# MayLeetCoding Challenge 2021 — Day 3: Minimum Operations to Make Array Equal

Today, we will solve the 3rd problem of the May LeetCoding Challenge 2021.

# Problem Statement

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]

# Solution

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 `0`

to `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.