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Find the multiple of x which is closest to a^b
  • Difficulty Level : Easy
  • Last Updated : 03 May, 2021

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

 

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