Check if a given number is sparse or not Last Updated : 14 Sep, 2024 Comments Improve Suggest changes Like Article Like Report Try it on GfG Practice 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 = 72Output: trueExplanation: Binary representation of 72 is 01001000. There are no two consecutive 1's in binary representationInput: x = 12Output: falseExplanation: 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. C++ #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; } Java // "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 # 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# // 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 JavaScript // 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. OutputNot SparseTime 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++ // 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 // 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. 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 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# // 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. 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> 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 // Function call echo checkSparse(72), "\n"; echo checkSparse(12), "\n"; echo checkSparse(2), "\n"; echo checkSparse(3), "\n"; // This code is contributed by Ajit. ?> Output1 0 1 0Time 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. Comment More infoAdvertise with us Next Article Check if a given number is sparse or not K kartik Improve Article Tags : Bit Magic DSA Practice Tags : Bit Magic Similar Reads Bitwise Algorithms Bitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit 4 min read Introduction to Bitwise Algorithms - Data Structures and Algorithms Tutorial Bit stands for binary digit. A bit is the basic unit of information and can only have one of two possible values that is 0 or 1. In our world, we usually with numbers using the decimal base. In other words. we use the digit 0 to 9 However, there are other number representations that can be quite use 15+ min read Bitwise Operators in C In C, bitwise operators are used to perform operations directly on the binary representations of numbers. These operators work by manipulating individual bits (0s and 1s) in a number.The following 6 operators are bitwise operators (also known as bit operators as they work at the bit-level). They are 6 min read Bitwise Operators in Java In Java, Operators are special symbols that perform specific operations on one or more than one operands. They build the foundation for any type of calculation or logic in programming.There are so many operators in Java, among all, bitwise operators are used to perform operations at the bit level. T 6 min read Python Bitwise Operators Python bitwise operators are used to perform bitwise calculations on integers. 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