Given two positive integers m and n, the task is to write a program that checks whether m^n is greater than, less than, or equal to n^m.
Examples :
Input: m = 3, n = 10
Output: m^n > n^m
Explanation : 3^10=59049 which is greater than 10^3=1000Input: m = 987654321, n = 123456987
Output: m^n < n^m
A naive approach is to compute m^n and n^m, which causes overflow when m and n are very large.
An efficient approach is to solve this problem using log.
Given LHS = m^n and RHS = n^m.
After taking log on both sides, LHS = n*log(m) and RHS = m*log(n)
Then compare the LHS and RHS.
C++
// CPP program to compare which is greater // m^n or n^m #include <bits/stdc++.h>using namespace std;// function to compare m^n and n^m void check(unsigned long long m, unsigned long long int n) { // m^n double RHS = m * (double)log(n); // n^m double LHS = n * (double)log(m); if ( LHS > RHS ) cout << "m^n > n^m"; else if ( LHS < RHS ) cout << "m^n < n^m"; else cout << "m^n = n^m"; }// Drivers Code int main() { unsigned long long m = 987654321, n = 123456987; // function call to compare m^n and n^m check(m, n); return 0;} |
Java
// Java program to compare which // is greater m^n or n^m import java .io.*;class GFG{// function to compare// m^n and n^m static void check(long m, long n){ // m^n double RHS = m * (double)Math.log(n); // n^m double LHS = n * (double)Math.log(m); if (LHS > RHS) System.out.print("m^n > n^m"); else if (LHS < RHS) System.out.print("m^n < n^m"); else System.out.print("m^n = n^m");}// Driver Code static public void main (String[] args){ long m = 987654321, n = 123456987; // function call to // compare m^n and n^m check(m, n);}}// This code is contributed by anuj_67. |
Python3
# Python3 program to compare # which is greater m^n or n^m import math# function to compare# m^n and n^m def check( m, n): # m^n RHS = m * math.log(n); # n^m LHS = n * math.log(m); if (LHS > RHS): print("m^n > n^m"); elif (LHS < RHS): print("m^n < n^m"); else: print("m^n = n^m");# Driver Code m = 987654321; n = 123456987;# function call to # compare m^n and n^mcheck(m, n);# This code is contributed by mits |
C#
// C# program to compare which // is greater m^n or n^m using System;class GFG{// function to compare// m^n and n^m static void check(ulong m, ulong n){ // m^n double RHS = m * (double)Math.Log(n); // n^m double LHS = n * (double)Math.Log(m); if (LHS > RHS) Console.Write("m^n > n^m"); else if (LHS < RHS) Console.Write("m^n < n^m"); else Console.Write("m^n = n^m");}// Driver Code static public void Main (){ ulong m = 987654321, n = 123456987; // function call to // compare m^n and n^m check(m, n);}}// This code is contributed by anuj_67. |
PHP
<?php// PHP program to compare // which is greater m^n or n^m // function to compare// m^n and n^m function check( $m, $n){ // m^n $RHS = $m * log($n); // n^m $LHS = $n * log($m); if ( $LHS > $RHS ) echo "m^n > n^m"; else if ( $LHS < $RHS ) echo "m^n < n^m"; else echo "m^n = n^m";}// Driver Code $m = 987654321; $n = 123456987;// function call to // compare m^n and n^mcheck($m, $n);// This code is contributed by anuj_67.?> |
Javascript
<script>// javascript program to compare which// is greater m^n or n^m// function to compare// m^n and n^mfunction check( m, n){ // m^n var RHS = m * Math.log(n); // n^m var LHS = n * Math.log(m); if (LHS > RHS){ document.write("m^n > n^m"); } else if (LHS < RHS) { document.write("m^n < n^m"); } else { document.write("m^n = n^m"); }} // Driver Code var m = 987654321 ; var n = 123456987; // function call to // compare m^n and n^m check(m, n); </script> |
Output
m^n < n^m
Time Complexity: O(max(log(m), log(n)))
Auxiliary Space: O(max(log(m), log(n))), due to the recursion call stack.
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