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.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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++

// 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; 
}  

Java

// 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 

Python3

# 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. 

C#

// 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 

PHP

<?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 
?> 

Output:

Time Complexity: O(1)

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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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