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Find n’th number in a number system with only 3 and 4

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  • Difficulty Level : Medium
  • Last Updated : 07 Jul, 2022
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Given a number system with only 3 and 4. Find the nth number in the number system. First few numbers in the number system are: 3, 4, 33, 34, 43, 44, 333, 334, 343, 344, 433, 434, 443, 444, 3333, 3334, 3343, 3344, 3433, 3434, 3443, 3444, …
Source: Zoho Interview

We can generate all numbers with i digits using the numbers with (i-1) digits. The idea is to first add a ‘3’ as prefix in all numbers with (i-1) digit, then add a ‘4’. For example, the numbers with 2 digits are 33, 34, 43 and 44. The numbers with 3 digits are 333, 334, 343, 344, 433, 434, 443 and 444 which can be generated by first adding a 3 as prefix, then 4. 
Following are detailed steps. 

1) Create an array 'arr[]' of strings size n+1. 
2) Initialize arr[0] as empty string. (Number with 0 digits)
3) Do following while array size is smaller than or equal to n
.....a) Generate numbers by adding a 3 as prefix to the numbers generated 
        in previous iteration.  Add these numbers to arr[]
.....a) Generate numbers by adding a 4 as prefix to the numbers generated 
        in previous iteration. Add these numbers to arr[]

Thanks to kaushik Lele for suggesting this idea in a comment here. Following is C++ implementation for the same.  

C++




// C++ program to find n'th number
// in a number system with
// only 3 and 4
#include <iostream>
using namespace std;
 
// Function to find n'th number
// in a number system with only
// 3 and 4
void find(int n)
{
   
    // An array of strings to
    // store first n numbers. arr[i]
    // stores i'th number
    string arr[n + 1];
   
    // arr[0] stores the empty string (String
    // with 0 digits)
    arr[0] = "";
 
    // size indicates number of
    // current elements in arr[]. m
    // indicates number of elements
    // added to arr[] in
    // previous iteration.
    int size = 1, m = 1;
 
    // Every iteration of following
    // loop generates and adds
    // 2*m numbers to arr[] using 
    // the m numbers generated in
    // previous iteration.
    while (size <= n) {
       
        // Consider all numbers added
        // in previous iteration,
        // add a prefix "3" to them and
        // add new numbers to
        // arr[]
        for (int i = 0; i < m && (size + i) <= n; i++)
            arr[size + i] = "3" + arr[size - m + i];
 
        // Add prefix "4" to numbers
        // of previous iteration
        // and add new numbers to arr[]
        for (int i = 0; i < m && (size + m + i) <= n; i++)
            arr[size + m + i] = "4" + arr[size - m + i];
 
        // Update no. of elements added in previous
        // iteration
        m = m << 1; // Or m = m*2;
 
        // Update size
        size = size + m;
    }
    cout << arr[n] << endl;
}
 
// Driver program to test above functions
int main()
{
    for (int i = 1; i < 16; i++)
        find(i);
    return 0;
}

Java




// Java program to find n'th number in a number system with
// only 3 and 4
import java.io.*;
 
class GFG {
    // Function to find n'th number in a number system with
    // only 3 and 4
    static void find(int n)
    {
        // An array of strings to store first n numbers.
        // arr[i] stores i'th number
        String[] arr = new String[n + 1];
 
        // arr[0] stores the empty string (String with 0
        // digits)
        arr[0] = "";
 
        // size indicates number of current elements in
        // arr[], m indicates number of elements added to
        // arr[] in previous iteration
        int size = 1, m = 1;
 
        // Every iteration of following loop generates and
        // adds 2*m numbers to arr[] using the m numbers
        // generated in previous iteration
        while (size <= n) {
            // Consider all numbers added in previous
            // iteration, add a prefix "3" to them and add
            // new numbers to arr[]
            for (int i = 0; i < m && (size + i) <= n; i++)
                arr[size + i] = "3" + arr[size - m + i];
 
            // Add prefix "4" to numbers of previous
            // iteration and add new numbers to arr[]
            for (int i = 0; i < m && (size + m + i) <= n;
                 i++)
                arr[size + m + i] = "4" + arr[size - m + i];
 
            // Update no. of elements added in previous
            // iteration
            m = m << 1; // Or m = m*2;
 
            // Update size
            size = size + m;
        }
        System.out.println(arr[n]);
    }
 
    // Driver program
    public static void main(String[] args)
    {
        for (int i = 0; i < 16; i++)
            find(i);
    }
}
 
