Minimum positive integer divisible by C and is not in range [A, B]

Given three positive integers A, B and C. The task is to find the minimum integer X > 0 such that: 

1. X % C = 0 and
2. X must not belong to the range [A, B]

Examples:

Input: A = 2, B = 4, C = 2 
Output: 6
Input: A = 5, B = 10, C = 4 
Output: 4 

Approach:

• If C doesn’t belong to [A, B] i.e. C < A or C > B then C is the required number.
• Else get the first multiple of C greater than B which is the required answer.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

// Function to return the required number
int getMinNum(int a, int b, int c)
{

    // If doesn't belong to the range
    // then c is the required number
    if (c < a || c > b)
        return c;

    // Else get the next multiple of c
    // starting from b + 1
    int x = ((b / c) * c) + c;

    return x;
}

// Driver code
int main()
{
    int a = 2, b = 4, c = 4;
    cout << getMinNum(a, b, c);
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
import java.math.*;
public class GFG
{
    // Function to return the required number
    int getMinNum(int a, int b, int c)
    {

        // If doesn't belong to the range
        // then c is the required number
        if (c < a || c > b)
        {
            return c;
        }

        // Else get the next multiple of c
        // starting from b + 1
        int x = ((b / c) * c) + c;

        return x;
    }

// Driver code
public static void main(String args[])
{
    int a = 2;
    int b = 4;
    int c = 4;
    GFG g = new GFG();
    System.out.println(g.getMinNum(a, b, c));
}
}

// This code is contributed by Shivi_Aggarwal

Python3

# Python3 implementation of the approach
# Function to return the required number
def getMinNum(a, b, c):

    # If doesn't belong to the range
    # then c is the required number
    if (c < a or c > b):
        return c

    # Else get the next multiple of c
    # starting from b + 1
    x = ((b // c) * c) + c

    return x

# Driver code
a, b, c = 2, 4, 4
print(getMinNum(a, b, c))

# This code is contributed by
# Mohit kumar 29

C#

// C# implementation of the approach
using System;

class GFG
{
    // Function to return the required number
    static int getMinNum(int a, int b, int c)
    {

        // If doesn't belong to the range
        // then c is the required number
        if (c < a || c > b)
        {
            return c;
        }

        // Else get the next multiple of c
        // starting from b + 1
        int x = ((b / c) * c) + c;

        return x;
    }

    // Driver code
    static public void Main ()
    {
        int a = 2, b = 4, c = 4;
        Console.WriteLine( getMinNum(a, b, c));
    }
}

// This Code is contributed by ajit..

PHP

<?php
// PHP implementation of the above approach

// Function to return the required number
function getMinNum($a, $b, $c)
{

    // If doesn't belong to the range
    // then c is the required number
    if ($c < $a || $c > $b)
        return $c;

    // Else get the next multiple of c
    // starting from b + 1
    $x = (floor(($b / $c)) * $c) + $c;

    return $x;
}

// Driver code
$a = 2;
$b = 4;
$c = 4;

echo getMinNum($a, $b, $c);

// This code is contributed by Ryuga
?>

Javascript

<script>
// Javascript implementation of the approach

// Function to return the required number
function getMinNum(a, b, c)
{

    // If doesn't belong to the range
    // then c is the required number
    if (c < a || c > b)
        return c;

    // Else get the next multiple of c
    // starting from b + 1
    let x = (parseInt(b / c) * c) + c;

    return x;
}

// Driver code
    let a = 2, b = 4, c = 4;
    document.write(getMinNum(a, b, c));

// This code is contributed by souravmahato348
</script>

Output:

8

Time Complexity: O(1)

Auxiliary Space: O(1)