Running Sum of 1d Array

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

🧠 Cpp

#include <numeric>

class Solution
{
public:
    vector<int> runningSum(vector<int>& nums)
    {
        vector<int> res;
//         res.reserve(nums.size());    
//         std::accumulate(nums.begin(), nums.end(), 0, [&res](int &a, int &b)
//                         {
//                             const int sum = a+b;
//                             res.emplace_back(sum);
//                             return sum;
//                         });
        std::partial_sum(nums.begin(), nums.end(), std::back_inserter(res));
        return res;
    }
};