HackerRank Drawing Book problem solution

In this Drawing Book problem you have Given n and p, find and print the minimum number of pages that must be turned in in order to arrive at page p.

Problem solution in Python programming.

#!/bin/python3

import sys


n = int(input().strip())
p = int(input().strip())
# your code goes here

page_in_book = p//2
total_pages = n//2

from_front = page_in_book
from_back = total_pages - page_in_book
print(min(from_front,from_back))

Problem solution in Java Programming.

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int p = in.nextInt();
        int frontdist = p / 2;
        int backdist = n % 2 == 0 ? (n - p + 1) / 2 : (n - p) / 2;
        System.out.println(Math.min(frontdist, backdist));
        // your code goes here
    }
}

Problem solution in C++ programming.

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;


int main(){
    int n;
    cin >> n;
    int p;
    cin >> p;
    int min_left = 0;
    int cur_left = 0, cur_right = 1;
    while (cur_left != p && cur_right != p) {
        ++min_left;
        cur_left += 2;
        cur_right += 2;
    }
    int min_right = 0;
    cur_left = n % 2 == 0 ? n : n - 1;
    cur_right = cur_left + 1;
    
    while (cur_left != p && cur_right != p) {
        ++min_right;
        cur_left -= 2;
        cur_right -= 2;
    }
    
    cout << std::min(min_left, min_right);
    return 0;
}

Problem solution in C programming.

#include <stdio.h>
int main()
    {
    int n,p,j=0,a[100000];
    scanf("%d",&n);
    scanf("%d",&p);
    for (int i=0;i<=n;i++)
        if (i%2==0)
        {
        j++;
        a[i]=j;
        a[i+1]=j;
    }
    if (a[p]-1<j-a[p]) printf("%d",a[p]-1);
    else printf("%d",j-a[p]);
    return 0;
}

Problem solution in JavaScript programming.

process.stdin.resume();
process.stdin.setEncoding('ascii');

var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;

process.stdin.on('data', function (data) {
    input_stdin += data;
});

process.stdin.on('end', function () {
    input_stdin_array = input_stdin.split("n");
    main();    
});

function readLine() {
    return input_stdin_array[input_currentline++];
}

/////////////// ignore above this line ////////////////////

function main() {
  var n = parseInt(readLine());
  var p = parseInt(readLine());
  var count = 0;
  var page = 1;
  while(page < p){
    count++;
    page += 2;
  }
  if(n % 2 !== 0){
    page = n - 1;
  } else {
    page = n;
  }
  var countTwo = 0;
  while(page > p){
    countTwo++;
    page -= 2;
  }
  if(count < countTwo){
    console.log(count);
  } else {
    console.log(countTwo);
  }
}