Given two numbers 
, find which is greater 
.
If 
, print a^b is greater
If 
, print b^a is greater
If 
, print Both are equal
Examples:
Input : 3 5 Output : a^b is greater 3^5 = 243, 5^3 = 125. Since, 243>125, therefore a^b > b^a. Input : 2 4 Output : Both are equal 2^4 = 16, 4^2 = 16. Since, 16=16, therefore a^b = b^a.
Brute Force solution would be to just compute and compare them. But since can be large enough that can not be stored even in long long int, so this solution is not feasible. Also computing to the power n would require at least 
time using the fast exponentiation technique.
Efficient approach would be to use logarithm. We have to compare . If we take log, the problem reduces to comparing 
.
Hence,
If 
, print a^b is greater
If 
, print b^a is greater
If 
, print Both are equal
Below is the implementation of the efficient approach discussed above.
C++
// C++ code for finding greater// between the a^b and b^a#include <bits/stdc++.h>using namespace std;// Function to find the greater valuevoid findGreater(int a, int b){ long double x = (long double)a * (long double)(log((long double)(b))); long double y = (long double)b * (long double)(log((long double)(a))); if (y > x) { cout << "a^b is greater" << endl; } else if (y < x) { cout << "b^a is greater" << endl; } else { cout << "Both are equal" << endl; }}// Driver codeint main(){ int a = 3, b = 5, c = 2, d = 4; findGreater(a, b); findGreater(c, d); return 0;} |
Java
// Java code for finding greater// between the a^b and b^apublic class GFG{ // Function to find the greater value static void findGreater(int a, int b) { double x = (double)a * (double)(Math.log((double)(b))); double y = (double)b * (double)(Math.log((double)(a))); if (y > x) { System.out.println("a^b is greater") ; } else if (y < x) { System.out.println("b^a is greater") ; } else { System.out.println("Both are equal") ; } } // Driver code public static void main(String []args) { int a = 3, b = 5, c = 2, d = 4; findGreater(a, b); findGreater(c, d); } // This code is contributed by Ryuga} |
Python 3
# Python 3 code for finding greater# between the a^b and b^aimport math# Function to find the greater valuedef findGreater(a, b): x = a * (math.log(b)); y = b * (math.log(a)); if (y > x): print ("a^b is greater"); elif (y < x): print("b^a is greater"); else : print("Both are equal"); # Driver codea = 3;b = 5;c = 2;d = 4;findGreater(a, b);findGreater(c, d);# This code is contributed # by Shivi_Aggarwal |
C#
// C# code for finding greater// between the a^b and b^a using System;public class GFG{ // Function to find the greater value static void findGreater(int a, int b) { double x = (double)a * (double)(Math.Log((double)(b))); double y = (double)b * (double)(Math.Log((double)(a))); if (y > x) { Console.Write("a^b is greater\n") ; } else if (y < x) { Console.Write("b^a is greater"+"\n") ; } else { Console.Write("Both are equal") ; } } // Driver code public static void Main() { int a = 3, b = 5, c = 2, d = 4; findGreater(a, b); findGreater(c, d); } } |
PHP
<?php// PHP code for finding greater// between the a^b and b^a// Function to find the greater valuefunction findGreater($a, $b){ $x = (double)$a * (double)(log((double)($b))); $y = (double)$b * (double)(log((double)($a))); if ($y > $x) { echo "a^b is greater", "\n"; } else if ($y < $x) { echo "b^a is greater", "\n" ; } else { echo "Both are equal", "\n" ; }}// Driver code$a = 3;$b = 5;$c = 2;$d = 4;findGreater($a, $b);findGreater($c, $d);// This code is contributed by ajit?> |
Javascript
<script>// javascript code for finding greater// between the a^b and b^a // Function to find the greater value function findGreater(a , b) { var x = a * (Math.log( (b))); var y = b * (Math.log( (a))); if (y > x) { document.write("a^b is greater<br/>"); } else if (y < x) { document.write("b^a is greater<br/>"); } else { document.write("Both are equal<br/>"); } } // Driver code var a = 3, b = 5, c = 2, d = 4; findGreater(a, b); findGreater(c, d);// This code is contributed by todaysgaurav </script> |
Output:
a^b is greater Both are equal
Time Complexity: O(logn), since it is using inbuilt log function
Auxiliary Space: O(1), since no extra space has been taken.
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!
