MayLeetCoding Challenge 2021 — Day 3: Minimum Operations to Make Array Equal
Today, we will solve the 3rd problem of the May LeetCoding Challenge 2021.
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]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].
Input: nums = [1,1,1,1,1]
Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].
Input: nums = [3,1,2,10,1]
This is a pretty easy problem. Here we have to find the running sum of an array. That means, for index
i we have to find the sum of all elements from
i-1 . It can be done by using
sum variable to store the summation of each element index by index.
The code is given below.
Time complexity: O(n)
Space complexity: O(n)
where n is the size of the array
For the whole repository click the link below.