Count divisors which generates same Quotient and Remainder

Given a positive integer N, the task is to find the count of all the numbers M such that when the number N is divided by M, the quotient is equal to its remainder i.e (?N/M? = N mod M) where ? ? denotes the floor value of a given number.

Examples:

Input: N = 8
Output: 2
Explanation: When 8 is divided by 3 and 7, it returns the same Quotient and Remainder.
8 / 3 = 8 % 3, 8 / 7 = 8 % 7. Therefore, the required answer is 2.

Input: N = 1000000000000
Output: 84

Naive Approach: The simplest approach is based on the fact that M can not be greater than N as for any number greater than N, the quotient would be zero. Whereas, its modulo with N will always be N. Therefore, iterate through all the numbers from 1 to N and count all such numbers satisfying the given condition. 

Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: The optimal idea is based on the observation stated below: 

If (?N/M? = N mod M), then M + 1 must be a divisor of N.

Proof for the observation:

If ?N/M? = N mod M, then let N / M be equal to K.
Therefore, N must be equal to K * M + K as K is the quotient as well as the remainder.
N = K * M + K
N = K * (M + 1)
Therefore, for M to be in our answer set, it is necessary that M + 1 is a divisor of N.
Note that M + 1 must be a divisor of N is a necessary condition but not a sufficient condition for ?N/M? = N mod M.

Follow the below steps to solve the problem: 

• Find all divisors of N and store them in an array. This can be computed in O(? N) complexity.
• Check for all divisors by iterating through the array, and if divisor minus 1 satisfy the given condition ?N / M? = N mod M, increase the count.

Below is the implementation of the above approach: 

C++

// C++ program of the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to calculate the
// count of numbers such that it
// gives same quotient and remainder
void countDivisors(long long int n)
{

    int count = 0;

    // Stores divisor of number N.
    vector<long long int> divisor;

    // Iterate through numbers from 2
    // to sqrt(N) and store divisors of N
    for (int i = 2; i <= sqrt(n); i++) {

        if (n % i == 0) {

            if (n / i == i)
                divisor.push_back(i);

            else {

                divisor.push_back(i);
                divisor.push_back(n / i);
            }
        }
    }

    // As N is also divisor of itself
    divisor.push_back(n);

    // Iterate through divisors
    for (auto x : divisor) {
        x -= 1;

        // Checking whether x satisfies the
        // required condition
        if ((n / x) == (n % x))
            count++;
    }

    // Print the count
    cout << count;
}

// Driver Code
int main()
{
    // Given N
    long long int N = 1000000000000;

    // Function Call
    countDivisors(N);

    return 0;
}

Java

// Java program of the above approach
import java.util.*;
import java.io.*;
class GFG {

    // Function to calculate the
    // count of numbers such that it
    // gives same quotient and remainder
    static void countDivisors(int n)
    {
        int count = 0;
        int j = 0;

        // Stores divisor of number N.
        int divisor[] =  new int[n];

        // Iterate through numbers from 2
        // to sqrt(N) and store divisors of N
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                if (n / i == i){
                    divisor[j] = i;
                     j += 1;
                }
                else {
                    divisor[j] = i;
                    divisor[j + 1] = n / i;
                    j += 2;
                }
            }
        }

        // As N is also divisor of itself
        divisor[j] = n;

        // Iterate through divisors
        for (int i = 0; i <= j; i++)
        {
            int x = divisor[i];
            x -= 1;

            // Checking whether x satisfies the
            // required condition
            if ((n / x) == (n % x))
                count++;
        }

        // Print the count
        System.out.print(count);
    }

    // Driver Code
    public static void main (String[] args)
    {

        // Given N
        int N = 10000000;

        // Function Call
        countDivisors(N);
    }
}

// This code is contributed by AnkThon

Python3

# Python3 program of the above approach
from math import floor, sqrt

# Function to calculate the
# count of numbers such that it
# gives same quotient and remainder
def countDivisors(n):

    count = 0

    # Stores divisor of number N.
    divisor = []

    # Iterate through numbers from 2
    # to sqrt(N) and store divisors of N
    for i in range(2, floor(sqrt(n))):
        if (n % i == 0):
            if (n // i == i):
                divisor.append(i)
            else:
                divisor.append(i)
                divisor.append(n // i)

    # As N is also divisor of itself
    divisor.append(n)

    # Iterate through divisors
    for x in divisor:
        x -= 1

        # Checking whether x satisfies the
        # required condition
        if ((n // x) == (n % x)):
            count += 1

    # Print the count
    print(count)

# Driver Code
if __name__ == '__main__':

    # Given N
    N = 1000000000000

    # Function Call
    countDivisors(N)

# This code is contributed by mohit kumar 29

C#

// C# program of the above approach
using System;
class GFG {

    // Function to calculate the
    // count of numbers such that it
    // gives same quotient and remainder
    static void countDivisors(int n)
    {
        int count = 0;
        int j = 0;

        // Stores divisor of number N.
        int []divisor =  new int[n];

        // Iterate through numbers from 2
        // to sqrt(N) and store divisors of N
        for (int i = 2; i <= Math.Sqrt(n); i++) {
            if (n % i == 0) {
                if (n / i == i){
                    divisor[j] = i;
                     j += 1;
                }
                else {
                    divisor[j] = i;
                    divisor[j + 1] = n / i;
                    j += 2;
                }
            }
        }

        // As N is also divisor of itself
        divisor[j] = n;

        // Iterate through divisors
        for (int i = 0; i <= j; i++)
        {
            int x = divisor[i];
            x -= 1;

            // Checking whether x satisfies the
            // required condition
            if ((n / x) == (n % x))
                count++;
        }

        // Print the count
        Console.Write(count);
    }

    // Driver Code
    public static void Main(String[] args)
    {

        // Given N
        int N = 10000000;

        // Function Call
        countDivisors(N);
    }
}

// This code contributed by shikhasingrajput

Javascript

<script>

// Javascript program of the above approach

// Function to calculate the
    // count of numbers such that it
    // gives same quotient and remainder
    function countDivisors(n)
    {
        var count = 0;
        var j = 0;

        // Stores divisor of number N.
        var divisor = Array(n).fill(0);

        // Iterate through numbers from 2
        // to sqrt(N) and store divisors of N
        for (var i = 2; i <= Math.sqrt(n); i++)
        {
            if (n % i == 0) {
                if (parseInt(n / i) == i)
                {
                    divisor[j] = i;
                    j += 1;
                } else {
                    divisor[j] = i;
                    divisor[j + 1] = parseInt(n / i);
                    j += 2;
                }
            }
        }

        // As N is also divisor of itself
        divisor[j] = n;

        // Iterate through divisors
        for (var i = 0; i <= j; i++) {
            var x = divisor[i];
            x -= 1;

            // Checking whether x satisfies the
            // required condition
            if (parseInt(n / x) == parseInt(n % x))
                count++;
        }

        // Print the count
        document.write(count);
    }

    // Driver Code


        // Given N
        var N = 10000000;

        // Function Call
        countDivisors(N);

// This code contributed by aashish1995

</script>

Output:

84

Time Complexity: O(sqrt(N))
Auxiliary Space: O(sqrt(N))