06_Running_sum.cpp

// 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]








class Solution {
public:
    vector<int> runningSum(vector<int>& nums) {
        int sum=0;
        int n=nums.size();
        vector<int> arr;
        for(int i=0;i<nums.size();i++){
            sum=sum+nums[i];
            arr.push_back(sum);
        }
        return arr;
    }
};