Running Sum of 1d Array

Author: anantkaushik

1480-running-sum-of-1d-array.py

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+"""
+Problem Link: https://leetcode.com/problems/running-sum-of-1d-array/
+
+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.
+
+Example 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].
+
+Example 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].
+
+Example 3:
+Input: nums = [3,1,2,10,1]
+Output: [3,4,6,16,17]
+ 
+Constraints:
+1 <= nums.length <= 1000
+-10^6 <= nums[i] <= 10^6
+"""
+class Solution:
+ def runningSum(self, nums: List[int]) -> List[int]:
+ for i in range(1, len(nums)):
+ nums[i] += nums[i-1]
+ 
+ return nums
+
+
+class Solution1:
+ def runningSum(self, nums: List[int]) -> List[int]:
+ cur_sum, res = 0, []
+ for num in nums:
+ cur_sum += num
+ res.append(cur_sum)
+ return res