Count the numbers which can convert N to 1 using given operation

Given a positive integer N (N ? 2), the task is to count the number of integers X in the range [2, N] such that X can be used to convert N to 1 using the following operation:

• If N is divisible by X, then update N’s value as N / X.
• Else, update N’s value as N – X.

Examples:

Input: N = 6 
Output: 3 
Explanation:
The following integers can be used to convert N to 1: 
X = 2 => N = 6 -> N = 3 -> N = 1 
X = 5 => N = 6 -> N = 1 
X = 6 => N = 6 -> N = 1

Input: N = 48 
Output: 4

Naive Approach: The naive approach for this problem is to iterate through all the integers from 2 to N and count the number of integers that can convert N to 1.

Below is the implementation of the above approach:

C++

// C++ program to count the numbers
// which can convert N to 1
// using the given operation

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

// Function to count the numbers
// which can convert N to 1
// using the given operation
int countValues(int n)
{

    int answer = 0;

    // Iterate through all the integers
    for (int i = 2; i <= n; i++) {
        int k = n;

        // Check if N can be converted
        // to 1
        while (k >= i) {
            if (k % i == 0)
                k /= i;
            else
                k -= i;
        }

        // Incrementing the count if it can
        // be converted
        if (k == 1)
            answer++;
    }
    return answer;
}

// Driver code
int main()
{
    int N = 6;

    cout << countValues(N);

    return 0;
}

Java

// Java program to count the numbers
// which can convert N to 1
// using the given operation
import java.io.*;
import java.util.*;
class GFG{

// Function to count the numbers
// which can convert N to 1
// using the given operation
static int countValues(int n)
{
    int answer = 0;

    // Iterate through all the integers
    for (int i = 2; i <= n; i++)
    {
        int k = n;

        // Check if N can be converted
        // to 1
        while (k >= i)
        {
            if (k % i == 0)
                k /= i;
            else
                k -= i;
        }

        // Incrementing the count if it can
        // be converted
        if (k == 1)
            answer++;
    }
    return answer;
}

// Driver code
public static void main(String args[])
{
    int N = 6;

    System.out.print(countValues(N));
}
}

// This code is contributed by shivanisinghss2110

Python3

# Python3 program to count the numbers
# which can convert N to 1
# using the given operation

# Function to count the numbers
# which can convert N to 1
# using the given operation
def countValues(n):
    answer = 0

    # Iterate through all the integers
    for i in range(2, n + 1, 1):
        k = n

        # Check if N can be converted
        # to 1
        while (k >= i):
            if (k % i == 0):
                k //= i
            else:
                k -= i

        # Incrementing the count if it can
        # be converted
        if (k == 1):
            answer += 1
    return answer

# Driver code
if __name__ == '__main__':

    N = 6
    print(countValues(N))

# This code is contributed by Samarth

C#

// C# program to count the numbers
// which can convert N to 1
// using the given operation
using System;
class GFG{

// Function to count the numbers
// which can convert N to 1
// using the given operation
static int countValues(int n)
{
    int answer = 0;

    // Iterate through all the integers
    for (int i = 2; i <= n; i++)
    {
        int k = n;

        // Check if N can be converted
        // to 1
        while (k >= i)
        {
            if (k % i == 0)
                k /= i;
            else
                k -= i;
        }

        // Incrementing the count if it can
        // be converted
        if (k == 1)
            answer++;
    }
    return answer;
}

// Driver code
public static void Main()
{
    int N = 6;

    Console.Write(countValues(N));
}
}

// This code is contributed by Nidhi_biet

Javascript

<script>

    // Javascript program to count the numbers
    // which can convert N to 1
    // using the given operation

    // Function to count the numbers
    // which can convert N to 1
    // using the given operation
    function countValues(n)
    {

        let answer = 0;

        // Iterate through all the integers
        for (let i = 2; i <= n; i++) {
            let k = n;

            // Check if N can be converted
            // to 1
            while (k >= i) {
                if (k % i == 0)
                    k /= i;
                else
                    k -= i;
            }

            // Incrementing the count if it can
            // be converted
            if (k == 1)
                answer++;
        }
        return answer;
    }

    let N = 6;

    document.write(countValues(N));

</script>

Output:

3

Time Complexity: O(N), where N is the given number.

Auxiliary Space: O(1)

Efficient Approach: The idea is to observe that if N is not divisible by X initially, then only the subtraction will be carried throughout as after each subtraction N still wouldn’t be divisible by N. Also these operations will stop when N ? X, and the final value of N will be equal to N mod X.

So, for all the numbers from 2 to N, there are only two possible cases :

1. No division operation occurs: For all these numbers, the final value will be equal to N mod X. N will become one in the end only if N mod X = 1. Clearly, for X = N – 1, and all divisors of N – 1, N mod X = 1 holds true.
2. Division operation occurs more than once: Division operation will only occur for divisors on N. For each divisor of N say d, perform the division till N mod d != 0. If finally N mod d = 1, then this will be included in the answer else not (using the property derived from Case 1).

