HackerRank Strange Counter Problem Solution

In this post, we will solve HackerRank Strange Counter Problem Solution.

There is a strange counter. At the first second, it displays the number 3. Each second, the
number displayed by decrements by 1 until it reaches 1. In next second, the timer resets to
2 × the initial number for the prior cycle and continues counting down. The diagram
below shows the counter values for each time t in the first three cycles:

Find and print the value displayed by the counter at time t.

Function Description

Complete the strangeCounter function in the editor below.

strangeCounter has the following parameter(s):

int t: an integer

Returns

int: the value displayed at time t

Input Format

A single integer, the value of t.

Sample Input

Sample Output

Explanation

Time t = 4 marks the beginning of the second cycle. It is double the number displayed at the beginning of the first cycle:2 x 3 = 6. This is shown in the diagram in the problem statement.

HackerRank Strange Counter Problem Solution

Strange Counter C Solution

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

int main() {

    long long unsigned int cnt, t, val, diff;
    
    val = cnt = t = diff = 0;
    scanf("%lld", &t);
    val = 3;
    cnt = 1;
    while(cnt <= t) {
        diff = val - 1;
        if ((cnt+diff) >= t) {
            printf ("%lld\n", val - (t - cnt));
            break;
        } else {
            cnt = cnt + diff + 1;
            val *= 2;
        }
    }
    
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    return 0;
}

Strange Counter C++ Solution

#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <sstream>
#include <cmath>
#include <set>
using namespace std;

#define REP(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define RREP(i,n) for(int (i)=(int)(n)-1;(i)>=0;(i)--)
#define REMOVE(Itr,n) (Itr).erase(remove((Itr).begin(),(Itr).end(),n),(Itr).end())
#define PB_VEC(Itr1,Itr2) (Itr1).insert((Itr1).end(),(Itr2).begin(),(Itr2).end())
#define UNIQUE(Itr) sort((Itr).begin(),(Itr).end()); (Itr).erase(unique((Itr).begin(),(Itr).end()),(Itr).end())

typedef long long ll;

int main(){
    
    ll t; cin>>t;
    
    ll ans=1;
    ll sum=0;
    while(true){
        if(sum+ans*3>=t)break;
        sum+=ans*3;
        ans*=2;
    }
    
    ll cnt=t-sum;
    
    cout<<(ans+1)*3-cnt-2<<endl;
    
    return 0;
}

Strange Counter C Sharp Solution

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
class Solution {

    private const int start = 3; 
        static void Main(string[] args)
        {
            long t = Convert.ToInt64(Console.ReadLine());
            var point = GetStartPoint(t);
            var distance = t + 2 - point;
            var res = point - distance; 
            Console.WriteLine(res);
        }

        private static long GetStartPoint(long time) {
            long res = start;
            var compare = time + 2;
            while (true)
            {                
                if ((compare >= res) && (compare < res * 2)) {
                    break;
                }
                res = res * 2;
            }
            
            return res;
        }
}

Strange Counter Java Solution

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {

    /*
     * Complete the 'strangeCounter' function below.
     *
     * The function is expected to return a LONG_INTEGER.
     * The function accepts LONG_INTEGER t as parameter.
     */

    public static long strangeCounter(long t) {
        long v=4;

        while(v<=t)
        {
            v=v*2+2;
        }
        return v-t;
    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        long t = Long.parseLong(bufferedReader.readLine().trim());

        long result = Result.strangeCounter(t);

        bufferedWriter.write(String.valueOf(result));
        bufferedWriter.newLine();

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Strange Counter JavaScript Solution


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 t = parseInt(readLine());
    console.log(kanter(t))
}

__find_kth_pivot_cache = []
function find_kth_pivot(k) {
	if (k==0) {
       return 1
	}
	if (__find_kth_pivot_cache[k]) {
		return __find_kth_pivot_cache[k]
	}
	var retVal = find_kth_pivot(k-1) + 3*Math.pow(2,k-1)
	__find_kth_pivot_cache[k] = retVal
	return retVal
}

//returns m <= n
function find_k_m_lte_n(n) {
   var k = 0;
   var m = find_kth_pivot(0);
   while(m <= n) {
   	k++
   	m = find_kth_pivot(k)

   	if (m > n) {
   	  k--
   	  m = find_kth_pivot(k);
   	  break
   	}
   };
   var retVal = {
   	m: m,
   	k: k
   };

   // console.log(n + '=>' + retVal.m + ',' + retVal.k)
   return retVal;
}

function calculate_pivot_value(k) {
	if (k==0) {
		return 3
	} else {
		return 2 * calculate_pivot_value(k-1)
	}
}

function kanter(n) {

  var k_and_m = find_k_m_lte_n(n);
  var k = k_and_m.k
  var m = k_and_m.m

  var val_m = calculate_pivot_value(k)
  retVal = val_m - (n - m)

  //console.log('n', n, 'm',m)

  //console.log('n = ', n)
  //console.log('f(m) = ' + val_m)
  //console.log('n = ', n, 'f(n) = ' + retVal)

  return retVal
}




/*
kanter(1)
kanter(2)
kanter(3)
kanter(4)
kanter(5)
kanter(6)
kanter(7)
kanter(8)
kanter(9)
kanter(10)
kanter(20)
kanter(21)
kanter(22)
*/

Strange Counter Python Solution

tf = int(input())

t = []

t.append(1)
t.append(4)
t.append(10)
t.append(22)
t.append(46)
t.append(94)
t.append(190)
t.append(382)
t.append(766)
t.append(1534)
t.append(3070)
t.append(6142)
t.append(12286)
t.append(24574)
t.append(49150)
t.append(98302)
t.append(196606)
t.append(393214)
t.append(786430)
t.append(1572862)
t.append(3145726)
t.append(6291454)
t.append(12582910)
t.append(25165822)
t.append(50331646)
t.append(100663294)
t.append(201326590)
t.append(402653182)
t.append(805306366)
t.append(1610612734)
t.append(3221225470)
t.append(6442450942)
t.append(12884901886)
t.append(25769803774)
t.append(51539607550)
t.append(103079215102)
t.append(206158430206)
t.append(412316860414)
t.append(824633720830)
t.append(1000000000001)

nv = next(x[0] for x in enumerate(t) if x[1] > tf) - 1
#print(nv)
sum = 2*t[nv] + 2 - tf
print(sum)