Given a number n, we need to check if it can be expressed as 2x + 2y or not . Here x and y can be equal.
Examples :
Input : 24 Output : Yes Explanation: 24 can be expressed as 24 + 23 Input : 13 output : No Explanation: It is not possible to express 13 as sum of two powers of 2.
If we take few examples, we can notice that a number can be expressed in the form of 2^x + 2^y if the number is already a power of 2 ( for n > 1 ) or the remainder we get after subtracting previous power of two from the number is also a power of 2.
Below is the implementation of above idea
C++
// CPP code to check if a number can be // expressed as 2^x + 2^y#include <bits/stdc++.h>using namespace std;// Utility function to check if// a number is power of 2 or notbool isPowerOfTwo(int n){ return (n && !(n & (n - 1)));}// Utility function to determine the// value of previous power of 2int previousPowerOfTwo(int n){ while (n & n - 1) { n = n & n - 1; } return n;}// function to check if n can be expressed // as 2^x + 2^y or notbool checkSum(int n){ // if value of n is 0 or 1 // it can not be expressed as // 2^x + 2^y if (n == 0 || n == 1) return false; // if a number is power of 2 // then it can be expressed as // 2^x + 2^y else if (isPowerOfTwo(n)) { cout << " " << n / 2 << " " << n / 2; return true; } else { // if the remainder after // subtracting previous power of 2 // is also a power of 2 then // it can be expressed as // 2^x + 2^y int x = previousPowerOfTwo(n); int y = n - x; if (isPowerOfTwo(y)) { cout << " " << x << " " << y; return true; } } return false;}// driver codeint main(){ int n1 = 20; if (checkSum(n1) == false) cout << "No"; cout << endl; int n2 = 11; if (checkSum(n2) == false) cout << "No"; return 0;} |
Java
// Java code to check if a number// can be expressed as 2^x + 2^yclass GFG { // Utility function to check if // a number is power of 2 or not static boolean isPowerOfTwo(int n) { return n != 0 && ((n & (n - 1)) == 0); } // Utility function to determine the // value of previous power of 2 static int previousPowerOfTwo(int n) { while ((n & n - 1) > 1) { n = n & n - 1; } return n; } // function to check if // n can be expressed as // 2^x + 2^y or not static boolean checkSum(int n) { // if value of n is 0 or 1 // it can not be expressed as // 2^x + 2^y if (n == 0 || n == 1) return false; // if a number is power of 2 // it can be expressed as // 2^x + 2^y else if (isPowerOfTwo(n)) { System.out.println(n / 2 + " " + n / 2); } else { // if the remainder after // subtracting previous power of 2 // is also a power of 2 then // it can be expressed as // 2^x + 2^y int x = previousPowerOfTwo(n); int y = n - x; if (isPowerOfTwo(y)) { System.out.println(x + " " + y); return true; } } return false; } // driver code public static void main(String[] argc) { int n1 = 20; if (checkSum(n1) == false) System.out.println("No"); System.out.println(); int n2 = 11; if (checkSum(n2) == false) System.out.println("No"); }} |
Python3
# Python3 code to check if a number# can be expressed as # 2 ^ x + 2 ^ y # Utility function to check if# a number is power of 2 or notdef isPowerOfTwo( n): return (n and (not(n & (n - 1))) ) # Utility function to determine the# value of previous power of 2 def previousPowerOfTwo( n ): while( n & n-1 ): n = n & n - 1 return n # function to check if# n can be expressed as# 2 ^ x + 2 ^ y or notdef checkSum(n): # if value of n is 0 or 1 # it can not be expressed as # 2 ^ x + 2 ^ y if (n == 0 or n == 1 ): return False # if n is power of two then # it can be expressed as # sum of 2 ^ x + 2 ^ y elif(isPowerOfTwo(n)): print(n//2, n//2) return True # if the remainder after # subtracting previous power of 2 # is also a power of 2 then # it can be expressed as # 2 ^ x + 2 ^ y else: x = previousPowerOfTwo(n) y = n-x; if (isPowerOfTwo(y)): print(x, y) return True else: return False # driver coden1 = 20if (checkSum(n1)): print("No")n2 = 11if (checkSum(n2)): print("No") |
C#
