Given an array
nums. We define a running sum of an array as
runningSum[i] = sum(nums…nums[i]). Return the running sum of
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].
From the example shared above, let's compare the numbers present in
nums in the input and the output.
You can see here, that we update each index in
nums by adding the value present at the previous index i.e.
nums[i] = nums[i] + nums[i-1] where
i is the index starting from
1 to len(nums).
Let's now implement this approach.
This solution has a time complexity of
O(N) since we iterate over the entire array and has a space complexity of
O(1) since we use no additional space.
If you have any questions, feel free to drop a comment and we'll get back to you as soon as possible.