Microsoft CoCubes Java Programming Questions – 4 Find the number closest to n and divisible by m

Microsoft CoCubes Java Programming Questions – 4

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

Program :

// Microsoft CoCubes Java Programming Questions – 4

import java.util.*;

class Main

{

// 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[])

{

Scanner sc=new Scanner (System.in);

int n=sc.nextInt();

int m=sc.nextInt();

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

}

For Output check the Video :

Comments

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