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

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 patterns 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 naive approach to check if a number has bits in alternate patterns:

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
#include <bits/stdc++.h>
using namespace std;

// Returns true if n has alternate bit pattern else false
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 code
int main()
{
    int n = 10;
    if (findPattern(n))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

C

// C program to find if a number has alternate bit pattern
#include <stdbool.h>
#include <stdio.h>
// Returns true if n has alternate bit pattern else false
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 code
int main()
{
    int n = 10;
    if (findPattern(n))
        printf("Yes");
    else
        printf("No");
    return 0;
}

// This code is contributed by Sania Kumari Gupta
// (kriSania804)

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");
    }
}

// This code is contributed by Sania Kumari Gupta
// (kriSania804)

Python3

# Python3 program to find if a number
# has alternate bit pattern

# Returns true if n has alternate
# bit pattern else false
def 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 code
n = 10
print("Yes") if(findPattern(n)) else print("No")

# This code is contributed by Anant Agarwal.

C#

// Program to find if a number
// has alternate bit pattern
using 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
// PHP program to find if a
// number has alternate
// bit pattern

// Returns true if n has
// alternate bit pattern
// else false
function 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.
?>

Javascript

<script>

// Javascript program to find if a number
// has alternate bit pattern

// Returns true if n has alternate
// bit pattern else false
function 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 code
let 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)

An efficient approach to check if a number has bits in alternate patterns:

The idea is to check bitwise AND of n with (n>>1) and check if it returns 0. When bits are in an alternate fashion, the result will be 0 else not.

Below is the implementation of the above idea:

C++

// Program to demonstrate tree traversal with
// ability to jump between nodes of same height
#include <iostream>
using namespace std;

// Returns true if n has alternate bit pattern else false
bool findPattern(int n)
{
    if (n > 1)
        return ((n & (n >> 1))) == 0;
    else
        return false;
}

// Driver code
int main()
{
    int n = 10; // equal to 10101
    if (findPattern(n))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}

// This code is contributed by Prajwal Kandekar

C

// C program to find if a number has alternate bit pattern
#include <stdbool.h>
#include <stdio.h>

// Returns 1 if n has alternate bit pattern else 0
bool findPattern(int n)
{
    if (n > 1)
        return (!(n & (n >> 1)));
    else
        return 0;
}

// Driver code
int main()
{
    int n = 10; // equal to 10101
    if (findPattern(n))
        printf("Yes");
    else
        printf("No");
    return 0;
}

// This code is contributed by "Madhukar Sharma"

Java

// Program to demonstrate tree traversal with
// ability to jump between nodes of same height
public class Solution {

    // Returns 1 if n has alternate bit pattern else 0
    static boolean findPattern(int n)
    {
        if (n > 1)
            return ((n & (n >> 1))) == 0;
        else
            return false;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 10; // equal to 10101
        if (findPattern(n))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by karandeep1234

Python3

# Program to demonstrate tree traversal with
# ability to jump between nodes of same height

# Returns True if n has alternate bit pattern else False


def findPattern(n):
    if n > 1:
        return ((n & (n >> 1))) == 0
    else:
        return False


# Driver code
if __name__ == '__main__':
    n = 10  # equal to 10101
    if findPattern(n):
        print("Yes")
    else:
        print("No")

Javascript

// Program to demonstrate tree traversal with
// ability to jump between nodes of same height


// Returns true if n has alternate bit pattern else false
function findPattern(n) {
    if (n > 1) {
        return ((n & (n >> 1))) == 0;
    } else {
        return false;
    }
}

let n = 21; // equal to 10101
if (findPattern(n)) {
    console.log("Yes");
} else {
    console.log("No");
}

C#

using System;

public class Solution {

    // Returns true if n has alternate bit pattern else false
    static bool FindPattern(int n) {
        if (n > 1)
            return ((n & (n >> 1))) == 0;
        else
            return false;
    }

    // Driver code
    public static void Main(string[] args) {
        int n = 10; // equal to 10101
        if (FindPattern(n))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

Output

Yes

Time Complexity: O(1)
Auxiliary Space: O(1)

Reference:
http://stackoverflow.com/questions/38690278/program-to-check-whether-the-given-integer-has-an-alternate-pattern

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