Count Pronic numbers from a given range

Last Updated : 02 Feb, 2022

Given two integers A and B, the task is to count the number of pronic numbers that are present in the range [A, B].

Examples:

Input: A = 3, B = 20
Output: 3
Explanation: The pronic numbers from the range [3, 20] are 6, 12, 20

Input: A = 5000, B = 990000000
Output: 31393

Naive Approach: Follow the given steps to solve the problem:

Initialize the count of pronic numbers to 0.
For every number in the range [A, B], check whether it is a pronic integer or not
If found to be true, increment the count.
Finally, print the count

Below is the implementation of the above approach:

C++14

// C++ program for
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if x
// is a Pronic Number or not
bool checkPronic(int x)
{

    for (int i = 0; i <= (int)(sqrt(x));
         i++) {

        // Check for Pronic Number by
        // multiplying consecutive
        // numbers
        if (x == i * (i + 1)) {
            return true;
        }
    }
    return false;
}

// Function to count pronic
// numbers in the range [A, B]
void countPronic(int A, int B)
{
    // Initialise count
    int count = 0;

    // Iterate from A to B
    for (int i = A; i <= B; i++) {

        // If i is pronic
        if (checkPronic(i)) {

            // Increment count
            count++;
        }
    }

    // Print count
    cout << count;
}

// Driver Code
int main()
{
    int A = 3, B = 20;

    // Function call to count pronic
    // numbers in the range [A, B]
    countPronic(A, B);

    return 0;
}

Java

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

class GFG{

// Function to check if x
// is a Pronic Number or not
static boolean checkPronic(int x)
{

    for (int i = 0; i <= (int)(Math.sqrt(x));
         i++) {

        // Check for Pronic Number by
        // multiplying consecutive
        // numbers
        if ((x == i * (i + 1)) != false) {
            return true;
        }
    }
    return false;
}

// Function to count pronic
// numbers in the range [A, B]
static void countPronic(int A, int B)
{
    // Initialise count
    int count = 0;

    // Iterate from A to B
    for (int i = A; i <= B; i++) {

        // If i is pronic
        if (checkPronic(i) != false) {

            // Increment count
            count++;
        }
    }

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

// Driver Code
public static void main(String args[])
{
    int A = 3, B = 20;

    // Function call to count pronic
    // numbers in the range [A, B]
    countPronic(A, B);
}
}

// This code is contributed by sanjoy_62.

Python3

# Python3 program for the above approach
import math

# Function to check if x
# is a Pronic Number or not
def checkPronic(x) :
    for i in range(int(math.sqrt(x)) + 1): 

        # Check for Pronic Number by
        # multiplying consecutive
        # numbers
        if (x == i * (i + 1)) :
            return True
    return False
  
# Function to count pronic
# numbers in the range [A, B]
def countPronic(A, B) :
    
    # Initialise count
    count = 0

    # Iterate from A to B
    for i in range(A, B + 1):

        # If i is pronic
        if (checkPronic(i) != 0) :

            # Increment count
            count += 1
        
    # Print count
    print(count)

# Driver Code
A = 3
B = 20

# Function call to count pronic
# numbers in the range [A, B]
countPronic(A, B)

# This code is contributed by susmitakundugoaldanga.

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

// Function to check if x
// is a Pronic Number or not
static bool checkPronic(int x)
{

    for (int i = 0; i <= (int)(Math.Sqrt(x));
         i++)
    {

        // Check for Pronic Number by
        // multiplying consecutive
        // numbers
        if ((x == i * (i + 1)) != false)
        {
            return true;
        }
    }
    return false;
}

// Function to count pronic
// numbers in the range [A, B]
static void countPronic(int A, int B)
{
  
    // Initialise count
    int count = 0;

    // Iterate from A to B
    for (int i = A; i <= B; i++)
    {

        // If i is pronic
        if (checkPronic(i) != false) 
        {

            // Increment count
            count++;
        }
    }

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


// Driver Code
public static void Main(String[] args)
{
    int A = 3, B = 20;

    // Function call to count pronic
    // numbers in the range [A, B]
    countPronic(A, B);
}
}

// This code is contributed by code_hunt.

JavaScript

<script>

// Javascript program for
// the above approach

// Function to check if x
// is a Pronic Number or not
function checkPronic(x)
{

    for (let i = 0; i <= parseInt(Math.sqrt(x));
         i++) {

        // Check for Pronic Number by
        // multiplying consecutive
        // numbers
        if (x == i * (i + 1)) {
            return true;
        }
    }
    return false;
}

// Function to count pronic
// numbers in the range [A, B]
function countPronic(A, B)
{
    // Initialise count
    let count = 0;

    // Iterate from A to B
    for (let i = A; i <= B; i++) {

        // If i is pronic
        if (checkPronic(i)) {

            // Increment count
            count++;
        }
    }

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

// Driver Code
let A = 3, B = 20;

