Find the number closest to n and divisible by m

Given two integers n and m. The problem is to find the number closest to n and divisible by m. If there are more than one such number, then output the one having maximum absolute value. If n is completely divisible by m, then output n only. Time complexity of O(1) is required.

Constraints: m != 0

Examples:



Input : n = 13, m = 4
Output : 12

Input : n = -15, m = 6
Output : -18
Both -12 and -18 are closest to -15, but
-18 has the maximum absolute value.

Source: Microsoft Interview experience | Set 125.

We find value of n/m. Let this value be q. Then we find closest of two possibilities. One is q * m other is (m * (q + 1)) or (m * (q – 1)) depending on whether one of the given two numbers is negative or not.

Algorithm:

closestNumber(n, m)
    Declare q, n1, n2
    q = n / m
    n1 = m * q

    if (n * m) > 0
        n2 = m * (q + 1)
    else
        n2 = m * (q - 1)

    if abs(n-n1) < abs(n-n2)
        return n1
    return n2

C++

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// C++ implementation to find the number closest to n
// and divisible by m
#include <bits/stdc++.h>
  
using namespace std;
  
// function to find the number closest to n
// and divisible by m
int closestNumber(int n, int m)
{
    // find the quotient
    int q = n / m;
      
    // 1st possible closest number
    int n1 = m * q;
      
    // 2nd possible closest number
    int n2 = (n * m) > 0 ? (m * (q + 1)) : (m * (q - 1));
      
    // if true, then n1 is the required closest number
    if (abs(n - n1) < abs(n - n2))
        return n1;
      
    // else n2 is the required closest number    
    return n2;    
}
  
// Driver program to test above
int main()
{
    int n = 13, m = 4;
    cout << closestNumber(n, m) << endl;
      
    n = -15; m = 6;
    cout << closestNumber(n, m) << endl;
      
    n = 0; m = 8;
    cout << closestNumber(n, m) << endl;
      
    n = 18; m = -7;
    cout << closestNumber(n, m) << endl;
      
    return 0;
} 

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Java














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// Java implementation to find the number closest to n 


// and divisible by m 


public class close_to_n_divisible_m { 


      


    // function to find the number closest to n 


    // and divisible by m 


    static int closestNumber(int n, int m) 


    { 


        // find the quotient 


        int q = n / m; 


           


        // 1st possible closest number 


        int n1 = m * q; 


           


        // 2nd possible closest number 


        int n2 = (n * m) > 0 ? (m * (q + 1)) : (m * (q - 1)); 


           


        // if true, then n1 is the required closest number 


        if (Math.abs(n - n1) < Math.abs(n - n2)) 


            return n1; 


           


        // else n2 is the required closest number     


        return n2;     


    } 


       


    // Driver program to test above 


    public static void main(String args[]) 


    { 


        int n = 13, m = 4; 


        System.out.println(closestNumber(n, m)); 


           


        n = -15; m = 6; 


        System.out.println(closestNumber(n, m)); 


           


        n = 0; m = 8; 


        System.out.println(closestNumber(n, m)); 


           


        n = 18; m = -7; 


        System.out.println(closestNumber(n, m)); 


    }  


} 


// This code is contributed by Sumit Ghosh 



















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Python3














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# Python 3 implementation to find 


# the number closest to n 


  


# Function to find the number closest  


# to n and divisible by m 


def closestNumber(n, m) : 


    # Find the quotient 


    q = int(n / m) 


      


    # 1st possible closest number 


    n1 = m * q 


      


    # 2nd possible closest number 


    if((n * m) > 0) : 


        n2 = (m * (q + 1))  


    else : 


        n2 = (m * (q - 1)) 


      


    # if true, then n1 is the required closest number 


    if (abs(n - n1) < abs(n - n2)) : 


        return n1 


      


    # else n2 is the required closest number  


    return n2 


      


      


# Driver program to test above 


n = 13; m = 4


print(closestNumber(n, m)) 


  


n = -15; m = 6


print(closestNumber(n, m)) 


  


n = 0; m = 8


print(closestNumber(n, m)) 


  


n = 18; m = -7


print(closestNumber(n, m)) 


  


# This code is contributed by Nikita tiwari. 



















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C#














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// C# implementation to find the 


// number closest to n and divisible by m 


using System; 


  


class GFG { 


  


    // function to find the number closest to n 


    // and divisible by m 


    static int closestNumber(int n, int m) 


    { 


        // find the quotient 


        int q = n / m; 


  


        // 1st possible closest number 


        int n1 = m * q; 


  


        // 2nd possible closest number 


        int n2 = (n * m) > 0 ? (m * (q + 1)) : (m * (q - 1)); 


  


        // if true, then n1 is the required closest number 


        if (Math.Abs(n - n1) < Math.Abs(n - n2)) 


            return n1; 


  


        // else n2 is the required closest number 


        return n2; 


    } 


  


    // Driver program to test above 


    public static void Main() 


    { 


        int n = 13, m = 4; 


        Console.WriteLine(closestNumber(n, m)); 


  


        n = -15; 


        m = 6; 


        Console.WriteLine(closestNumber(n, m)); 


  


        n = 0; 


        m = 8; 


        Console.WriteLine(closestNumber(n, m)); 


  


        n = 18; 


        m = -7; 


        Console.WriteLine(closestNumber(n, m)); 


    } 


} 


  


// This code is contributed by Sam007 



















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PHP














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<?php 


// PHP implementation to find  


// the number closest to n and 


// divisible by m  


  


// function to find the number  


// closest to n and divisible by m  


function closestNumber($n, $m)  


{  


    // find the quotient  


    $q = (int) ($n / $m);  


      


    // 1st possible closest number  


    $n1 = $m * $q;  


      


    // 2nd possible closest number  


    $n2 = ($n * $m) > 0 ?  


        ($m * ($q + 1)) : ($m * ($q - 1));  


      


    // if true, then n1 is the  


    // required closest number  


    if (abs($n - $n1) < abs($n - $n2))  


        return $n1;  


      


    // else n2 is the required  


    // closest number  


    return $n2;  


}  


  


// Driver Code 


$n = 13; 


$m = 4;  


echo closestNumber($n, $m), "\n";  


  


$n = -15; 


$m = 6;  


echo closestNumber($n, $m), "\n";  


  


$n = 0;  


$m = 8;  


    echo closestNumber($n, $m), "\n";  


  


$n = 18; 


$m = -7;  


    echo closestNumber($n, $m), "\n";  


  


// This code is contributed by jit_t 


?> 



















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Output:


12
-18
0
21



Time Complexity: O(1)


This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


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