// C# code to check if a number// can be expressed as// 2^x + 2^yusing System;class GFG { // Utility function to check if // a number is power of 2 or not static bool isPowerOfTwo(int n) { return n != 0 && ((n & (n - 1)) == 0); } // Utility function to determine the // value of previous power of 2 static int previousPowerOfTwo(int n) { while ((n & n - 1) > 1) { n = n & n - 1; } return n; } // function to check if // n can be expressed as // 2^x + 2^y or not static bool checkSum(int n) { // if value of n is 0 or 1 // it can not be expressed as // 2^x + 2^y if (n == 0 || n == 1) { Console.WriteLine("No"); return false; } // if a number is power of // it can be expressed as // 2^x + 2^y else if (isPowerOfTwo(n)) { Console.WriteLine(n / 2 + " " + n / 2); return true; } else { // if the remainder after // subtracting previous power of 2 // is also a power of 2 then // it can be expressed as // 2^x + 2^y int x = previousPowerOfTwo(n); int y = n - x; if (isPowerOfTwo(y)) { Console.WriteLine(x + " " + y); return true; } else { return false; } } } // driver code public static void Main() { int n1 = 20; if (checkSum(n1) == false) Console.WriteLine("No"); Console.WriteLine(); int n2 = 11; if (checkSum(n2) == false) Console.WriteLine("No"); }} |
PHP
<?php// PHP code to check if a number // can be expressed as 2^x + 2^y// Utility function to check if// a number is power of 2 or notfunction isPowerOfTwo($n){ return ($n and !($n & ($n - 1)));}// Utility function to determine // the value of previous power of 2function previousPowerOfTwo($n){ while ($n & $n - 1) { $n = $n & $n - 1; } return $n;}// function to check if // n can be expressed // as 2^x + 2^y or notfunction checkSum($n){ // if value of n is 0 or 1 // it can not be expressed // as 2^x + 2^y if ($n == 0 or $n == 1) return false; // if a number is power of 2 // then it can be expressed // as 2^x + 2^y else if (isPowerOfTwo($n)) { echo " " , $n / 2 , " " , $n / 2; return true; } else { // if the remainder after // subtracting previous power // of 2 is also a power of 2 // then it can be expressed // as 2^x + 2^y $x = previousPowerOfTwo($n); $y = $n - $x; if (isPowerOfTwo($y)) { echo $x, " ", $y; return true; } } return false;}// Driver code$n1 = 20;if (checkSum($n1) == false) echo "No";echo"\n";$n2 = 11;if (checkSum($n2) == false) echo "No"; // This code is contributed // by chandan_jnu.?> |
Javascript
<script>// javascript code to check if a number// can be expressed as 2^x + 2^y // Utility function to check if // a number is power of 2 or not function isPowerOfTwo(n) { return n != 0 && ((n & (n - 1)) == 0); } // Utility function to determine the // value of previous power of 2 function previousPowerOfTwo(n) { while ((n & n - 1) > 1) { n = n & n - 1; } return n; } // function to check if // n can be expressed as // 2^x + 2^y or not function checkSum(n) { // if value of n is 0 or 1 // it can not be expressed as // 2^x + 2^y if (n == 0 || n == 1) return false; // if a number is power of 2 // it can be expressed as // 2^x + 2^y else if (isPowerOfTwo(n)) { document.write(n / 2 + " " + n / 2+"<br/>"); } else { // if the remainder after // subtracting previous power of 2 // is also a power of 2 then // it can be expressed as // 2^x + 2^y var x = previousPowerOfTwo(n); var y = n - x; if (isPowerOfTwo(y)) { document.write(x + " " + y + "<br/>"); return true; } } return false; } // driver code var n1 = 20; if (checkSum(n1) == false) document.write("No"); document.write(); var n2 = 11; if (checkSum(n2) == false) document.write("No");// This code is contributed by gauravrajput1.</script> |
Output:
16 4 No
Time Complexity: O(logn)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!