# 1480. Running Sum of 1d Array

## 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 $O(1)$.

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

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

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


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

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


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


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