// Contributed by Pramod Kumar

Python3




# Python3 program to find n'th
# number in a number system
# with only 3 and 4
 
# Function to find n'th number in a
# number system with only 3 and 4
def find(n):
     
    # An array of strings to store
    # first n numbers. arr[i] stores
    # i'th number
    arr = [''] * (n + 1);
     
    # arr[0] = ""; # arr[0] stores
    # the empty string (String with 0 digits)
 
    # size indicates number of current
    # elements in arr[]. m indicates
    # number of elements added to arr[]
    # in previous iteration.
    size = 1;
    m = 1;
 
    # Every iteration of following
    # loop generates and adds 2*m
    # numbers to arr[] using the m
    # numbers generated in previous
    # iteration.
    while (size <= n):
         
        # Consider all numbers added
        # in previous iteration, add
        # a prefix "3" to them and
        # add new numbers to arr[]
        i = 0;
        while(i < m and (size + i) <= n):
            arr[size + i] = "3" + arr[size - m + i];
            i += 1;
 
        # Add prefix "4" to numbers of
        # previous iteration and add
        # new numbers to arr[]
        i = 0;
        while(i < m and (size + m + i) <= n):
            arr[size + m + i] = "4" + arr[size - m + i];
            i += 1;
 
        # Update no. of elements added
        # in previous iteration
        m = m << 1; # Or m = m*2;
 
        # Update size
        size = size + m;
    print(arr[n]);
 
# Driver Code
for i in range(1, 16):
    find(i);
 
# This code is contributed by mits

C#




// C# program to find n'th number in a
// number system with only 3 and 4
using System;
 
class GFG {
     
    // Function to find n'th number in a
    // number system with only 3 and 4
    static void find(int n)
    {
         
        // An array of strings to store first
        // n numbers. arr[i] stores i'th number
        String[] arr = new String[n + 1];
         
        // arr[0] stores the empty string
        // (String with 0 digits)
        arr[0] = "";
 
        // size indicates number of current
        // elements in arr[], m indicates
        // number of elements added to arr[]
        // in previous iteration
        int size = 1, m = 1;
 
        // Every iteration of following loop
        // generates and adds 2*m numbers to
        // arr[] using the m numbers generated
        // in previous iteration
        while (size <= n)
        {
            // Consider all numbers added in
            // previous iteration, add a prefix
            // "3" to them and add new numbers
            // to arr[]
            for (int i = 0; i < m &&
                             (size + i) <= n; i++)
                              
                arr[size + i] = "3" +
                               arr[size - m + i];
 
            // Add prefix "4" to numbers of
            // previous iteration and add new
            // numbers to arr[]
            for (int i = 0; i < m &&
                          (size + m + i) <= n; i++)
                           
                arr[size + m + i] = "4" +
                                  arr[size - m + i];
 
            // Update no. of elements added
            // in previous iteration
            m = m << 1; // Or m = m*2;
 
            // Update size
            size = size + m;
        }
         
        Console.WriteLine(arr[n]);
    }
     
    // Driver program
    public static void Main ()
    {
        for (int i = 0; i < 16; i++)
            find(i);
    }
}
 
// This code is contributed by Sam007.

PHP




<?php
// PHP program to find n'th
// number in a number system
// with only 3 and 4
 
// Function to find n'th number in a
// number system with only 3 and 4
function find($n)
{
    // An array of strings to store
    // first n numbers. arr[i] stores
    // i'th number
    $arr = array_fill(0, $n + 1, "");
     
    // $arr[0] = ""; // arr[0] stores
    // the empty string (String with 0 digits)
 
    // size indicates number of current
    // elements in arr[]. m indicates
    // number of elements added to arr[]
    // in previous iteration.
    $size = 1;
    $m = 1;
 
    // Every iteration of following
    // loop generates and adds 2*m
    // numbers to arr[] using the m
    // numbers generated in previous
    // iteration.
    while ($size <= $n)
    {
        // Consider all numbers added
        // in previous iteration, add
        // a prefix "3" to them and
        // add new numbers to arr[]
        for ($i = 0; $i < $m &&
            ($size + $i) <= $n; $i++)
            $arr[$size + $i] = "3" .
            $arr[$size - $m + $i];
 
        // Add prefix "4" to numbers of
        // previous iteration and add
        // new numbers to arr[]
        for ($i = 0; $i < $m &&
            ($size + $m + $i) <= $n; $i++)
            $arr[$size + $m + $i] = "4" .
            $arr[$size - $m + $i];
 
        // Update no. of elements added
        // in previous iteration
        $m = $m << 1; // Or m = m*2;
 
        // Update size
        $size = $size + $m;
    }
    echo $arr[$n] . "\n";
}
 
// Driver Code
for ($i = 1; $i < 16; $i++)
    find($i);
 
