Given a positive 32-bit integer N, the task is to find the maximum between the value of N and the number obtained by decimal representation of reversal of binary representation of N in a 32-bit integer.
Examples:
Input: N = 6
Output: 1610612736
Explanation:
Binary representation of 6 in a 32-bit integer is 00000000000000000000000000000110 i.e., (00000000000000000000000000000110)2 = (6)10.
Reversing this binary string gives (01100000000000000000000000000000)2 = (1610612736)10.
The maximum between N and the obtained number is 1610612736. Therefore, print 1610612736.Input: N = 1610612736
Output: 1610612736
Approach: Follow the steps below to solve the problem:
- Calculate the binary string of the number N and store it in a string S.
- Reverse the string S.
- Calculate the decimal value of the newly reversed string S in a variable, say M.
- After completing the above steps, print the maximum value of N and M as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function that obtains the number// using said operations from Nint reverseBin(int N){ // Stores the binary representation // of the number N string S = ""; int i; // Find the binary representation // of the number N for (i = 0; i < 32; i++) { // Check for the set bits if (N & (1LL << i)) S += '1'; else S += '0'; } // Reverse the string S reverse(S.begin(), S.end()); // Stores the obtained number int M = 0; // Calculating the decimal value for (i = 0; i < 32; i++) { // Check for set bits if (S[i] == '1') M += (1LL << i); } return M;}// Function to find the maximum value// between N and the obtained numberint maximumOfTwo(int N){ int M = reverseBin(N); return max(N, M);}// Driver Codeint main(){ int N = 6; cout << maximumOfTwo(N); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;public class GFG { // Function that obtains the number // using said operations from N static int reverseBin(int N) { // Stores the binary representation // of the number N String S = ""; int i; // Find the binary representation // of the number N for (i = 0; i < 32; i++) { // Check for the set bits if ((N & (1L << i)) != 0) S += '1'; else S += '0'; } // Reverse the string S S = (new StringBuilder(S)).reverse().toString(); // Stores the obtained number int M = 0; // Calculating the decimal value for (i = 0; i < 32; i++) { // Check for set bits if (S.charAt(i) == '1') M += (1L << i); } return M; } // Function to find the maximum value // between N and the obtained number static int maximumOfTwo(int N) { int M = reverseBin(N); return Math.max(N, M); } // Driver Code public static void main(String[] args) { int N = 6; System.out.print(maximumOfTwo(N)); }}// This code is contributed by Kingash. |
Python3
# Python3 program for the above approach# Function that obtains the number# using said operations from Ndef reverseBin(N): # Stores the binary representation # of the number N S = "" i = 0 # Find the binary representation # of the number N for i in range(32): # Check for the set bits if (N & (1 << i)): S += '1' else: S += '0' # Reverse the string S S = list(S) S = S[::-1] S = ''.join(S) # Stores the obtained number M = 0 # Calculating the decimal value for i in range(32): # Check for set bits if (S[i] == '1'): M += (1 << i) return M# Function to find the maximum value# between N and the obtained numberdef maximumOfTwo(N): M = reverseBin(N) return max(N, M)# Driver Codeif __name__ == '__main__': N = 6 print(maximumOfTwo(N))# This code is contributed by SURENDRA_GANGWAR |
C#
// C# program for the above approachusing System;class GFG{static string ReverseString(string s) { char[] array = s.ToCharArray(); Array.Reverse(array); return new string(array); } // Function that obtains the number// using said operations from Npublic static int reverseBin(int N){ // Stores the binary representation // of the number N string S = ""; int i; // Find the binary representation // of the number N for (i = 0; i < 32; i++) { // Check for the set bits if ((N & (1L << i)) != 0) S += '1'; else S += '0'; } // Reverse the string S S = ReverseString(S); // Stores the obtained number int M = 0; // Calculating the decimal value for (i = 0; i < 32; i++) { // Check for set bits if (S[i] == '1') M += (1 << i); } return M;}// Function to find the maximum value// between N and the obtained numberstatic int maximumOfTwo(int N){ int M = reverseBin(N); return Math.Max(N, M);}// Driver Codestatic void Main(){ int N = 6; Console.Write(maximumOfTwo(N));}}// This code is contributed by SoumikMondal |
Javascript
<script>// JavaScript program for the above approach// Function that obtains the number// using said operations from Nfunction reverseBin(N){ // Stores the binary representation // of the number N let S = ""; let i; // Find the binary representation // of the number N for (i = 0; i < 32; i++) { // Check for the set bits if (N & (1 << i)) S += '1'; else S += '0'; } // Reverse the string S S = S.split("").reverse().join(""); // Stores the obtained number let M = 0; // Calculating the decimal value for (i = 0; i < 32; i++) { // Check for set bits if (S[i] == '1') M += (1 << i); } return M;}// Function to find the maximum value// between N and the obtained numberfunction maximumOfTwo(N){ let M = reverseBin(N); return Math.max(N, M);}// Driver Code let N = 6; document.write(maximumOfTwo(N));</script> |
Output:
1610612736
Time Complexity: O(32)
Auxiliary Space: O(1)
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