Minimum value exceeding X whose count of divisors has different parity with count of divisors of X

Given an integer X, the task is to determine the minimum value of Y greater than X, such that count of divisors of X and Y have different parities.

Examples:

Input: X = 5
Output: 9
Explanation: The count of divisors of 5 and 9 are 2 and 3 respectively, which are of different parities.

Input: X = 9
Output: 10
Explanation: The counts of divisors of 9 and 10 are 3 and 4, which are of different parities.

Naive Approach: The simplest approach to solve the problem is to iterate each number starting from X + 1 until an element with count of the divisors with parity opposite to that of X is obtained.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

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

// Function to count divisors of n
int divisorCount(int n)
{
    int x = 0;
    for (int i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
            if (i == n / i)
                x++;
            else
                x += 2;
        }
    }
    return x;
}

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
int minvalue_y(int x)
{
    // Divisor count of x
    int a = divisorCount(x);
    int y = x + 1;

    // Iterate from x + 1 and
    // check for each element
    while ((a & 1)
           == (divisorCount(y) & 1))
        y++;
    return y;
}

// Driver Code
int main()
{
    // Given X
    int x = 5;

    // Function call
    cout << minvalue_y(x) << endl;
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to count divisors of n
static int divisorCount(int n)
{
    int x = 0;
    for(int i = 1; i <= Math.sqrt(n); i++)
    {
        if (n % i == 0)
        {
            if (i == n / i)
                x++;
            else
                x += 2;
        }
    }
    return x;
}

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
static int minvalue_y(int x)
{

    // Divisor count of x
    int a = divisorCount(x);
    int y = x + 1;

    // Iterate from x + 1 and
    // check for each element
    while ((a & 1) == (divisorCount(y) & 1))
        y++;

    return y;
}

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

    // Given X
    int x = 5;

    // Function call
    System.out.println(minvalue_y(x));
}
}

// This code is contributed by chitranayal

C#

// C# program for the above approach
using System;

class GFG{

// Function to count divisors of n
static int divisorCount(int n)
{
    int x = 0;
    for(int i = 1; i <= Math.Sqrt(n); i++)
    {
        if (n % i == 0)
        {
            if (i == n / i)
                x++;
            else
                x += 2;
        }
    }
    return x;
}

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
static int minvalue_y(int x)
{

    // Divisor count of x
    int a = divisorCount(x);
    int y = x + 1;

    // Iterate from x + 1 and
    // check for each element
    while ((a & 1) == (divisorCount(y) & 1))
        y++;

    return y;
}

// Driver Code
public static void Main()
{

    // Given X
    int x = 5;

    // Function call
    Console.WriteLine(minvalue_y(x));
}
}

// This code is contributed by susmitakundugoaldanga

Python3

# Python program for the above approach

# Function to count divisors of n
def divisorCount(n):
    x = 0;
    for i in range(1, n):
        if (n % i == 0):
            if (i == n // i):
                x += 1;
            else:
                x += 2;
        if(i * i > n):
            break;

    return x;

# Function to find the minimum
# value exceeding x whose count
# of divisors has different parity
# with count of divisors of X
def minvalue_y(x):

    # Divisor count of x
    a = divisorCount(x);
    y = x + 1;

    # Iterate from x + 1 and
    # check for each element
    while ((a & 1) == (divisorCount(y) & 1)):
        y += 1;

    return y;

# Driver Code
if __name__ == '__main__':

    # Given X
    x = 5;

    # Function call
    print(minvalue_y(x));

# This code is contributed by 29AjayKumar

Javascript

<script>

// javascript program of the above approach

// Function to count divisors of n
function divisorCount(n)
{
    let x = 0;
    for(let i = 1; i <= Math.sqrt(n); i++)
    {
        if (n % i == 0)
        {
            if (i == n / i)
                x++;
            else
                x += 2;
        }
    }
    return x;
}

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
function minvalue_y(x)
{

    // Divisor count of x
    let a = divisorCount(x);
    let y = x + 1;

    // Iterate from x + 1 and
    // check for each element
    while ((a & 1) == (divisorCount(y) & 1))
        y++;

    return y;
}

    // Driver Code

     // Given X
    let x = 5;

    // Function call
    document.write(minvalue_y(x));

// This code is contributed by target_2.
</script>

Output:

9

Time Complexity: O((1+√X)²)
Auxiliary Space: O(1)

Efficient Approach: The problem can be solved based on the following observations:

• If X is a perfect square, then X + 1 will be the answer.
• Otherwise, (1 + √X)² will be the answer.

Follow the steps below to solve the problem:

1. Check if X is a perfect square. If found to be true, print X + 1.
2. Otherwise, print (1 + floor(√X))²).

Below is the implementation of the above approach:

C++

// C++ program for the above approach

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

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
int minvalue_y(int x)
{
    // Check if x is
    // perfect square
    int n = sqrt(x);
    if (n * n == x)
        return x + 1;

    return pow(n + 1, 2);
}

// Driver Code
int main()
{
    int x = 5;
    cout << minvalue_y(x) << endl;
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
static int minvalue_y(int x)
{

    // Check if x is
    // perfect square
    int n = (int)Math.sqrt(x);
    if (n * n == x)
        return x + 1;

    return (int)Math.pow(n + 1, 2);
}

// Driver Code
public static void main(String[] args)
{
    int x = 5;

    System.out.print(minvalue_y(x));
}
}

// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach

# Function to find the minimum
# value exceeding x whose count
# of divisors has different parity
# with count of divisors of X
def minvalue_y(x):

    # Check if x is
    # perfect square
    n = int(pow(x, 1 / 2))

    if (n * n == x):
        return x + 1

    return(pow(n + 1, 2))

# Driver Code
if __name__ == '__main__':

    x = 5

    print(minvalue_y(x))

# This code is contributed by Rajput-Ji

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
static int minvalue_y(int x)
{

    // Check if x is
    // perfect square
    int n = (int)Math.Sqrt(x);
    if (n * n == x)
        return x + 1;

    return (int)Math.Pow(n + 1, 2);
}

// Driver Code
public static void Main()
{
    int x = 5;

    Console.WriteLine(minvalue_y(x));
}
}

// This code is contributed by code_hunt

Javascript

<script>

// JavaScript program to implement
// the above approach

// Function to find the minimum
// value exceeding x whose count
// of divisors has different parity
// with count of divisors of X
function minvalue_y(x)
{

    // Check if x is
    // perfect square
    let n = Math.floor(Math.sqrt(x));
    if (n * n == x)
        return x + 1;

    return Math.floor(Math.pow(n + 1, 2));
}

// Driver code

    let x = 5;

    document.write(minvalue_y(x));

</script>

Output:

9

Time Complexity: O(logx) for input x because using inbuilt sqrt function
Auxiliary Space: O(1)