Merge pull request #539 from sidparashar2001/issue#439

RunningSumOf1dArray

Author: smv1999

Arrays/RunningSumOf1dArray.cpp

@@ -0,0 +1,47 @@
+/*Question:-
+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].
+Constraints:
+
+1 <= nums.length <= 1000
+-10^6 <= nums[i] <= 10^6
+*/
+#include<bits/stdc++.h>
+using namespace std;
+vector<int> runningSum(vector<int>& nums) {
+        for(int i=1;i<nums.size();i++){
+            nums[i]+=nums[i-1];
+        }
+        return nums;//returning the array
+    }
+int main(){
+vector<int> nums;//to store elements
+int n,element;
+cin>>n;
+ for (int i = 0; i < n; i++)
+    {
+        cin>>element;
+       nums.push_back(element);//storeing the elements
+    }
+vector<int> result=runningSum(nums);//calling the fuction to return the running sum of array
+for (int i = 0; i < result.size(); i++)
+{
+    cout<<result[i]<<",";//print the elements of running sum array
+}
+
+return 0;
+}