Closest Number

Problem

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.

Examples:

Example 1:

Input:
N = 13 , M = 4
Output:
12
Explanation:
12 is the Closest Number to 13 which is divisible by 4.

Example 2:

Input:
N = -15 , M = 6
Output:
-18
Explanation:
-12 and -18 are both similarly close to -15 and divisible by 6. but -18 has the maximum absolute value. So, Output is -18

Your task:

You don't need to read input or print anything. Your task is to complete the function closestNumber() which takes an Integer n as input and returns the answer.

• Expected Time Complexity: $O(1)$
• Expected Auxiliary Space: $O(1)$

Constraints:

• $-10^5 <= N <= 10^5$

Solution

Python

def closestNumber(self, N , M):
    if M == 0:
        return None  
    quotient = N // M
    lower_multiple = M * quotient
    upper_multiple = M * (quotient + 1)
    if abs(N - lower_multiple) < abs(N - upper_multiple):
        return lower_multiple
    elif abs(N - upper_multiple) < abs(N - lower_multiple):
        return upper_multiple
    else:
        return max(lower_multiple, upper_multiple, key=abs) 

Java

static int closestNumber(int N , int M) {
    int i=0;
    while(true){
        if((N-i)%M==0)
            return N-i;
    else if((N+i)%M==0)
            return N+i;
        else   
            i++;    
    }
}

C++

int closestNumber(int N, int M) {
    int r = N % M;
    if (r == 0) {
        return N; 
    } 
    else if (r <= M / 2) {
        return N - r; 
    } 
    else {
        return N + (M - r); 
    }
}

C

int closestNumber(int N, int M) {
    int r = N % M;
    if (r == 0) {
        return N; 
    } 
    else if (r <= M / 2) {
        return N - r; 
    } 
    else {
        return N + (M - r); 
    }
}

• Time Complexity: $O(1)$
• Auxiliary Space: $O(1)$