Print numbers in the range 1 to n having bits in an alternate pattern

Alternate bit pattern implies the positioning of 0’s and 1’s in a number at an alternate position i.e. no two 0s or 1’s are together. For example, 10 in binary representation is (1010)₂ which has an alternate bit pattern as 0’s and 1’s are separated by each other.

Problem Statement

Given an integer, N. Find all the integers in the range 1 to N where the bit pattern of the integer is alternating.

Example 1

Input: 10

Output: 1, 2, 5, 10

Explanation

$\mathrm{(1)_{10} = (1)_2, (2)_{10} = (10)_2, (5)_{10} = (101)_2, (10)_{10} = (1010)_2}$

Example 2

Input: 31

Output: 1, 2, 5, 10, 21

Explanation

$\mathrm{(1)_{10} = (1)_2, (2)_{10} = (10)_2, (5)_{10} = (101)_2, (10)_{10} = (1010)_2, (21)_{10} = (10101)_2}$

Approach 1: Check for Alternate Pattern

The naive approach to solve this problem would be to check for every number in the range 1 to n if it has bits in an alternate pattern or not.

This can be done as follows,

For checking if a number has an alternate bit pattern, we can use the truth table of the AND operation to check if consecutive bits are different.

Pseudocode −

procedure checkAltPattern (num)
   temp = num
   count = 0
   while (temp)
      count = count +1
      num = num >> 1
   end while
   for i = 0 to count - 2
      iTh = num >> i
      i2Th = num >> (i + 2)
      if ((iTh & 1) != (i2Th & 1))
         ans = FALSE
      end if
   end for
   ans = TRUE
end procedure

procedure numbers (num)
   ans []
   for i = 1 to num
      if checkAltPattern = TRUE
         ans.add(i)
      end if
   end for
end procedure

Example

In the following program, we check for every number in the range 1 to N if they have an alternate bit pattern by checking if, at every alternate position, the same bits are present or not.

#include <bits/stdc++.h>
using namespace std;
// Function to check if the number has alternate pattern
bool checkAltPattern(int num){
   int temp = num, count = 0;
   
   // Counting total bits of the number by using the right shift operator
   while (temp)    {
      count++;
      
      // Right shift of a number shift all the bits one position right
      temp >>= 1;
   }
   
   // For every alternate position checking if bits are same
   for (int i = 0; i < (count - 1); i++)    {
      int iTh = num >> i, i2Th = num >> (i + 2);
      
      // iTh & 1 gives the bit at ith position
      // i2Th &N1 gives the bit at (i+2)th position
      // If both bits are not same then bits pattern is not alternate
      if ((iTh & 1) != (i2Th & 1))        {
         return false;
      }
   }
   return true;
}
vector<int> numbers(int num){

   // Initializing an array to add numbers with alternate bit pattern
   vector<int> ans;
   for (int i = 1; i <= num; i++){
      if (checkAltPattern(i)){
         ans.push_back(i);
      }
   }
   return ans;
}
int main(){
   int N = 50;
   cout << "Numbers in range 1 to " << N << " having alternate bit pattern : ";
   vector<int> res = numbers(N);
   for (int i = 0; i < res.size(); i++){
      cout << res[i] << " ";
   }
   return 0;
}

Output

Numbers in range 1 to 50 having alternate bit pattern : 1 2 5 10 21 42

Time Complexity − O(nlog_n) as the for each number we check for the alternate pattern which takes O(log_n) time as logn is the number of bits used to represent that number. Thus total time is O(log_n + log_n + … n times) = O(nlog_n)

Space Complexity − O(1) as no extra space is used.

Approach 2: Generate the Numbers From 1

An optimized and better approach to the problem would be to start with the first number having an alternate pattern and then consecutively alternately add 0s and 1s.

For adding 0 on the right side, use the left shift operator.

For adding 1 on the right side, use the left shift first and then add 1 using the XOR operator.

Pseudocode −

procedure addZero (pass)
   pass = pass << 1
   ans = pass
end procedure

procedure addOne (pass)
   pass = pass << 1
   pass = pass ^ 1
   ans = pass
end procedure

procedure numbers (N)
   ans []
   int num = 1
   ans.add (num)
   while (1)
      num = addZero (num)
      if (num > N)
         break
      else
         ans.add (num)
      end if
      num = addOne (num)
      if (num > N)
         break
      else
         ans.add (num)
      end if
   end while
end procedure

Example

In the following program, addZero() function adds 0 at the LSB position and addOne() function adds 1 at the LSB position.

#include <bits/stdc++.h>
using namespace std;
int addZero(int pass){
   // Left shift puts 0 at LSB position
   pass = pass << 1;
   return pass;
}
int addOne(int pass){

   // Left shift puts 0 at LSB position
   pass = pass << 1;
   
   // XOR with 1 sets the LSB position
   pass = pass ^ 1;
   return pass;
}
vector<int> numbers(int N){

   // Initializing an array to add numbers with alternate bit pattern
   vector<int> ans;
   
   // First number with alternate bit pattern is 1
   int num = 1;
   ans.push_back(num);
   while (1)    {
      num = addZero(num);
      
      // If on adding 0, num is greater than N then leave loop
      if (num > N)        {
         break;
      }
      else{
         ans.push_back(num);
      }
      num = addOne(num);
      
      // If on adding 1, num is greater than N then leave loop
      if (num > N){
         break;
      }
      else{
         ans.push_back(num);
      }
   }
   return ans;
}

int main(){
   int N = 50;
   cout << "Numbers in range 1 to " << N << " having alternate bit pattern : ";
   vector<int> res = numbers(N);
   for (int i = 0; i < res.size(); i++){
      cout << res[i] << " ";
   }
   return 0;
}

Output

Numbers in range 1 to 50 having alternate bit pattern : 1 2 5 10 21 42

Time Complexity − O(log_n)

Space Complexity − O(1)

Conclusion

In conclusion, for finding the numbers with alternate bit pattern an optimised approach will be to produce numbers itself from 1 by left shifting and then setting and unsetting the LSB position. This approach is the most optimized and optimal.