Running time of algorithms hackerrank solution in python

Hello Programmers/Coders, Today we are going to share solutions of Programming problems of HackerRank, Algorithm Solutions of Problem Solving Section in Java. At Each Problem with Successful submission with all Test Cases Passed, you will get an score or marks. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for Running Time of Algorithms in Java-HackerRank Problem. We are providing the correct and tested solutions of coding problems present on HackerRank. If you are not able to solve any problem, then you can take help from our Blog/website.

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Introduction To Algorithm

The word Algorithm means “a process or set of rules to be followed in calculations or other problem-solving operations”. Therefore Algorithm refers to a set of rules/instructions that step-by-step define how a work is to be executed upon in order to get the expected results. 

Advantages of Algorithms:

• It is easy to understand.
• Algorithm is a step-wise representation of a solution to a given problem.
• In Algorithm the problem is broken down into smaller pieces or steps hence, it is easier for the programmer to convert it into an actual program.

Link for the Problem – Running Time of Algorithms – Hacker Rank Solution

Running Time of Algorithms – Hacker Rank Solution

Problem:

In a previous challenge you implemented the Insertion Sort algorithm. It is a simple sorting algorithm that works well with small or mostly sorted data. However, it takes a long time to sort large unsorted data. To see why, we will analyze its running time.

Running Time of Algorithms
The running time of an algorithm for a specific input depends on the number of operations executed. The greater the number of operations, the longer the running time of an algorithm. We usually want to know how many operations an algorithm will execute in proportion to the size of its input, which we will call .

What is the ratio of the running time of Insertion Sort to the size of the input? To answer this question, we need to examine the algorithm.

Analysis of Insertion Sort
For each element  in an array of  numbers, Insertion Sort compares the number to those to its left until it reaches a lower value element or the start. At that point it shifts everything to the right up one and inserts  into the array.

How long does all that shifting take?

In the best case, where the array was already sorted, no element will need to be moved, so the algorithm will just run through the array once and return the sorted array. The running time would be directly proportional to the size of the input, so we can say it will take  time.

However, we usually focus on the worst-case running time (computer scientists are pretty pessimistic). The worst case for Insertion Sort occurs when the array is in reverse order. To insert each number, the algorithm will have to shift over that number to the beginning of the array. Sorting the entire array of  numbers will therefore take  operations, which is  (almost ). Computer scientists just round that up (pick the dominant term) to  and say that Insertion Sort is an “ time” algorithm.

What this means
The running time of the algorithm against an array of  elements is . For  elements, it will be . Insertion Sort can work well for small inputs or if you know the data is likely to be nearly sorted, like check numbers as they are received by a bank. The running time becomes unreasonable for larger inputs.

Challenge
Can you modify your previous Insertion Sort implementation to keep track of the number of shifts it makes while sorting? The only thing you should print is the number of shifts made by the algorithm to completely sort the array. A shift occurs when an element’s position changes in the array. Do not shift an element if it is not necessary.

Function Description

Complete the runningTime function in the editor below.

runningTime has the following parameter(s):

• int arr[n]: an array of integers

Returns

• int: the number of shifts it will take to sort the array

Input Format

The first line contains the integer , the number of elements to be sorted.
The next line contains  integers of .

Constraints

Sample Input

STDIN Function ----- -------- 5 arr[] size n =5 2 1 3 1 2 arr = [2, 1, 3, 1, 2]

Sample Output

4

Explanation

Iteration Array Shifts 0 2 1 3 1 2 1 1 2 3 1 2 1 2 1 2 3 1 2 0 3 1 1 2 3 2 2 4 1 1 2 2 3 1 Total 4 Running Time of Algorithms – Hacker Rank Solution import java.util.Scanner; /** * @author Techno-RJ * */ public class RunningTimeOfAlgorithms { public static int shiftCount(int[] a) { int count = 0; for (int i = 0; i < a.length; i++) { for (int j = i; j > 0; j--) { if (a[j] < a[j - 1]) { int temp = a[j - 1]; a[j - 1] = a[j]; a[j] = temp; count++; } } } return count; } public static void main(String[] args) { Scanner sc = new Scanner(System.in); int N = sc.nextInt(); int[] a = new int[N]; for (int i = 0; i < N; i++) { a[i] = sc.nextInt(); } System.out.println(shiftCount(a)); sc.close(); } }

Hello coders, today we are going to solve Day 25: Running Time and Complexity HackerRank Solution in C++, Java and Python.

Objective

Today we will learn about running time, also known as time complexity.

Task

A prime is a natural number greater than 1 that has no positive divisors other than 1 and itself. Given a number, n, determine and print whether it is Prime or Not prime.

Note: If possible, try to come up with a O(n1/2) primality algorithm, or see what sort of optimizations you come up with for an  algorithm. Be sure to check out the Editorial after submitting your code.

Input Format

The first line contains an integer, T, the number of test cases.
Each of the T subsequent lines contains an integer, n, to be tested for primality.

Constraints

• 1 <= T <= 30
• 1 <= n <= 2 x 109

Output Format

For each test case, print whether n is Prime or Not Prime on a new line.

Sample Input

3 12 5 7

Sample Output

Not prime Prime Prime

Explanation

Test Case 0: n = 12.
12 is divisible by numbers other than 1 and itself (i.e.: 2, 3, 4, 6), so we print Not Prime on a new line.

Test Case 1: n = 5.
5 is only divisible 1 and itself, so we print Prime on a new line.

Test Case 2: n = 7.
7 is only divisible 1 and itself, so we print Prime on a new line.

Solution – Day 25: Running Time and Complexity

C++

#include #include #include #include #include using namespace std; int main() { /* Enter your code here. Read input from STDIN. Print output to STDOUT */ int T; cin >> T; for (size_t t = 0 ; t < T ; ++t) { int n; bool prime = true; cin >> n; if (n > 1) { for (size_t i = pow(M_E, log(n)/2) ; i > 1 ; --i) { if ( !(n % i) ) { prime = false; break; } } } else { prime = false; } if ( prime ) { cout << "Prime" << endl; } else { cout << "Not prime" << endl; } } return 0; }

Java

import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { static void prime(int n) { boolean flag=false; for(int i=2;i<=Math.sqrt(n);i++) { if(n%i==0) { flag=true; break; } } if(n==1){ System.out.println("Not prime"); } else if(!flag){ System.out.println("Prime"); } else{ System.out.println("Not prime"); } } public static void main(String args[]) { Scanner sc=new Scanner(System.in); int t=sc.nextInt(); while(t--!=0) { int n=sc.nextInt(); prime(n); } } }

Python

from math import sqrt from sys import stdin def checkPrime(n): for i in range(2, int(sqrt(n))+1): if n % i is 0: return False return True n = int(input()) for line in stdin: val = int(line) if (val >= 2 and checkPrime(val)): print("Prime") else: print("Not prime")

Disclaimer: The above Problem (Day 25: Running Time and Complexity) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.