// Function call to count pronic
// numbers in the range [A, B]
countPronic(A, B);

</script>

Output:

Time Complexity: O((B - A) * ?(B - A))
Auxiliary Space: O(1)

Efficient Approach: Follow the below steps to solve the problem:

Define a function pronic(N) to find the count of pronic integers which are ? N.
Calculate the square root of N, say X.
If product of X and X - 1 is less than or equal to N, return N.
Otherwise, return N - 1.
Print pronic(B) - pronic(A - 1) to get the count of pronic integers in the range [A, B]

Below is the implementation of the above approach:

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count pronic
// numbers in the range [A, B]
int pronic(int num)
{
    // Check upto sqrt N
    int N = (int)sqrt(num);

    // If product of consecutive
    // numbers are less than equal to num
    if (N * (N + 1) <= num) {
        return N;
    }

    // Return N - 1
    return N - 1;
}

// Function to count pronic
// numbers in the range [A, B]
int countPronic(int A, int B)
{
    // Subtract the count of pronic numbers
    // which are <= (A - 1) from the count
    // f pronic numbers which are <= B
    return pronic(B) - pronic(A - 1);
}

// Driver Code
int main()
{
    int A = 3;
    int B = 20;

    // Function call to count pronic
    // numbers in the range [A, B]
    cout << countPronic(A, B);

    return 0;
}

Java

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

// Function to count pronic
// numbers in the range [A, B]
static int pronic(int num)
{
  
    // Check upto sqrt N
    int N = (int)Math.sqrt(num);

    // If product of consecutive
    // numbers are less than equal to num
    if (N * (N + 1) <= num)
    {
        return N;
    }

    // Return N - 1
    return N - 1;
}

// Function to count pronic
// numbers in the range [A, B]
static int countPronic(int A, int B)
{
  
    // Subtract the count of pronic numbers
    // which are <= (A - 1) from the count
    // f pronic numbers which are <= B
    return pronic(B) - pronic(A - 1);
}

// Driver Code
public static void main(String[] args)
{
    int A = 3;
    int B = 20;

    // Function call to count pronic
    // numbers in the range [A, B]
    System.out.print(countPronic(A, B));

}
}

// This code is contributed by shikhasingrajput

Python3

# Python3 program for the above approach

# Function to count pronic
# numbers in the range [A, B]
def pronic(num) : 

    # Check upto sqrt N
    N = int(num ** (1/2));

    # If product of consecutive
    # numbers are less than equal to num
    if (N * (N + 1) <= num) :
        return N;

    # Return N - 1
    return N - 1;

# Function to count pronic
# numbers in the range [A, B]
def countPronic(A, B) :
    
    # Subtract the count of pronic numbers
    # which are <= (A - 1) from the count
    # f pronic numbers which are <= B
    return pronic(B) - pronic(A - 1);

# Driver Code
if __name__ == "__main__" :

    A = 3;
    B = 20;

    # Function call to count pronic
    # numbers in the range [A, B]
    print(countPronic(A, B));

    # This code is contributed by AnkThon

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

// Function to count pronic
// numbers in the range [A, B]
static int pronic(int num)
{
  
    // Check upto sqrt N
    int N = (int)Math.Sqrt(num);

    // If product of consecutive
    // numbers are less than equal to num
    if (N * (N + 1) <= num)
    {
        return N;
    }

    // Return N - 1
    return N - 1;
}

// Function to count pronic
// numbers in the range [A, B]
static int countPronic(int A, int B)
{
  
    // Subtract the count of pronic numbers
    // which are <= (A - 1) from the count
    // f pronic numbers which are <= B
    return pronic(B) - pronic(A - 1);
}

// Driver Code
public static void Main(String[] args)
{
    int A = 3;
    int B = 20;

    // Function call to count pronic
    // numbers in the range [A, B]
    Console.Write(countPronic(A, B));
}
}

// This code is contributed by 29AjayKumar

JavaScript

<script>

// Javascript program for the above approach

// Function to count pronic
// numbers in the range [A, B]
function pronic(num)
{

    // Check upto sqrt N
    var N = parseInt(Math.sqrt(num));

    // If product of consecutive
    // numbers are less than equal to num
    if (N * (N + 1) <= num) {
        return N;
    }

    // Return N - 1
    return N - 1;
}

// Function to count pronic
// numbers in the range [A, B]
function countPronic(A, B)
{

    // Subtract the count of pronic numbers
    // which are <= (A - 1) from the count
    // f pronic numbers which are <= B
    return pronic(B) - pronic(A - 1);
}

// Driver Code
var A = 3;
var B = 20;

// Function call to count pronic
// numbers in the range [A, B]
document.write(countPronic(A, B));

// This code is contributed by noob2000.
</script>

Output:

Time Complexity: O(log₂(B))
Auxiliary Space: O(1)

Count of Double Prime numbers in a given range L to R

koulick_sadhu

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Count Pronic numbers from a given range

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