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, n2q= n / mn1= m * q if (n * m) > 0n2= m * (q + 1) elsen2= 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` `?>` |

## Javascript

`<script>` `// Javascript 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` ` ` `let q = parseInt(n / m);` ` ` ` ` `// 1st possible closest number` ` ` `let n1 = m * q;` ` ` ` ` `// 2nd possible closest number` ` ` `let 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 Code` `let n = 13;` `let m = 4;` `document.write(closestNumber(n, m) + ` `"<br>"` `);` `n = -15;` `m = 6;` `document.write(closestNumber(n, m) + ` `"<br>"` `);` `n = 0;` `m = 8;` `document.write(closestNumber(n, m) + ` `"<br>"` `);` `n = 18;` `m = -7;` `document.write(closestNumber(n, m) + ` `"<br>"` `);` `// This code is contributed by gfgking` `</script>` |

**Output:**

12 -18 0 21

**Time Complexity: **O(1)

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