Check if a number has bits in alternate pattern | Set 1

• Difficulty Level : Easy
• Last Updated : 25 Nov, 2021

Given an integer n > 0, the task is to find whether this integer has an alternate pattern in its bits representation. For example- 5 has an alternate pattern i.e. 101.
Print “Yes” if it has an alternate pattern otherwise “No”. Here alternate pattern can be like 0101 or 1010.

Examples:

Input :  15
Output :  No
Explanation: Binary representation of 15 is 1111.

Input :  10
Output :  Yes
Explanation: Binary representation of 10 is 1010.

A simple approach is to find its binary equivalent and then check its bits.

C++

 // C++ program to find if a number has alternate// bit pattern#includeusing namespace std; // Returns true if n has alternate bit pattern// else falsebool findPattern(int n){    // Store last bit    int prev = n % 2;    n = n/2;     // Traverse through remaining bits    while (n > 0)    {        int curr = n % 2;         // If current bit is same as previous        if (curr == prev)           return false;         prev = curr;        n = n / 2;    }     return true;} // Driver codeint main(){    int n = 10;    if (findPattern(n))        cout << "Yes";    else        cout << "No";     return 0;}

Java

 // Java program to find if a number has alternate// bit pattern class Test{    // Returns true if n has alternate bit pattern    // else false    static boolean findPattern(int n)    {        // Store last bit        int prev = n % 2;        n = n/2;              // Traverse through remaining bits        while (n > 0)        {            int curr = n % 2;                  // If current bit is same as previous            if (curr == prev)               return false;                  prev = curr;            n = n / 2;        }              return true;    }         // Driver method    public static void main(String args[])    {        int n = 10;        System.out.println(findPattern(n) ? "Yes" : "No");          }}

Python3

 # Python3 program to find if a number# has alternate bit pattern # Returns true if n has alternate# bit pattern else falsedef findPattern(n):     # Store last bit    prev = n % 2    n = n // 2     # Traverse through remaining bits    while (n > 0):             curr = n % 2         # If current bit is same as previous        if (curr == prev):            return False         prev = curr        n = n // 2         return True # Driver coden = 10print("Yes") if(findPattern(n)) else print("No") # This code is contributed by Anant Agarwal.

C#

 // Program to find if a number// has alternate bit patternusing System; class Test {     // Returns true if n has alternate    // bit pattern else returns false    static bool findPattern(int n)    {        // Store last bit        int prev = n % 2;        n = n / 2;         // Traverse through remaining bits        while (n > 0) {            int curr = n % 2;             // If current bit is same as previous            if (curr == prev)                return false;             prev = curr;            n = n / 2;        }         return true;    }     // Driver method    public static void Main()    {        int n = 10;        Console.WriteLine(findPattern(n) ? "Yes" : "No");    }} // This code is contributed by Anant Agarwal.

PHP

 0)    {        \$curr = \$n % 2;         // If current bit is        // same as previous        if (\$curr == \$prev)        return false;         \$prev = \$curr;        \$n = floor(\$n / 2);    }     return true;}     // Driver code    \$n = 10;    if (findPattern(\$n))        echo "Yes";    else        echo "No";             return 0; // This code is contributed by nitin mittal.?>

Javascript



Output:

Yes

Time Complexity: O(log2n)

Auxiliary Space: O(1)

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