Tarkibga o'tish

1480 - Running Sum of 1d Array

🇬🇧 Inglizcha: Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).

Return the running sum of nums.

🇺🇿 O'zbekcha: nums array berilgan. Biz nums array yig'indisini quyidagicha hisoblab runningSum[i] = sum(nums[0]…nums[i]).

nums ning yig'indisi qaytarilsin.

🟢 Misol 1

$ Input: nums = [1,2,3,4]
$ Output: [1,3,6,10]
$ Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].

🟢 Misol 2

$ Input: nums = [1,1,1,1,1]
$ Output: [1,2,3,4,5]
$ Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].

🟢 Misol 3

$ Input: nums = [3,1,2,10,1]
$ Output: [3,4,6,16,17]

🔴 Cheklovlar

  • 1 <= nums.length <= 1000
  • -10^6 <= nums[i] <= 10^6

1⃣ - Yechim: LINK

Masala shartiga ko'ra nums[i] elementi o'zidan oldingi elementlar yig'indisiga tenglab chiqishimiz kerak. Shu sababli 1 dan n gacha sikl ochamiz. Sababi 0 dan boshlashimiz shart emas 0 - elementdan oldin boshqa qiymat yo'q va har bir yurishida biz nums[i] = nums[i] + nums[i - 1] ga tenglab boramiz.

Natija sifatida nums ni o'zini qaytarib yuboramiz.

⏰ Time Complexity: O(n - 1)
📦 Space Complexity: O(1)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
# Author: Abdulaminkhon Khaydarov
# Date: 06/11/22 
# Problem URL: https://leetcode.com/problems/running-sum-of-1d-array/

from typing import List


class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        for i in range(1, len(nums)):
            nums[i] = nums[i] + nums[i - 1]
        return nums


if __name__ == '__main__':
    solution = Solution()

    # Example 1
    print(solution.runningSum([1, 2, 3, 4]) == [1, 3, 6, 10])

    # Example 2
    print(solution.runningSum([1, 1, 1, 1, 1]) == [1, 2, 3, 4, 5])

    # Example 3
    print(solution.runningSum([3, 1, 2, 10, 1]) == [3, 4, 6, 16, 17])
Oxirgi yangilanish: November 13, 2022 09:15:23