Below is the implementation of the above approach:

C++

// C++ program to count the numbers
// which can convert N to 1
// using the given operation

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

// Function to count the numbers
// which can convert N to 1
// using the given operation
int countValues(int N)
{

    vector<int> div;

    // Store all the divisors of N
    for (int i = 2; i * i <= N; i++) {

        // If i is a divisor
        if (N % i == 0) {
            div.push_back(i);

            // If i is not equal to N / i
            if (N != i * i) {
                div.push_back(N / i);
            }
        }
    }

    int answer = 0;

    // Iterate through all the divisors of
    // N - 1 and count them in answer
    for (int i = 1; i * i <= N - 1; i++) {

        // Check if N - 1 is a divisor
        // or not
        if ((N - 1) % i == 0) {
            if (i * i == N - 1)
                answer++;
            else
                answer += 2;
        }
    }

    // Iterate through all divisors and check
    // for N mod d = 1 or (N-1) mod d = 0
    for (auto d : div) {
        int K = N;
        while (K % d == 0)
            K /= d;
        if ((K - 1) % d == 0)
            answer++;
    }
    return answer;
}

// Driver code
int main()
{
    int N = 6;

    cout << countValues(N);

    return 0;
}

Java

// Java program to count the numbers
// which can convert N to 1
// using the given operation
import java.util.*;

class GFG{

// Function to count the numbers
// which can convert N to 1
// using the given operation
static int countValues(int N)
{
    Vector<Integer> div = new Vector<>();

    // Store all the divisors of N
    for(int i = 2; i * i <= N; i++)
    {

        // If i is a divisor
        if (N % i == 0)
        {
            div.add(i);

            // If i is not equal to N / i
            if (N != i * i)
            {
                div.add(N / i);
            }
        }
    }

    int answer = 0;

    // Iterate through all the divisors of
    // N - 1 and count them in answer
    for(int i = 1; i * i <= N - 1; i++)
    {

        // Check if N - 1 is a divisor
        // or not
        if ((N - 1) % i == 0)
        {
            if (i * i == N - 1)
                answer++;
            else
                answer += 2;
        }
    }

    // Iterate through all divisors and check
    // for N mod d = 1 or (N-1) mod d = 0
    for(int d : div)
    {
        int K = N;
        while (K % d == 0)
            K /= d;

        if ((K - 1) % d == 0)
            answer++;
    }
    return answer;
}

// Driver code
public static void main(String[] args)
{
    int N = 6;

    System.out.print(countValues(N));
}
}

// This code is contributed by gauravrajput1

Python3

# Python3 program to count the numbers
# which can convert N to 1
# using the given operation

# Function to count the numbers
# which can convert N to 1
# using the given operation
def countValues(N):

    div = []
    i = 2

    # Store all the divisors of N
    while ((i * i) <= N):

        # If i is a divisor
        if (N % i == 0):
            div.append(i)

            # If i is not equal to N / i
            if (N != i * i):
                div.append(N // i)

        i += 1

    answer = 0
    i = 1

    # Iterate through all the divisors of
    # N - 1 and count them in answer
    while((i * i) <= N - 1):

        # Check if N - 1 is a divisor
        # or not
        if ((N - 1) % i == 0):
            if (i * i == N - 1):
                answer += 1
            else:
                answer += 2

        i += 1

    # Iterate through all divisors and check
    # for N mod d = 1 or (N-1) mod d = 0
    for d in div:
        K = N

        while (K % d == 0):
            K //= d
        if ((K - 1) % d == 0):
            answer += 1

    return answer

# Driver code
if __name__=="__main__":

    N = 6

    print(countValues(N))

# This code is contributed by rutvik_56

C#

// C# program to count the numbers
// which can convert N to 1
// using the given operation
using System;
using System.Collections.Generic;

class GFG{

// Function to count the numbers
// which can convert N to 1
// using the given operation
static int countValues(int N)
{
    List<int> div = new List<int>();

    // Store all the divisors of N
    for(int i = 2; i * i <= N; i++)
    {

        // If i is a divisor
        if (N % i == 0)
        {
            div.Add(i);

            // If i is not equal to N / i
            if (N != i * i)
            {
                div.Add(N / i);
            }
        }
    }

    int answer = 0;

    // Iterate through all the divisors of
    // N - 1 and count them in answer
    for(int i = 1; i * i <= N - 1; i++)
    {

        // Check if N - 1 is a divisor
        // or not
        if ((N - 1) % i == 0)
        {
            if (i * i == N - 1)
                answer++;
            else
                answer += 2;
        }
    }

    // Iterate through all divisors and check
    // for N mod d = 1 or (N-1) mod d = 0
    foreach(int d in div)
    {
        int K = N;
        while (K % d == 0)
            K /= d;

        if ((K - 1) % d == 0)
            answer++;
    }
    return answer;
}

// Driver code
public static void Main(String[] args)
{
    int N = 6;

    Console.Write(countValues(N));
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

// JavaScript program to count the numbers
// which can convert N to 1
// using the given operation

// Function to count the numbers
// which can convert N to 1
// using the given operation
function countValues(N)
{

    var div = [];

    // Store all the divisors of N
    for (var i = 2; i * i <= N; i++) {

        // If i is a divisor
        if (N % i == 0) {
            div.push(i);

            // If i is not equal to N / i
            if (N != i * i) {
                div.push(N / i);
            }
        }
    }

    var answer = 0;

    // Iterate through all the divisors of
    // N - 1 and count them in answer
    for (var i = 1; i * i <= N - 1; i++) {

        // Check if N - 1 is a divisor
        // or not
        if ((N - 1) % i == 0) {
            if (i * i == N - 1)
                answer++;
            else
                answer += 2;
        }
    }

    // Iterate through all divisors and check
    // for N mod d = 1 or (N-1) mod d = 0
    div.forEach(d => {

        var K = N;
        while (K % d == 0)
            K /= d;
        if ((K - 1) % d == 0)
            answer++;
    });
    return answer;
}

// Driver code
var N = 6;
document.write( countValues(N));


</script>

Output:

3

Time Complexity:

Auxiliary Space: O(sqrt(N))