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)
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