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++
// 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
//Java implementation of the approachimport 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 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#
// 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 |
Javascript
<script>// 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); </script> |
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// 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
// 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
# 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 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// 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 ?> |
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> |
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.
Python3
num = 86bin_num = bin(num)[2:] # get binary representation without '0b' prefixfirst_three_bits = bin_num[:3] # get first three bitslast_three_bits = bin_num[-3:] # get last three bitsprint(int(first_three_bits, 2), int(last_three_bits, 2)) # convert binary strings to decimal integers and print |
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).
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
