First and Last Three Bits - GeeksforGeeks

25 Mar, 2021
Given an integer N. The task is to print the decimal equivalent of the first three bits and the last three bits in the binary representation of N.Examples:  
Input: 86 Output: 5 6 The binary representation of 86 is 1010110. The decimal equivalent of the first three bits (101) is 5. The decimal equivalent of the last three bits (110) is 6. Hence the output is 5 6.Input: 7 Output: 7 7  
 
Simple Approach:  
Convert N into binary and store the bits in an array.
Convert the first three values from the array into decimal equivalent and print it.
Similarly, convert the last three values from the array into decimal equivalent and print it.
Below is the implementation of the above approach:  
C++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to print the first// and last 3 bits equivalent decimal numbervoid binToDecimal3(int n){    // Converting n to binary    int a[64] = { 0 };    int x = 0, i;    for (i = 0; n > 0; i++) {        a[i] = n % 2;        n /= 2;    }     // Length of the array has to be at least 3    x = (i < 3) ? 3 : i;     // Convert first three bits to decimal    int d = 0, p = 0;    for (int i = x - 3; i < x; i++)        d += a[i] * pow(2, p++);     // Print the decimal    cout << d << " ";     // Convert last three bits to decimal    d = 0;    p = 0;    for (int i = 0; i < 3; i++)        d += a[i] * pow(2, p++);     // Print the decimal    cout << d;} // Driver codeint main(){    int n = 86;     binToDecimal3(n);    return 0;}
//Java implementation of the approach import java.math.*;public class GFG {     //Function to print the first    //and last 3 bits equivalent decimal number    static void binToDecimal3(int n)    {     // Converting n to binary     int a[] = new int[64] ;     int x = 0, i;     for (i = 0; n > 0; i++) {         a[i] = n % 2;         n /= 2;     }      // Length of the array has to be at least 3     x = (i < 3) ? 3 : i;      // Convert first three bits to decimal     int d = 0, p = 0;     for (int j = x - 3; j < x; j++)         d += a[j] * Math.pow(2, p++);      // Print the decimal     System.out.print( d + " ");      // Convert last three bits to decimal     d = 0;     p = 0;     for (int k = 0; k < 3; k++)         d += a[k] * Math.pow(2, p++);      // Print the decimal     System.out.print(d);    }     //Driver code    public static void main(String[] args) {                 int n = 86;          binToDecimal3(n);     } }
# Python 3 implementation of the approachfrom math import pow # Function to print the first and last 3# bits equivalent decimal numberdef binToDecimal3(n):         # Converting n to binary    a = [0 for i in range(64)]    x = 0    i = 0    while(n > 0):        a[i] = n % 2        n = int(n / 2)        i += 1     # Length of the array has to    # be at least 3    if (i < 3):        x = 3    else:        x = i     # Convert first three bits to decimal    d = 0    p = 0    for i in range(x - 3, x, 1):        d += a[i] * pow(2, p)        p += 1     # Print the decimal    print(int(d), end =" ")     # Convert last three bits to decimal    d = 0    p = 0    for i in range(0, 3, 1):        d += a[i] * pow(2, p)        p += 1     # Print the decimal    print(int(d),end = " ") # Driver codeif __name__ == '__main__':    n = 86     binToDecimal3(n)     # This code is contributed by# Sanjit_Prasad
// C# implementation of the approachusing System; class GFG{ // Function to print the first and last// 3 bits equivalent decimal numberstatic void binToDecimal3(int n){         // Converting n to binary    int [] a= new int[64] ;    int x = 0, i;    for (i = 0; n > 0; i++)    {        a[i] = n % 2;        n /= 2;    }     // Length of the array has to be    // at least 3    x = (i < 3) ? 3 : i;         // Convert first three bits to decimal    int d = 0, p = 0;    for (int j = x - 3; j < x; j++)        d += a[j] *(int)Math.Pow(2, p++);         // Print the decimal    int d1 = d;         // Convert last three bits to decimal    d = 0;    p = 0;    for (int k = 0; k < 3; k++)        d += a[k] * (int)Math.Pow(2, p++);         // Print the decimal    Console.WriteLine(d1 + " " + d);} // Driver codestatic void Main(){    int n = 86;     binToDecimal3(n);}} // This code is contributed by Mohit kumar 29
<script> // Javascript implementation of the approach // Function to print the first// and last 3 bits equivalent decimal numberfunction binToDecimal3(n){    // Converting n to binary    var a = Array(64).fill(0);    var x = 0, i;    for (i = 0; n > 0; i++) {        a[i] = n % 2;        n = parseInt(n/2);    }     // Length of the array has to be at least 3    x = (i < 3) ? 3 : i;     // Convert first three bits to decimal    var d = 0, p = 0;    for (var i = x - 3; i < x; i++)        d += a[i] * parseInt(Math.pow(2, p++));     // Print the decimal    document.write(d + " ");     // Convert last three bits to decimal    d = 0;    p = 0;    for (var i = 0; i < 3; i++)        d += a[i] * parseInt(Math.pow(2, p++));     // Print the decimal    document.write(d);} // Driver codevar n = 86;binToDecimal3(n);     </script>
5 6
 
