## 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

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));
}
}