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.

sksaikia/LeetCode