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

Solution

Vector - O(n) time - O(n) space;

class Solution {
public:
    vector<int> runningSum(vector<int>& nums) {
        int currentSum = 0;
        vector<int> result(nums.size(), 0);
        for (int i = 0; i < nums.size(); i++) {
            currentSum += nums[i];
            result[i]= currentSum;
        } 
		return result;
    }
};

Overwritten Input - O(n) time - O(1) space

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