Left Number Twice Right

 Left Number Twice Right 

PROBLEM STATEMENT :

A set of N numbers (separated by one or more spaces) is passed as input to the program. The program must identify the count of numbers C where the number on the left is twice the number on the right. 

Boundary Conditions: 

3 <= N <= 50 

The value of the numbers can be from -99999999 to 99999999 

Input Format: The first line will contain the N numbers separated by one or more spaces. 

Output Format: The first line contains the value of C. 

Example Input/Output 1: 

Input: 10 20 5 40 15 -2 -30 -1 60 

Output: 2

Explanation: The numbers meeting the criteria are 20, -30 

Example Input/Output 2: 

Input: 5 90 10 2 5 -4 10 6 5 3 

Output: 3

Explanation: The numbers meeting the criteria are 2, 6, 5

SOLUTION :

C (Programming Language)

       
#include<stdio.h>
#include<stdlib.h>

int main()
{
    int a[50],n=-1,i,cnt=0;
    while(scanf("%d",&a[++n])==1); //scans the array
    for(i=1;i<n;++i) //Here n is the size of the array
    {
        if(a[i+1]*2==a[i-1]) cnt++;
    }
    printf("%d",cnt);
}
 

C++ (CPP)

       
#include <iostream>
 
using namespace std;

int main(int argc, char** argv)
{
    int a[50],n=-1,i,cnt=0;
    while(cin>>a[++n]);
    for(i=1;i<n-1;++i)
    {
        if(a[i+1]*2==a[i-1]) cnt++;
    }
    cout<<cnt;
}
 

JAVA

       
import java.util.*;
public class Hello {

    public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int[] a=new int[50];
		int n=0,i,cnt=0;
		while(sc.hasNext())
		{
		    a[n++]=sc.nextInt();
		}
		for(i=1;i<n-1;++i)
		{
		    if(a[i+1]*2==a[i-1]) cnt++;
		}
		System.out.println(cnt);
	}
}
 

PYTHON

       
l=list(map(int,input().split()))
print(len([i for i in range(len(l)-1) if l[i+1]*2==l[i-1]]))
 

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