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Sourav
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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].
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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].
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Example 3:

**Input:** nums = [3,1,2,10,1]
**Output:** [3,4,6,16,17]
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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.

LeetCode

I have been solving Leetcode problems for around a year or so. Suddenly I have developed a passion of writing tutorials for these problems. I am starting with Leetcode problem, in future I will try to make tutorials on Spring,Android,Java,Algorithms and many more.

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The following table contains all the problems with their respective solution tutorials. I will try to add more articles to it soon.

March LeetCoding Challenge

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