Find the multiple of x which is closest to a^b

Given three integers a, b, and x, the task is to get the multiple of x which is closest to a^b.

Examples:

Input: a = 5, b = 4, x = 3 
Output: 624 
5⁴ = 625 and 624 is the multiple of 3 which is closest to 625

Input: a = 349, b = 1, x = 4 
Output: 348 

Approach:

• Calculate a^b and store it in a variable say num.
• Then, calculate ?num / x? and store it in a variable floor.
• Now the closest element at the left will be closestLeft = x * floor.
• And the closest element on the right will be closestRight = x * (floor + 1).
• Finally, the closest number among them will be min(num – closestLeft, closestRight – num).

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int

// Function to return the multiple of x
// which is closest to a^b
ll getClosest(int a, int b, int x)
{
    ll num = pow(a, b);

    int floor = num / x;

    // Closest element on the left
    ll numOnLeft = x * floor;

    // Closest element on the right
    ll numOnRight = x * (floor + 1);

    // If numOnLeft is closer than numOnRight
    if ((num - numOnLeft) < (numOnRight - num))
        return numOnLeft;

    // If numOnRight is the closest
    else
        return numOnRight;
}

// Driver code
int main()
{
    int a = 349, b = 1, x = 4;
    cout << getClosest(a, b, x) << endl;
    return 0;
}

Java

//Java implementation of the approach

public class GFG {

// Function to return the multiple of x
// which is closest to a^b
    static long getClosest(int a, int b, int x) {
        long num = (long) Math.pow(a, b);

        int floor = (int) (num / x);

        // Closest element on the left
        long numOnLeft = x * floor;

        // Closest element on the right
        long numOnRight = x * (floor + 1);

        // If numOnLeft is closer than numOnRight
        if ((num - numOnLeft) < (numOnRight - num)) {
            return numOnLeft;
        } // If numOnRight is the closest
        else {
            return numOnRight;
        }
    }

    public static void main(String[] args) {
        int a = 349, b = 1, x = 4;
        System.out.println(getClosest(a, b, x));

    }
}

Python3

# Python3 implementation of the approach

# Function to return the multiple of x
# which is closest to a^b
def getClosest(a, b, x) :
    num = pow(a, b)

    floor = num // x

    # Closest element on the left
    numOnLeft = x * floor

    # Closest element on the right
    numOnRight = x * (floor + 1)

    # If numOnLeft is closer than numOnRight
    if ((num - numOnLeft) <
        (numOnRight - num)):
        return numOnLeft

    # If numOnRight is the closest
    else :
        return numOnRight

# Driver code
if __name__ == "__main__" :

    a, b, x = 349, 1, 4
    print(getClosest(a, b, x))

# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;

// #define ll long long int

class GFG
{
// Function to return the multiple of x
// which is closest to a^b
static long getClosest(int a, int b, int x)
{
    int num = (int)Math.Pow(a, b);

    int floor = (int)(num / x);

    // Closest element on the left
    int numOnLeft = (int)(x * floor);

    // Closest element on the right
    int numOnRight = (int)(x * (floor + 1));

    // If numOnLeft is closer than numOnRight
    if ((num - numOnLeft) < (numOnRight - num))
        return numOnLeft;

    // If numOnRight is the closest
    else
        return numOnRight;
}

// Driver code
public static void Main()
{
    int a = 349, b = 1, x = 4;
    Console.WriteLine(getClosest(a, b, x));
}
}

// This code is contributed
// by Akanksha Rai

PHP

<?php
// PHP implementation of the above approach

// Function to return the multiple of x
// which is closest to a^b
function getClosest($a, $b, $x)
{
    $num = pow($a, $b);

    $floor = (int)($num / $x);

    // Closest element on the left
    $numOnLeft = $x * $floor;

    // Closest element on the right
    $numOnRight = $x * ($floor + 1);

    // If numOnLeft is closer than numOnRight
    if (($num - $numOnLeft) <
        ($numOnRight - $num))
        return $numOnLeft;

    // If numOnRight is the closest
    else
        return ceil($numOnRight);
}

// Driver code
$a = 349;
$b = 1;
$x = 4;
echo getClosest($a, $b, $x);

// This code is contributed by jit_t
?>

Javascript

<script>

// Javascript implementation of the approach


// Function to return the multiple of x
// which is closest to a^b
function getClosest( a,  b,  x)
{
    let num = Math.pow(a, b);

    let floor = Math.floor(num / x);

    // Closest element on the left
    let numOnLeft = x * floor;

    // Closest element on the right
    let numOnRight = x * (floor + 1);

    // If numOnLeft is closer than numOnRight
    if ((num - numOnLeft) < (numOnRight - num))
        return numOnLeft;

    // If numOnRight is the closest
    else
        return numOnRight;
}


    // Driver Code

    let a = 349, b = 1, x = 4;
    document.write(getClosest(a, b, x) + "</br>");

</script>

Output:

348

Time Complexity: O(log(b))
Auxiliary Space: O(1)