// This code is contributed by mits
?>

Javascript




<script>
// javascript program to find n'th number in a number system with
// only 3 and 4
 
   // Function to find n'th number in a number system with
    // only 3 and 4
function find(n)
{
 
    // An array of strings to store first n numbers.
    // arr[i] stores i'th number
    var arr = Array.from({length: n + 1}, (_, i) => " ");
 
    // arr[0] stores the empty string (String with 0
    // digits)
    arr[0] = "";
 
    // size indicates number of current elements in
    // arr, m indicates number of elements added to
    // arr in previous iteration
    var size = 1, m = 1;
 
    // Every iteration of following loop generates and
    // adds 2*m numbers to arr using the m numbers
    // generated in previous iteration
    while (size <= n)
    {
     
        // Consider all numbers added in previous
        // iteration, add a prefix "3" to them and add
        // new numbers to arr
        for (var i = 0; i < m && (size + i) <= n; i++)
            arr[size + i] = "3" + arr[size - m + i];
 
        // Add prefix "4" to numbers of previous
        // iteration and add new numbers to arr
        for (var i = 0; i < m && (size + m + i) <= n;
             i++)
            arr[size + m + i] = "4" + arr[size - m + i];
 
        // Update no. of elements added in previous
        // iteration
        m = m << 1; // Or m = m*2;
 
        // Update size
        size = size + m;
    }
    document.write(arr[n]+"<br>");
}
 
// Driver program
for (i = 0; i < 16; i++)
    find(i);
 
// This code is contributed by Amit Katiyar
</script>

Output: 
 

3
4
33
34
43
44
333
334
343
344
433
434
443
444
3333

Better Approach (using bits) :

This idea was suggested by Arjun J (https://auth.geeksforgeeks.org/user/camsboyfriend/profile).

The idea here is, as we are going to deal with only two numbers, i.e., 3 and 4, so we can just compare them with binary numbers.

Explanation :

1)  3   -  0     (0)
2)  4   -  1     (1)

3)  33  -  00    (0)
4)  34  -  01    (1)
5)  43  -  10    (2)  
6)  44  -  11    (3)

7)  333 -  000   (0)
8)  334 -  001   (1)
9)  343 -  010   (2) 
10) 344 -  011   (3)
11) 433 -  100   (4)
12) 434 -  101   (5)
13) 443 -  110   (6)
14) 444 -  111   (7)
15) 3333 - 1000  (8)

Here we can note that 

  1. Every (n – 1)’th number gets a new digit where n is a power of 2
  2. Whenever a new digit is added we start the counting binary numbers from 0.
  3. 0 in binary form corresponds to 3 in our number system and similarly corresponds to 4.

Below is the C++ implementation for the same :

C++




// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// function to find highest power of 2
// less than or equal to n
int highestPowerof2(unsigned int n)
{
    if (n < 1)
        return 0;
 
    int res = 1;
 
    for (int i = 0; i < 8 * sizeof(unsigned int); i++) {
        int curr = 1 << i;
 
        if (curr > n)
            break;
 
        res = curr;
    }
 
    return res;
}
 
// function to convert decimal to binary form
vector<int> decToBinary(int n, int size)
{
    vector<int> binaryNum(size + 1);
 
    int i = 0;
    while (n > 0) {
        binaryNum[i] = n % 2;
        n = (n >> 1);
        i++;
    }
 
    return binaryNum;
}
 
// Driver Code
signed main()
{
    for (int n = 1; n < 16; n++) {
        int hp2 = highestPowerof2(n + 1);
 
        int howMany = n - hp2 + 1;
 
        vector<int> arr
            = decToBinary(howMany, log2(hp2 - 1));
 
        for (int i = log2(hp2 - 1); i >= 0; i--) {
            if (arr[i])
                cout << 4;
            else
                cout << 3;
        }
        cout << '\n';
    }
}

Java




/*package whatever //do not write package name here */
import java.io.*;
 
class GFG
{
 
  // Java program for the above approach
 
  // function to find highest power of 2
  // less than or equal to n
  static long highestPowerof2(int n)
  {
    if (n < 1)
      return 0;
 
    long res = 1;
 
    for (int i = 0; i < 32; i++) {
      long curr = 1 << i;
 
      if (curr > n)
        break;
 
      res = curr;
    }
 
    return res;
  }
 
  // function to convert decimal to binary form
  static long[] decToBinary(long n, int size)
  {
    long[] binaryNum = new long[size+1];
 
    int i = 0;
    while (n > 0) {
      binaryNum[i] = n % 2;
      n = (n >> 1);
      i = i + 1;
    }
    return binaryNum;
  }
 
