# Next greater number than N with exactly one bit different in binary representation of N

Given a number N. The task is to find the smallest number which is greater than N and has only one bit different in the binary representation of N.

Note: Here N can be very large 10^9 < N < 10^15.

Examples:

Input : N = 11
Output : The next number is 15
The binary representation of 11 is 1011
So the smallest number greater than 11
with one bit different is 1111 which is 15.

Input : N = 17
Output : The next number is 19

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Simple Approach: We will run a loop from N+1 until we find a number which has exactly one bit different than N. This process might take a long time to process in case of large numbers

Efficient Approach: In the efficient approach we have to represent the number in its binary form. Now a number greater than N with only 1 bit different is only possible when we keep all the set bits of the number N intact and switch the lowest possible bit which is zero to 1.

Let us take the example, 1001 which is binary form of 9,
If we switch the set bits of 1001 to either 1000 or 0001 then we find that the numbers are 8 and 1 which are less than N. While if we turn the bits which are zero to 1 we find the numbers are 11 (1011) or 13 (1101) so if switch the lowest possible bit which is zero to 1 then we get the smallest possible number greater than N with only 1 bit different, in this case it is 11 (1011).

Note: It is guaranteed that the input number N does not have all of its bits as set. There exists at least one unset bit in the binary representation of N.

Below is the implementation of the above approach:

## C++

 // CPP program to find next greater number than N  // with exactly one bit different in binary // representation of N    #include using namespace std;    // Function to find next greater number than N  // with exactly one bit different in binary // representation of N long long nextGreater(long long N) {     long long power_of_2 = 1, shift_count = 0;        // It is guaranteed that there      // is a bit zero in the number     while (true) {         // If the shifted bit is zero then break         if (((N >> shift_count) & 1) % 2 == 0)             break;            // increase the bit shift         shift_count++;            // increase the power of 2         power_of_2 = power_of_2 * 2;     }        // set the lowest bit of the number     return (N + power_of_2); }    // Driver code int main() {     long long N = 11;        // display the next number     cout << "The next number is = " << nextGreater(N);        return 0; }

## Java

 // Java program to find next greater number than N  // with exactly one bit different in binary  // representation of N     class GFG{ // Function to find next greater number than N  // with exactly one bit different in binary  // representation of N  static int nextGreater(int N)  {      int power_of_2 = 1, shift_count = 0;         // It is guaranteed that there      // is a bit zero in the number      while (true) {          // If the shifted bit is zero then break          if (((N >> shift_count) & 1) % 2 == 0)              break;             // increase the bit shift          shift_count++;             // increase the power of 2          power_of_2 = power_of_2 * 2;      }         // set the lowest bit of the number      return (N + power_of_2);  }     // Driver code  public static void main(String[]a)  {      int N = 11;         // display the next number      System.out.println("The next number is = " + nextGreater(N));     } }

## Python3

 # Python3 program to find next greater  # number than N with exactly one # bit different in binary # representation of N    # Function to find next greater  # number than N with exactly  # one bit different in binary # representation of N def nextGreater(N):        power_of_2 = 1;      shift_count = 0;        # It is guaranteed that there      # is a bit zero in the number     while (True):                    # If the shifted bit is         # zero then break         if (((N >> shift_count) & 1) % 2 == 0):             break;            # increase the bit shift         shift_count += 1;            # increase the power of 2         power_of_2 = power_of_2 * 2;        # set the lowest bit      # of the number     return (N + power_of_2);    # Driver code N = 11;    # display the next number print("The next number is =",               nextGreater(N));    # This code is contributed by mits

## C#

 // C# program to find next  // greater number than N with   // exactly one bit different in // binary representation of N  using System;    class GFG { // Function to find next greater  // number than N with exactly  // one bit different in binary  // representation of N  static int nextGreater(int N)  {      int power_of_2 = 1,          shift_count = 0;         // It is guaranteed that there      // is a bit zero in the number      while (true)     {          // If the shifted bit is          // zero then break          if (((N >> shift_count) & 1) % 2 == 0)              break;             // increase the bit shift          shift_count++;             // increase the power of 2          power_of_2 = power_of_2 * 2;      }         // set the lowest bit      // of the number      return (N + power_of_2);  }     // Driver code  public static void Main()  {      int N = 11;         // display the next number      Console.WriteLine("The next number is = " +                                 nextGreater(N));  } }     // This code is contributed  // by anuj_67

## PHP

 > \$shift_count) & 1) % 2 == 0)             break;            // increase the bit shift         \$shift_count++;            // increase the power of 2         \$power_of_2 = \$power_of_2 * 2;     }        // set the lowest bit of the number     return (\$N + \$power_of_2); }    // Driver code \$N = 11;    // display the next number echo "The next number is = " ,                nextGreater(\$N);    // This code is contributed  // by anuj_67 ?>

Output:

The next number is = 15

Time Complexity: O(log(N))

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Second year Department of Information Technology Jadavpur University

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