Check if a number has bits in alternate pattern | Set 1
Last Updated :
18 Mar, 2025
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 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 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 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.
// 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.
<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>
Time Complexity: O(log2n)
Auxiliary Space: O(1)
An efficient approach to check if a number has bits in alternate patterns:
1) Right shift the number by 1 (x = n >> 1) so that all 1s become aligned with all 0s for an alternate bit pattern number.
2) Now do an XOR of x with n (y = x ^ n). This will have a number with all 1s. Note that XOR of two bits is 0 when they are same, otherwise 1.
3) Now we mainly need to check if all bits are set or not. To check this, we simply check if y & (y+1) is 0 or not.
C++
#include <iostream>
using namespace std;
// Returns true if n has alternate bit pattern else false
bool findPattern(int n)
{
if (n <= 1) return false;
int x = n >> 1;
int y = x ^ n;
return (y & (y+1)) == 0;
}
// Driver code
int main()
{
int n = 10; // equal to 10101
if (findPattern(n))
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}
// 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 false;
int x = n >> 1;
int y = x ^ n;
return (y & (y+1)) == 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"
public class Solution {
// Returns 1 if n has alternate bit pattern else 0
static boolean findPattern(int n)
{
if (n <= 1) return false;
int x = n >> 1;
int y = x ^ n;
return (y & (y + 1)) == 0;
}
// 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
# Returns True if n has alternate bit pattern else False
def findPattern(n):
if (n <= 1):
return False
x = n >> 1;
y = x ^ n;
return (y & (y+1)) == 0;
# Driver code
if __name__ == '__main__':
n = 10 # equal to 10101
if findPattern(n):
print("Yes")
else:
print("No")
using System;
public class Solution {
// Returns true if n has alternate bit pattern else
// false
static bool FindPattern(int n)
{
if (n <= 1) return false;
int x = n >> 1;
int y = x ^ n;
return (y & (y + 1)) == 0;
}
// Driver code
public static void Main(string[] args)
{
int n = 10; // equal to 10101
if (FindPattern(n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// Returns true if n has alternate bit pattern else false
function findPattern(n) {
if (n <= 1) return false;
let x = n >> 1;
let y = x ^ n;
return (y & (y+1)) == 0;
}
let n = 21; // equal to 10101
if (findPattern(n)) {
console.log("Yes");
} else {
console.log("No");
}
Time Complexity: O(1)
Auxiliary Space: O(1)
Explore
DSA Fundamentals
Data Structures
Algorithms
Advanced
Interview Preparation
Practice Problem