Efficient Approach: We can use bitwise operators to find the required numbers.  
C++
Java
Python3
C#
PHP
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to print the first// and last 3 bits equivalent decimal// numbervoid binToDecimal3(int n){    // Number formed from last three    // bits    int last_3 = ((n & 4) + (n & 2) + (n & 1));     // Let us get first three bits in n    n = n >> 3;    while (n > 7)        n = n >> 1;     // Number formed from first three    // bits    int first_3 = ((n & 4) + (n & 2) + (n & 1));     // Printing result    cout << first_3 << " " << last_3;} // Driver codeint main(){    int n = 86;    binToDecimal3(n);    return 0;}
// Java implementation of the approachimport java.util.*; class GFG{ // Function to print the first// and last 3 bits equivalent// decimal numberstatic void binToDecimal3(int n){    // Number formed from last three    // bits    int last_3 = ((n & 4) +                  (n & 2) + (n & 1));     // Let us get first three bits in n    n = n >> 3;    while (n > 7)        n = n >> 1;     // Number formed from first    // three bits    int first_3 = ((n & 4) +                   (n & 2) + (n & 1));     // Printing result    System.out.println(first_3 + " " + last_3);} // Driver codepublic static void main(String args[]){    int n = 86;    binToDecimal3(n);}} // This code is contributed by// Surendra_Gangwar
# Python3 implementation of the approach # Function to print the first and# last 3 bits equivalent decimal# numberdef binToDecimal3(n) :         # Number formed from last three    # bits    last_3 = ((n & 4) + (n & 2) + (n & 1));     # Let us get first three bits in n    n = n >> 3    while (n > 7) :        n = n >> 1     # Number formed from first three    # bits    first_3 = ((n & 4) + (n & 2) + (n & 1))     # Printing result    print(first_3,last_3) # Driver codeif __name__ == "__main__" :     n = 86    binToDecimal3(n) # This code is contributed by Ryuga
// C# implementation of the approachusing System; class GFG{     // Function to print the first// and last 3 bits equivalent// decimal numberstatic void binToDecimal3(int n){    // Number formed from last three    // bits    int last_3 = ((n & 4) +                (n & 2) + (n & 1));     // Let us get first three bits in n    n = n >> 3;    while (n > 7)        n = n >> 1;     // Number formed from first    // three bits    int first_3 = ((n & 4) +                (n & 2) + (n & 1));     // Printing result    Console.WriteLine(first_3 + " " + last_3);} // Driver codestatic public void Main (){    int n = 86;    binToDecimal3(n);}} // This code is contributed by akt_mit..
<?php// PHP implementation of the approach // Function to print the first and last// 3 bits equivalent decimal numberfunction binToDecimal3($n){    // Number formed from last three    // bits    $last_3 = (($n & 4) + ($n & 2) + ($n & 1));     // Let us get first three bits in n    $n = $n >> 3;    while ($n > 7)        $n = $n >> 1;     // Number formed from first three    // bits    $first_3 = (($n & 4) + ($n & 2) + ($n & 1));     // Printing result    echo($first_3);    echo(" ");    echo($last_3);} // Driver code$n = 86;binToDecimal3($n); // This code is contributed// by Shivi_Aggarwal?>
  <script>     // Javascript implementation of the approach     // Function to print the first    // and last 3 bits equivalent decimal    // number    function binToDecimal3(n)    {           // Number formed from last three      // bits      var last_3 = ((n & 4) + (n & 2) + (n & 1));       // Let us get first three bits in n      n = n >> 3;      while (n > 7)        n = n >> 1;       // Number formed from first three      // bits      var first_3 = ((n & 4) + (n & 2) + (n & 1));       // Printing result      document.write(first_3 + " " + last_3);    }     // Driver code    var n = 86;    binToDecimal3(n); // This code is contributed by rrrtnx.  </script>
5 6
 
Sanjit_Prasad
ankthon
ukasp
Shivi_Aggarwal
SURENDRA_GANGWAR
jit_t
mohit kumar 29
noob2000
rrrtnx
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