Given three positive integers A, B and C. The task is to find the minimum integer X > 0 such that:
- X % C = 0 and
- 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 numberint 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 codeint main(){ int a = 2, b = 4, c = 4; cout << getMinNum(a, b, c); return 0;} |
Java
// Java implementation of the approachimport 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 codepublic 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 numberdef 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 codea, b, c = 2, 4, 4print(getMinNum(a, b, c))# This code is contributed by# Mohit kumar 29 |
C#
// C# implementation of the approachusing 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 numberfunction 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)
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