1480. Running Sum of 1d Array

Problem

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

Problem Summary

Given an array nums. The formula for calculating the "running sum" of the array is: runningSum[i] = sum(nums[0]…nums[i]). Please return the running sum of nums.

Solution Approach

• Easy problem; just loop through in order according to the problem statement to calculate the prefix sum.

Code

package leetcode

func runningSum(nums []int) []int {
	dp := make([]int, len(nums)+1)
	dp[0] = 0
	for i := 1; i <= len(nums); i++ {
		dp[i] = dp[i-1] + nums[i-1]
	}
	return dp[1:]
}