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First and Last Three Bits

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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:
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++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the first
// and last 3 bits equivalent decimal number
void 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 code
int main()
{
    int n = 86;
 
    binToDecimal3(n);
    return 0;
}


Java




//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);
 
    }
 
}


Python3




# Python 3 implementation of the approach
from math import pow
 
# Function to print the first and last 3
# bits equivalent decimal number
def 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 code
if __name__ == '__main__':
    n = 86
 
    binToDecimal3(n)
     
# This code is contributed by
# Sanjit_Prasad


C#




// C# implementation of the approach
using System;
 
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] *(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 code
static void Main()
{
    int n = 86;
 
    binToDecimal3(n);
}
}
 
// This code is contributed by Mohit kumar 29


Javascript




// Javascript implementation of the approach
 
// Function to print the first
// and last 3 bits equivalent decimal number
function 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 code
var n = 86;
binToDecimal3(n);
    


Output

5 6








Time Complexity: O(logn)

Auxiliary Space: O(1)

Efficient Approach: 
We can use bitwise operators to find the required numbers. 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the first
// and last 3 bits equivalent decimal
// number
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
    cout << first_3 << " " << last_3;
}
 
// Driver code
int main()
{
    int n = 86;
    binToDecimal3(n);
    return 0;
}


Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to print the first
// and last 3 bits equivalent
// decimal number
static 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 code
public static void main(String args[])
{
    int n = 86;
    binToDecimal3(n);
}
}
 
// This code is contributed by
// Surendra_Gangwar


Python3




# Python3 implementation of the approach
 
# Function to print the first and
# last 3 bits equivalent decimal
# number
def 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 code
if __name__ == "__main__" :
 
    n = 86
    binToDecimal3(n)
 
# This code is contributed by Ryuga


C#




// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to print the first
// and last 3 bits equivalent
// decimal number
static 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 code
static public void Main ()
{
    int n = 86;
    binToDecimal3(n);
}
}
 
// This code is contributed by akt_mit..


Javascript




  <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>


PHP




<?php
// PHP 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
    $last_3 = (($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
?>


Output

5 6








Time Complexity: O(1)

Auxiliary Space: O(1)

Approach#3: Using slicing

Read an integer input from the user. Convert the integer to its binary representation using the bin() function. Remove the ‘0b’ prefix from the binary representation using string slicing. Get the first three bits of the binary representation using string slicing. Get the last three bits of the binary representation using string slicing. Convert the binary strings of the first and last three bits to decimal integers using the int() function with the second argument set to 2 (which means the input string is interpreted as a binary string). Print the decimal values of the first and last three bits.

Algorithm

1. Start the program.
2. Read an integer input from the user and store it in a variable num.
3. Convert the integer to its binary representation using the bin() function and store it in a variable bin_num.
4. Remove the ‘0b’ prefix from the binary representation using string slicing and store the result in bin_num.
5. Get the first three bits of the binary representation using string slicing and store the result in a variable first_three_bits.
6. Get the last three bits of the binary representation using string slicing and store the result in a variable last_three_bits.
7. Convert the binary strings of the first and last three bits to decimal integers using the int() function with the second argument set to 2 and store the results in variables first_decimal and last_decimal.
8. Print the decimal values of the first and last three bits.
9. End the program.

C++




#include <iostream>
#include <bitset>
 
int main() {
    int num = 86;
    std::bitset<8> bin_num(num); // create a bitset with the binary representation of num
    std::string bin_str = bin_num.to_string(); // convert the bitset to a string
 
    std::string first_three_bits = bin_str.substr(3, 3); // get first three bits starting from the 6th position
    std::string last_three_bits = bin_str.substr(bin_str.size() - 3); // get last three bits
 
    int first_three_decimal = std::stoi(first_three_bits, nullptr, 2); // convert binary string to decimal integer
    int last_three_decimal = std::stoi(last_three_bits, nullptr, 2); // convert binary string to decimal integer
 
    std::cout << first_three_decimal << " " << last_three_decimal << std::endl;
 
    return 0;
}


Java




public class BinaryConversion {
    public static void main(String[] args) {
        int num = 86;
        String binNum = Integer.toBinaryString(num); // Get binary representation as a string
        String binNumWithoutPrefix = binNum.substring(2); // Remove the '0b' prefix
        String firstThreeBits = binNumWithoutPrefix.substring(0, 3); // Get the first three bits
        // Get the last three bits
        String lastThreeBits = binNumWithoutPrefix.substring(binNumWithoutPrefix.length() - 3);
 
        // Convert the first three bits to decimal
        int firstThreeBitsDecimal = Integer.parseInt(firstThreeBits, 2);
        // Convert the last three bits to decimal
        int lastThreeBitsDecimal = Integer.parseInt(lastThreeBits, 2);
        // Print the decimal integers
        System.out.println(firstThreeBitsDecimal + " " + lastThreeBitsDecimal);
    }
}


Python3




num = 86
bin_num = bin(num)[2:]  # get binary representation without '0b' prefix
first_three_bits = bin_num[:3# get first three bits
last_three_bits = bin_num[-3:]  # get last three bits
print(int(first_three_bits, 2), int(last_three_bits, 2))  # convert binary strings to decimal integers and print


C#




using System;
 
class BinaryConversion {
    static void Main()
    {
        int num = 86;
        string binNum = Convert.ToString(
            num,
            2); // Get binary representation as a string
        string binNumWithoutPrefix
            = binNum.Substring(2); // Remove the '0b' prefix
 
        // Get the first three bits
        string firstThreeBits
            = binNumWithoutPrefix.Substring(0, 3);
        // Get the last three bits
        string lastThreeBits
            = binNumWithoutPrefix.Substring(
                binNumWithoutPrefix.Length - 3);
 
        // Convert the first three bits to decimal
        int firstThreeBitsDecimal
            = Convert.ToInt32(firstThreeBits, 2);
        // Convert the last three bits to decimal
        int lastThreeBitsDecimal
            = Convert.ToInt32(lastThreeBits, 2);
 
        // Print the decimal integers
        Console.WriteLine(firstThreeBitsDecimal + " "
                          + lastThreeBitsDecimal);
    }
}


Javascript




// Function to convert an integer to its binary representation
function decToBinary(num) {
    return (num >>> 0).toString(2);
}
 
// Main function
function main() {
    const num = 86;
    const binNum = decToBinary(num); // Get binary representation as a string
 
    const binNumWithoutPrefix = binNum.substring(2); // Remove the '0b' prefix
    const firstThreeBits = binNumWithoutPrefix.substring(0, 3); // Get the first three bits
 
    // Get the last three bits
    const lastThreeBits = binNumWithoutPrefix.substring(binNumWithoutPrefix.length - 3);
 
    // Convert the first three bits to decimal
    const firstThreeBitsDecimal = parseInt(firstThreeBits, 2);
 
    // Convert the last three bits to decimal
    const lastThreeBitsDecimal = parseInt(lastThreeBits, 2);
 
    // Print the decimal integers
    console.log(firstThreeBitsDecimal + " " + lastThreeBitsDecimal);
}
 
// Call the main function
main();


Output

5 6









Time complexity: O(1) The time complexity of this program is constant because it performs a fixed number of operations that do not depend on the input size.

Space complexity: O(log n) The space complexity of this program depends on the size of the input integer. The bin() function returns a string representation of the binary number, which requires O(log n) space to store, where n is the input integer. The other variables created in the program (first_three_bits, last_three_bits, first_decimal, and last_decimal) all require constant space. Therefore, the overall space complexity is O(log n).



Last Updated : 09 Dec, 2023
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