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

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++
Java
Python3
C#
PHP
Javascript
// C++ program to find if a number has alternate// bit pattern#include<bits/stdc++.h>using 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 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 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.
// 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// PHP program to find if a// number has alternate// bit pattern // Returns true if n has// alternate bit pattern// else falsefunction 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;        $n = floor($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.?>
<script> // Javascript program to find if a number// has alternate bit pattern // Returns true if n has alternate// bit pattern else falsefunction findPattern(n){         // Store last bit    let prev = n % 2;    n = Math.floor(n / 2);     // Traverse through remaining bits    while (n > 0)    {        let curr = n % 2;         // If current bit is        // same as previous        if (curr == prev)            return false;         prev = curr;        n = Math.floor(n / 2);    }    return true;} // Driver codelet n = 10;if (findPattern(n))    document.write("Yes");else    document.write("No");     // This code is contributed by gfgking </script>
Output: 
Yes
Time Complexity: O(log2n)
Auxiliary Space: O(1)
Reference: http://stackoverflow.com/questions/38690278/program-to-check-whether-the-given-integer-has-an-alternate-pattern
This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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