README.md

1480. 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

Solutions (Ruby)

1. Prefix Sum

# @param {Integer[]} nums
# @return {Integer[]}
def running_sum(nums)
    for i in 1...nums.length
        nums[i] += nums[i - 1]
    end

    return nums
end

Solutions (Rust)

1. Prefix Sum

impl Solution {
    pub fn running_sum(nums: Vec<i32>) -> Vec<i32> {
        let mut running_sums = nums;

        for i in 1..running_sums.len() {
            running_sums[i] += running_sums[i - 1];
        }

        running_sums
    }
}