Check if a given number is sparse or not
Last Updated :
14 Sep, 2024
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Write a function to check if a given number is Sparse or not.
A number is said to be a sparse number if in the binary representation of the number no two or more consecutive bits are set.
Example:
Input: x = 72
Output: true
Explanation: Binary representation of 72 is 01001000.
There are no two consecutive 1's in binary representationInput: x = 12
Output: false
Explanation: Binary representation of 12 is 1100.
Third and fourth bits (from end) are set.
Naive Approach: The idea is to check the consecutive bits of the number until the number becomes 0.
#include <iostream>
using namespace std;
// Function to check if the number
// is sparse or not.
bool isSparse(int n)
{
int prev;
if (n == 1)
return true;
while (n > 0) {
prev = n & 1;
n = n >> 1;
int curr = n & 1;
if (prev == curr && prev == 1)
return false;
prev = curr;
}
return true;
}
// Driver Code
int main()
{
int n = 100;
if (isSparse(n)) {
cout << "Sparse";
}
else {
cout << "Not Sparse";
}
return 0;
}
// "static void main" must be defined in a public class.
public class GFG {
// Function to check if the number
// is sparse or not.
static boolean isSparse(int n)
{
int prev;
if (n == 1)
return true;
while (n > 0) {
prev = n & 1;
n = n >> 1;
int curr = n & 1;
if (prev == curr && prev == 1)
return false;
prev = curr;
}
return true;
}
public static void main(String[] args)
{
int n = 100;
if (isSparse(n)) {
System.out.println("Sparse");
}
else {
System.out.println("Not Sparse");
}
}
}
// This code is contributed by garg28harsh.
# Python Code to Check if a given
# number is sparse or not
def isSparse(n):
if (n == 1):
return true
global prev
while(n > 0):
prev = n & 1
n = n >> 1
curr = n & 1
if(prev == curr and prev == 1):
return False
prev = curr
return True
# Driver code
n = 100
if (isSparse(n)):
print("Sparse")
else:
print("Not Sparse")
# This code is contributed by karandeep1234
// C# Code to Check if a given
// number is sparse or not
using System;
public class GFG {
// Function to check if the number
// is sparse or not.
static bool isSparse(int n)
{
int prev;
if (n == 1)
return true;
while (n > 0) {
prev = n & 1;
n = n >> 1;
int curr = n & 1;
if (prev == curr && prev == 1)
return false;
prev = curr;
}
return true;
}
public static void Main(string[] args)
{
int n = 100;
if (isSparse(n)) {
Console.WriteLine("Sparse");
}
else {
Console.WriteLine("Not Sparse");
}
}
}
// This code is contributed by karandeep1234
// Function to check if the number
// is sparse or not.
function isSparse(n)
{
let prev;
if (n == 1)
return true;
while (n > 0) {
prev = n & 1;
n = n >> 1;
let curr = n & 1;
if (prev == curr && prev == 1)
return false;
prev = curr;
}
return true;
}
// Driver Code
let n = 100;
if (isSparse(n)) {
console.log( "Sparse");
}
else {
console.log( "Not Sparse");
}
// This code is contributed by garg28harsh.
Output
Not Sparse
Time Complexity: O(Log2n)
Auxiliary Space: O(1)
Efficient Approach: To solve the problem follow the below idea:
If we observe carefully, then we can notice that if we can use bitwise AND of the binary representation of the "given number, then it's "right-shifted number"(i.e., half the given number) to figure out whether the number is sparse or not. The result of AND operator would be 0 if the number is sparse and non-zero if not sparse.
Below is the implementation of the above approach:
// C++ program to check if n is sparse or not
#include <bits/stdc++.h>
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 code
int main()
{
// Function call
cout << checkSparse(72) << endl;
cout << checkSparse(12) << endl;
cout << checkSparse(2) << endl;
cout << checkSparse(3) << endl;
return 0;
}
// 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)
{
// Function call
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.
# 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
if __name__ == "__main__":
# Function call
print(checkSparse(72))
print(checkSparse(12))
print(checkSparse(2))
print(checkSparse(30))
# This code is contributed
# by Anant Agarwal.
// C# Code to Check if a given
// 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()
{
// Function call
Console.WriteLine(checkSparse(72));
Console.WriteLine(checkSparse(12));
Console.WriteLine(checkSparse(2));
Console.WriteLine(checkSparse(3));
}
}
// This code is contributed by Sam007.
<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>
<?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
// Function call
echo checkSparse(72), "\n";
echo checkSparse(12), "\n";
echo checkSparse(2), "\n";
echo checkSparse(3), "\n";
// This code is contributed by Ajit.
?>
Output
1 0 1 0
Time Complexity: O(1)
Auxiliary Space: O(1)
Note: Instead of the right shift, we could have used the left shift also, but the left shift might lead to an overflow in some cases.
