 104
-1000 <= numbers[i] <= 1000
numbers is sorted in non-decreasing order.
-1000 <= target <= 1000
The tests are generated such that there is exactly one solution.
## тест кейсы
```ts
describe("twoSum", () => {
it("[2,7,11,15] 9 => [1,2]", () => {
expect(twoSum([2, 7, 11, 15], 9)).toEqual([1, 2]);
});
it("[2,3,4] 6 => [1,3]", () => {
expect(twoSum([2, 3, 4], 6)).toEqual([1, 3]);
});
it("[-1, 0] -1 => [1, 2]", () => {
expect(twoSum([-1, 0], -1)).toEqual([1, 2]);
});
it("[5,25,75] 100 => [2,3]", () => {
expect(twoSum([5, 25, 75], 100)).toEqual([2, 3]);
})
});
```
## текстовое описание решения
- заводим два указателя, в начале и конце
- в цикле получаем по указателям значение и складываем
- если сумма больше таргет двигаем правый указатель
- если сумма меньше таргет двигаем левый указатель
- если сумма равна таргет возвращаем значение указателей +1
- если указатели сошлись а сумма не равна таргет выходим из цикла и возвращаем [-1, -1]
## ассимптотическая оценка
| Description | Estimation |
| ----------- | ---------- |
| time: | O(n) |
| mem: | O(n) |
## time
| Description | Time |
| ------------------------------------------- | ----- |
| анализ и сбор информации | 11:50 |
| обдумываение решения и формулировка решения | 08:47 |
| имплементация | 20:28 |
| исправление ошибок | 20:00 |
| полное время затраченое на решение | 01:01:07 |
## журнал ошибок
- не разобрался с правилами для движения указателя, какой в какой момент двигать
- не учел условие возврата индексов + 1
## code
### typescript
```ts
function twoSum(numbers: number[], target: number): number[] {
let L = 0;
let R = numbers.length - 1;
while (L <= R) {
const curSum = numbers[L] + numbers[R];
if (curSum === target) return [L + 1, R + 1];
curSum > target ? R-- : L++;
}
return [-1, -1];
};
```

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import { describe, it, expect } from "vitest";
import { twoSum } from "./twoSum.ts";
describe("twoSum", () => {
it("[2,7,11,15] 9 => [1,2]", () => {
expect(twoSum([2, 7, 11, 15], 9)).toEqual([1, 2]);
});
it("[2,3,4] 6 => [1,3]", () => {
expect(twoSum([2, 3, 4], 6)).toEqual([1, 3]);
});
it("[-1, 0] -1 => [1, 2]", () => {
expect(twoSum([-1, 0], -1)).toEqual([1, 2]);
});
it("[5,25,75] 100 => [2,3]", () => {
expect(twoSum([5, 25, 75], 100)).toEqual([2, 3]);
})
});

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export function twoSum(numbers: number[], target: number): number[] {
let L = 0;
let R = numbers.length - 1;
while (L <= R) {
const curSum = numbers[L] + numbers[R];
if (curSum === target) return [L + 1, R + 1];
curSum > target ? R-- : L++;
}
return [-1, -1];
};
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