LeetCode 1480. Running Sum of 1d Array Solution in Java, C++, Python & More | Explanation + Code

Description

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

Solutions

Solution 1: Prefix Sum

We directly traverse the array. For the current element nums[i], we add it with the prefix sum nums[i-1] to get the prefix sum nums[i] of the current element.

The time complexity is O(n), where n is the length of the array. The space complexity is O(1).

PythonJavaC++GoTypeScriptC#PHP

class Solution: def runningSum(self, nums: List[int]) -> List[int]: return list(accumulate(nums))(code-box)

class Solution { public int[] runningSum(int[] nums) { for (int i = 1; i < nums.length; ++i) { nums[i] += nums[i - 1]; } return nums; } }(code-box)

class Solution { public: vector<int> runningSum(vector<int>& nums) { for (int i = 1; i < nums.size(); ++i) nums[i] += nums[i - 1]; return nums; } };(code-box)

func runningSum(nums []int) []int { for i := 1; i < len(nums); i++ { nums[i] += nums[i-1] } return nums }(code-box)

function runningSum(nums: number[]): number[] { for (let i = 1; i < nums.length; ++i) { nums[i] += nums[i - 1]; } return nums; }(code-box)

public class Solution { public int[] RunningSum(int[] nums) { for (int i = 1; i < nums.Length; ++i) { nums[i] += nums[i - 1]; } return nums; } }(code-box)

class Solution { /** * @param Integer[] $nums * @return Integer[] */ function runningSum($nums) { for ($i = 1; $i < count($nums); $i++) { $nums[$i] += $nums[$i - 1]; } return $nums; } }(code-box)