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.

HackerRank Drawing Book problem solution


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);
  }
}