Find the maximum between N and the number formed by reversing 32-bit binary representation of N Last Updated : 08 Nov, 2021 Summarize Comments Improve Suggest changes Share Like Article Like Report 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 = 6Output: 1610612736Explanation: 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 = 1610612736Output: 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 N 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 & (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 number int maximumOfTwo(int N) { int M = reverseBin(N); return max(N, M); } // Driver Code int main() { int N = 6; cout << maximumOfTwo(N); return 0; } Java // Java program for the above approach import 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 N def 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 number def maximumOfTwo(N): M = reverseBin(N) return max(N, M) # Driver Code if __name__ == '__main__': N = 6 print(maximumOfTwo(N)) # This code is contributed by SURENDRA_GANGWAR C# // C# program for the above approach using 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 N public 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 number static int maximumOfTwo(int N) { int M = reverseBin(N); return Math.Max(N, M); } // Driver Code static 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 N function 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 number function 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) Comment More infoAdvertise with us Next Article Find the maximum between N and the number formed by reversing 32-bit binary representation of N T tmprofessor Follow Improve Article Tags : Misc Bit Magic Mathematical DSA binary-representation +1 More Practice Tags : Bit MagicMathematicalMisc Similar Reads DSA Tutorial - Learn Data Structures and Algorithms DSA (Data Structures and Algorithms) is the study of organizing data efficiently using data structures like arrays, stacks, and trees, paired with step-by-step procedures (or algorithms) to solve problems effectively. 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