Skip to content
Related Articles

Related Articles

Count of distinct remainders when N is divided by all the numbers from the range [1, N]
  • Last Updated : 03 May, 2021

Given an integer N, the task is to find the count of total distinct remainders which can be obtained when N is divided by every element from the range [1, N].
Examples: 
 

Input: N = 5 
Output:
5 % 1 = 0 
5 % 2 = 1 
5 % 3 = 2 
5 % 4 = 1 
5 % 5 = 0 
The distinct remainders are 0, 1 and 2.
Input: N = 44 
Output: 22 
 

 

Approach: It can be easily observed that for even values of N the number of distinct remainders will be N / 2 and for odd values of N it will be 1 + ⌊N / 2⌋.
Below is the implementation of the above approach:
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of distinct
// remainders that can be obtained when
// n is divided by every element
// from the range [1, n]
int distinctRemainders(int n)
{
 
    // If n is even
    if (n % 2 == 0)
        return (n / 2);
 
    // If n is odd
    return (1 + (n / 2));
}
 
// Driver code
int main()
{
    int n = 5;
 
    cout << distinctRemainders(n);
 
    return 0;
}

Java




// Java implementation of the above approach
class GFG
{
 
// Function to return the count of distinct
// remainders that can be obtained when
// n is divided by every element
// from the range [1, n]
static int distinctRemainders(int n)
{
 
    // If n is even
    if (n % 2 == 0)
        return (n / 2);
 
    // If n is odd
    return (1 + (n / 2));
}
 
// Driver code
public static void main(String[] args)
{
    int n = 5;
    System.out.println(distinctRemainders(n));
}
}
 
// This code is contributed by Mohit Kumar

Python3




# Python3 implementation of the approach
 
# Function to return the count of distinct
# remainders that can be obtained when
# n is divided by every element
# from the range [1, n]
def distinctRemainders(n):
     
    # If n is even
    if n % 2 == 0:
        return n//2
     
    # If n is odd
    return ((n//2)+1)
 
# Driver code
if __name__=="__main__":
 
    n = 5
    print(distinctRemainders(n))

C#




// C# implementation of the above approach
using System;
 
class GFG
{
     
    // Function to return the count of distinct
    // remainders that can be obtained when
    // n is divided by every element
    // from the range [1, n]
    static int distinctRemainders(int n)
    {
     
        // If n is even
        if (n % 2 == 0)
            return (n / 2);
     
        // If n is odd
        return (1 + (n / 2));
    }
     
    // Driver code
    public static void Main()
    {
        int n = 5;
        Console.WriteLine(distinctRemainders(n));
    }
}
 
// This code is contributed by AnkitRai01

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the count of distinct
// remainders that can be obtained when
// n is divided by every element
// from the range [1, n]
function distinctRemainders(n)
{
 
    // If n is even
    if (n % 2 == 0)
        return parseInt(n / 2);
 
    // If n is odd
    return (1 + parseInt(n / 2));
}
 
// Driver code
    let n = 5;
 
    document.write(distinctRemainders(n));
 
</script>
Output: 
3

 

Time Complexity: O(1)
 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :