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Check if a given number is sparse or not

  • Difficulty Level : Medium
  • Last Updated : 07 Jul, 2021

A number is said to be a sparse number if in binary representation of the number no two or more consecutive bits are set. Write a function to check if a given number is Sparse or not.
Example : 

Input:  x  = 72
Output: true
Explanation:  Binary representation of 72 is 01001000. 
There are no two consecutive 1's in binary representation

Input:  x  = 12
Output: false
Explanation:  Binary representation of 12 is 1100. 
Third and fourth bits (from end) are set.

 

We strongly recommend that you click here and practice it, before moving on to the solution.

If we observer carefully, then we can notice that if we can use bitwise AND of binary representation of the “given number its “right shifted number”(i.e., half the given number) to figure out whether the number is sparse or not. Result of AND operator would be 0 if number is sparse and non-zero if not sparse.
 

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Below is the implementation of above idea. 
 



C++




// C++ program to check if n is sparse or not
#include<iostream>
using namespace std;
 
// Return true if n is sparse, else false
bool checkSparse(int n)
{
    // n is not sparse if there is set
    // in AND of n and n/2
    if (n & (n>>1))
        return false;
 
    return true;
}
 
// Driver program
int main()
{
    cout << checkSparse(72) << endl;
    cout << checkSparse(12) << endl;
    cout << checkSparse(2) << endl;
    cout << checkSparse(3) << endl;
    return 0;
}

Java




// JAVA Code to Check if a
// given number is sparse or not
import java.util.*;
 
class GFG {
     
    // Return true if n is
    // sparse,else false
    static int checkSparse(int n)
    {
 
        // n is not sparse if there
        // is set in AND of n and n/2
        if ((n & (n>>1)) >=1)
            return 0;
      
        return 1;
    }
     
    // Driver code
    public static void main(String[] args)
    {
        System.out.println(checkSparse(72)) ;
        System.out.println(checkSparse(12)) ;
        System.out.println(checkSparse(2)) ;
        System.out.println(checkSparse(3)) ;
        }
    }
 
//This code is contributed by Arnav Kr. Mandal.

Python3




# Python program to check
# if n is sparse or not
 
# Return true if n is
# sparse, else false
def checkSparse(n):
 
    # n is not sparse if there is set
    # in AND of n and n/2
    if (n & (n>>1)):
        return 0
  
    return 1
     
# Driver code
print(checkSparse(72))
print(checkSparse(12))
print(checkSparse(2))
print(checkSparse(30))
 
# This code is contributed
# by Anant Agarwal.

C#




// C# Code to Check if a guven
// number is sparse or not
using System;
 
class GFG {
     
    // Return true if n is
    // sparse,else false
    static int checkSparse(int n)
    {
 
        // n is not sparse if there
        // is set in AND of n and n/2
        if ((n & (n >> 1)) >= 1)
            return 0;
     
        return 1;
    }
     
    // Driver code
    public static void Main()
    {
        Console.WriteLine(checkSparse(72));
        Console.WriteLine(checkSparse(12));
        Console.WriteLine(checkSparse(2));
        Console.WriteLine(checkSparse(3));
    }
}
 
// This code is contributed by Sam007.

PHP




<?php
// PHP program to check if
// n is sparse or not
// Return true if n is sparse,
// else false
 
function checkSparse($n)
{
    // n is not sparse if
    // there is set in AND
    // of n and n/2
    if ($n & ($n >> 1))
        return 0;
 
    return 1;
}
 
// Driver Code
echo checkSparse(72), "\n";
echo checkSparse(12), "\n";
echo checkSparse(2), "\n";
echo checkSparse(3), "\n";
 
// This code is contributed by Ajit.
?>

Javascript




<script>
 
    // Javascript program to check if n is sparse or not
     
    // Return true if n is sparse, else false
    function checkSparse(n)
    {
        // n is not sparse if there is set
        // in AND of n and n/2
        if ((n & (n>>1)) > 0)
            return 0;
 
        return 1;
    }
     
    document.write(checkSparse(72) + "</br>");
    document.write(checkSparse(12) + "</br>");
    document.write(checkSparse(2) + "</br>");
    document.write(checkSparse(3) + "</br>");
     
</script>

Output : 

1
0
1
0

Note: Instead of right shift, we could have used left shift also, but left shift might lead to overflow in some cases.
This article is contributed by Vimal Vestron. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

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