###### tags: `LeetCode Study Group - Array` `LeetCode Easy` `LeetCode Study Group`
# 1480. Running Sum of 1d Array
### Easy
Link: https://leetcode.com/problems/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

## Solution-1

Time complexity: O(n*n)

class Solution {
    public int[] runningSum(int[] nums) {
        int[] results = new int[nums.length];
        for(int i = 0; i < results.length; i++) {
            int sum = 0;
            for(int j = 0; j <= i; j++){
                sum = sum + nums[j];
            }
            results[i] = sum;
        }
        return results;
    }
}

This is my first solution.
Although I solve it, it is brute force solotion.
I use for loop twice to calculate the sum of the nums array.

## Solution-2

Time complexity: O(n)

Evolve from brute force to optimal​. The better solution is adding the value of previous index in array.

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