Find Next Sparse Number

A number is Sparse if there are no two adjacent 1s in its binary representation. For example 5 (binary representation: 101) is sparse, but 6 (binary representation: 110) is not sparse.
Given a number x, find the smallest Sparse number which greater than or equal to x.

Examples:

Input: x = 6
Output: Next Sparse Number is 8

Input: x = 4
Output: Next Sparse Number is 4

Input: x = 38
Output: Next Sparse Number is 40

Input: x = 44
Output: Next Sparse Number is 64

We strongly recommend that you click here and practice it, before moving on to the solution.

A Simple Solution is to do following:

1) Write a utility function isSparse(x) that takes a number
and returns true if x is sparse, else false.  This function
can be easily written by traversing the bits of input number.
2) Start from x and do following
while(1)
{
if (isSparse(x))
return x;
else
x++
}

Time complexity of isSparse() is O(Log x). Time complexity of this solution is O(x Log x). The next sparse number can be at most O(x) distance away.

Thanks to kk_angel for suggesting above solution.

An Efficient Solution can solve this problem without checking all numbers on by one. Below are steps.

1) Find binary of the given number and store it in a
boolean array.
2) Initialize last_finalized bit position as 0.
2) Start traversing the binary from least significant bit.
a) If we get two adjacent 1's such that next (or third)
bit is not 1, then
(i)  Make all bits after this 1 to last finalized
bit (including last finalized) as 0.
(ii) Update last finalized bit as next bit.

For example, let binary representation be 01010001011101, we change it to 01010001100000 (all bits after highlighted 11 are set to 0). Again two 1’s are adjacent, so change binary representation to 01010010000000. This is our final answer.

Below is the implementation of above solution.

C++

 // C++ program to find next sparse number #include using namespace std;    int nextSparse(int x) {     // Find binary representation of x and store it in bin[].     // bin contains least significant bit (LSB), next     // bit is in bin, and so on.     vector bin;     while (x != 0)     {         bin.push_back(x&1);         x >>= 1;     }        // There my be extra bit in result, so add one extra bit     bin.push_back(0);     int n = bin.size();  // Size of binary representation        // The position till which all bits are finalized     int last_final = 0;        // Start from second bit (next to LSB)     for (int i=1; i=last_final; j--)                 bin[j] = 0;                // Store position of the bit set so that this bit             // and bits before it are not changed next time.             last_final = i+1;         }     }        // Find decimal equivalent of modified bin[]     int ans = 0;     for (int i =0; i

Java

 // Java program to find next sparse number  import java.util.*;     class GFG{ static int nextSparse(int x)  {      // Find binary representation of x and store it in bin.get(].      // bin.get(0] contains least significant bit (LSB), next      // bit is in bin.get(1], and so on.      ArrayList bin = new ArrayList();      while (x != 0)      {          bin.add(x&1);          x >>= 1;      }         // There my be extra bit in result, so add one extra bit      bin.add(0);      int n = bin.size(); // Size of binary representation         // The position till which all bits are finalized      int last_final = 0;         // Start from second bit (next to LSB)      for (int i=1; i=last_final; j--)                  bin.set(j,0);                 // Store position of the bit set so that this bit              // and bits before it are not changed next time.              last_final = i+1;          }      }         // Find decimal equivalent of modified bin.get(]      int ans = 0;      for (int i =0; i

Python3

 # Python3 program to find next  # sparse number     def nextSparse(x):            # Find binary representation of      # x and store it in bin[].      # bin contains least significant      # bit (LSB), next bit is in bin,      # and so on.      bin = []      while (x != 0):         bin.append(x & 1)          x >>= 1        # There my be extra bit in result,      # so add one extra bit      bin.append(0)      n = len(bin) # Size of binary representation             # The position till which all      # bits are finalized      last_final = 0        # Start from second bit (next to LSB)      for i in range(1,n - 1):                    # If current bit and its previous          # bit are 1, but next bit is not 1.         if ((bin[i] == 1 and bin[i - 1] == 1              and bin[i + 1] != 1)):                                # Make the next bit 1             bin[i + 1] = 1                             # Make all bits before current             # bit as 0 to make sure that              # we get the smallest next number              for j in range(i,last_final-1,-1):                 bin[j] = 0                            # Store position of the bit set             # so that this bit and bits             # before it are not changed next time.             last_final = i + 1         # Find decimal equivalent      # of modified bin[]      ans = 0      for i in range(n):          ans += bin[i] * (1 << i)      return ans     # Driver Code if __name__=='__main__':     x = 38      print("Next Sparse Number is",nextSparse(x))     # This code is contributed by  # mits

C#

 // C# program to find next sparse number  using System; using System.Collections;       class GFG{ static int nextSparse(int x)  {      // Find binary representation of x and store it in bin.get(].      // bin.get(0] contains least significant bit (LSB), next      // bit is in bin.get(1], and so on.      ArrayList bin = new ArrayList();      while (x != 0)      {          bin.Add(x&1);          x >>= 1;      }         // There my be extra bit in result, so add one extra bit      bin.Add(0);      int n = bin.Count; // Size of binary representation         // The position till which all bits are finalized      int last_final = 0;         // Start from second bit (next to LSB)      for (int i = 1; i < n-1; i++)      {      // If current bit and its previous bit are 1, but next      // bit is not 1.      if ((int)bin[i] == 1 && (int)bin[i-1] == 1 && (int)bin[i+1] != 1)      {              // Make the next bit 1              bin[i+1]=1;                 // Make all bits before current bit as 0 to make              // sure that we get the smallest next number              for (int j = i; j >= last_final; j--)                  bin[j]=0;                 // Store position of the bit set so that this bit              // and bits before it are not changed next time.              last_final = i + 1;          }      }         // Find decimal equivalent of modified bin.get(]      int ans = 0;      for (int i = 0; i < n; i++)          ans += (int)bin[i]*(1<

PHP

 >= 1;     }        // There my be extra bit in result,     // so add one extra bit     array_push(\$bin, 0);     \$n = count(\$bin); // Size of binary representation            // The position till which all      // bits are finalized     \$last_final = 0;        // Start from second bit (next to LSB)     for (\$i = 1; \$i < \$n - 1; \$i++)     {     // If current bit and its previous      // bit are 1, but next bit is not 1.     if (\$bin[\$i] == 1 &&          \$bin[\$i - 1] == 1 &&          \$bin[\$i + 1] != 1)     {         // Make the next bit 1         \$bin[\$i + 1] = 1;            // Make all bits before current          // bit as 0 to make sure that          // we get the smallest next number         for (\$j = \$i; \$j >= \$last_final; \$j--)             \$bin[\$j] = 0;            // Store position of the bit set          // so that this bit and bits          // before it are not changed next time.         \$last_final = \$i + 1;     }     }        // Find decimal equivalent     // of modified bin[]     \$ans = 0;     for (\$i = 0; \$i < \$n; \$i++)         \$ans += \$bin[\$i] * (1 << \$i);     return \$ans; }    // Driver Code \$x = 38; echo "Next Sparse Number is " .                  nextSparse(\$x);    // This code is contributed by mits ?>

Output:

Next Sparse Number is 40

Time complexity of this solution is O(Log x).

Thanks to gccode for suggesting above solution.

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Improved By : Mithun Kumar