Program to find greater value between a^n and b^n

Given three integers a, b and n, the task is to find out greater value between aⁿ and bⁿ.
Examples:

Input: a = 3, b = 4, n = 5 
Output: b^n is greater than a^n 
Value of aⁿ is 243 and the value of bⁿ is 1024. 
So, bⁿ is greater than aⁿ.
Input: a = -3, b = 2, n = 4 
Output: a^n is greater than b^n 
Value of aⁿ is 243 and the value of bⁿ is 16. 
So, aⁿ is greater than bⁿ. 

Basic Approach: For every value of a, b and n, calculate the values of aⁿ and bⁿ. Then compare the result obtained and display the according to output.
The problem with this approach arises when there are large values of a, b and n. For large values of a, n, calculating aⁿ can exceed the limit of integer which will cause integer overflow.
Better approach is to check the value of n. 

• If n is even then calculate the absolute value of a and b.
• If n is odd then take the given value as it is.
• Now check if a is equal to b. If yes, print 0.
• If a is greater than b, print 1.
• Otherwise, print 2.

C++

// C++ code for finding greater
// between the a^n and b^n
#include <bits/stdc++.h>
using namespace std;

// Function to find the greater value
void findGreater(int a, int b, int n)
{
    // If n is even
    if (!(n & 1)) {

        a = abs(a);
        b = abs(b);
    }
    if (a == b)
        cout << "a^n is equal to b^n";

    else if (a > b)
        cout << "a^n is greater than b^n";

    else
        cout << "b^n is greater than a^n";
}

// Driver code
int main()
{
    int a = 12, b = 24, n = 5;
    findGreater(a, b, n);
    return 0;
}

Java

// JAVA code for finding greater
// between the a^n and b^n
import java.io.*;

class GFG
{
    // Function to find
    // the greater value
    static void findGreater(int a,
                            int b, int n)
{
    // If n is even
    if (!((n & 1) > 0))
    {

        a = Math.abs(a);
        b = Math.abs(b);
    }
    if (a == b)
        System.out.println("a^n is " +
                           "equal to b^n");

    else if (a > b)
        System.out.println("a^n is greater " +
                                  "than b^n");

    else
        System.out.println("b^n is greater " +
                                  "than a^n");
}

// Driver code
public static void main (String[] args)
{
int a = 12, b = 24, n = 5;
findGreater(a, b, n);
}
}

// This code is contributed
// by shiv_bhakt.

Python3

# Python3 code for finding greater
# between the a^n and b^n
import math

# Function to find the greater value
def findGreater(a, b, n):

    # If n is even
    if ((n & 1) > 0):

        a = abs(a);
        b = abs(b);
    if (a == b):
        print("a^n is equal to b^n");

    elif (a > b):
        print("a^n is greater than b^n");

    else:
        print("b^n is greater than a^n");

# Driver code
a = 12;
b = 24;
n = 5;
findGreater(a, b, n);

# This code is contributed by mits

C#

// C# code for finding greater
// between the a^n and b^n
using System;

class GFG
{
    // Function to find
    // the greater value
    static void findGreater(int a,
                            int b,
                            int n)
    {
    // If n is even
    if (!((n & 1) > 0))
    {

        a = Math.Abs(a);
        b = Math.Abs(b);
    }

    if (a == b)
        Console.WriteLine("a^n is " +
                          "equal to b^n");

    else if (a > b)
        Console.WriteLine("a^n is greater " +
                                 "than b^n");

    else
        Console.WriteLine("b^n is greater " +
                                 "than a^n");
}

// Driver code
public static void Main()
{
    int a = 12, b = 24, n = 5;
    findGreater(a, b, n);
}
}

// This code is contributed
// by shiv_bhakt.

PHP

<?php
// PHP code for finding greater
// between the a^n and b^n

// Function to find
// the greater value
function findGreater($a, $b, $n)
{
    // If n is even
    if (!($n & 1))
    {

        $a = abs($a);
        $b = abs($b);
    }
    if ($a == $b)
        echo "a^n is equal to b^n";

    else if ($a > $b)
        echo "a^n is greater than b^n";

    else
        echo "b^n is greater than a^n";
}

// Driver code
$a = 12; $b = 24; $n = 5;
findGreater($a, $b, $n);

// This code is contributed by ajit
?>

Javascript

<script>

// Javascript code for finding greater
// between the a^n and b^n

    // Function to find
    // the greater value
    function findGreater(a , b , n)
    {
        // If n is even
        if (!((n & 1) > 0)) {

            a = Math.abs(a);
            b = Math.abs(b);
        }
        if (a == b)
            document.write("a^n is " + "equal to b^n");

        else if (a > b)
            document.write("a^n is greater " + "than b^n");

        else
            document.write("b^n is greater " + "than a^n");
    }

    // Driver code

        var a = 12, b = 24, n = 5;
        findGreater(a, b, n);

// This code is contributed by todaysgaurav

</script>

Output:

b^n is greater than a^n

Time Complexity:O(1)

Auxiliary Space: O(1)