# First and Last Three Bits

• Difficulty Level : Medium
• Last Updated : 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:
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 ``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

 ``

Output:

`5 6`

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

## C++

 `// C++ implementation of the approach``#include ``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..`

## PHP

 `> 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``?>`

## Javascript

 `  ```

Output:

`5 6`

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