  /* Driver program to test above function*/
  public static void main(String args[])
  {
    for (int n = 1; n < 16; n++) {
      long hp2 = highestPowerof2(n + 1);
 
      long howMany = n - hp2 + 1;
 
      long[] arr = decToBinary(howMany, (int)Math.floor(Math.log(hp2 - 1)/Math.log(2)));
 
      for (int i = (int)Math.floor(Math.log(hp2 - 1)/Math.log(2)); i >= 0; i--) {
        if (arr[i] > 0)
          System.out.print(4);
        else
          System.out.print(3);
      }
      System.out.println();
    }
  }
}
 
// This code is contributed by shinjanpatra.

Python3




# Python3 program for the above approach
 
# function to find highest power of 2
# less than or equal to n
from math import floor, log2
 
def highestPowerof2(n):
 
    if (n < 1):
        return 0
 
    res = 1
 
    for i in range(32):
        curr = 1 << i
 
        if (curr > n):
            break
 
        res = curr
 
    return res
 
# function to convert decimal to binary form
def decToBinary(n, size):
 
    binaryNum = [0 for i in range(size+1)]
 
    i = 0
    while (n > 0):
        binaryNum[i] = n % 2
        n = (n >> 1)
        i = i + 1
     
    return binaryNum
 
# Driver Code
for n in range(1,16):
    hp2 = highestPowerof2(n + 1)
 
    howMany = n - hp2 + 1
 
    arr = decToBinary(howMany, floor(log2(hp2 - 1)))
 
    for i in range(floor(log2(hp2 - 1)),-1,-1):
        if (arr[i]):
            print(4,end="")
        else:
            print(3,end="")
     
    print()
 
# This code is contributed by Shinjanpatra

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
  // function to find highest power of 2
  // less than or equal to n
  static long highestPowerof2(int n)
  {
    if (n < 1)
      return 0;
 
    long res = 1;
 
    for (int i = 0; i < 32; i++) {
      long curr = 1 << i;
 
      if (curr > n)
        break;
 
      res = curr;
    }
 
    return res;
  }
 
  // function to convert decimal to binary form
  static long[] decToBinary(long n, int size)
  {
    long[] binaryNum = new long[size+1];
 
    int i = 0;
    while (n > 0) {
      binaryNum[i] = n % 2;
      n = (n >> 1);
      i = i + 1;
    }
    return binaryNum;
  }
 
  /* Driver program to test above function*/
  public static void Main(string[] args)
  {
    for (int n = 1; n < 16; n++) {
      long hp2 = highestPowerof2(n + 1);
 
      long howMany = n - hp2 + 1;
 
      long[] arr = decToBinary(howMany, (int)Math.Floor(Math.Log(hp2 - 1)/Math.Log(2)));
 
      for (int i = (int)Math.Floor(Math.Log(hp2 - 1)/Math.Log(2)); i >= 0; i--) {
        if (arr[i] > 0)
          Console.Write(4);
        else
          Console.Write(3);
      }
      Console.WriteLine();
    }
  }
}
 
// This code is contributed by  phasing17

Javascript




// JavaScript program for the above approach
 
// function to find highest power of 2
// less than or equal to n
function highestPowerof2(n)
{
    if (n < 1)
        return 0;
 
    let res = 1;
 
    for (let i = 0; i < 32; i++) {
        let curr = 1 << i;
 
        if (curr > n)
            break;
 
        res = curr;
    }
 
    return res;
}
 
// function to convert decimal to binary form
function decToBinary(n, size)
{
    let binaryNum = new Array(size+1);
 
    let i = 0;
    while (n > 0) {
        binaryNum[i] = n % 2;
        n = (n >> 1);
        i = i + 1;
    }
    return binaryNum;
}
 
// Driver Code
for (let n = 1; n < 16; n++) {
    let hp2 = highestPowerof2(n + 1);
 
    let howMany = n - hp2 + 1;
 
    let arr = decToBinary(howMany, Math.floor(Math.log2(hp2 - 1)));
 
    for (let i = Math.floor(Math.log2(hp2 - 1)); i >= 0; i--) {
        if (arr[i])
            document.write(4);
        else
            document.write(3);
    }
    document.write("\n");
}
 
// The code is contributed by Nidhi goel

Output: 

3
4
33
34
43
44
333
334
343
344
433
434
443
444
3333

This article is contributed by Raman. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Time Complexity: O(log(n))

Auxiliary Space: O(Size), where size represents the size of the decimal number in its binary representation.


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