Count ordered pairs with product less than N

Given an integer N. The task is to count the number of ordered pairs (a, b) such that a * b < N.

Examples:

Input: N = 5
Output: 8
Ordered Pairs are = (1, 1), (1, 2), (1, 3),
(1, 4), (2, 1), (2, 2), (3, 1), (4, 1).

Input: N = 1000
Output: 7053



Naive Approach: Run two for loops upto N – 1 and count ordered pairs whose product are less than N.

Efficient Approach: Let’s considered an ordered pair (a, b). Then, if the product of two numbers is less than n i:e a * b < n, then at least one of them must be less then square root of n. We can proof by contradiction that if both numbers are greater then square root of n the their product is not less than n.

So, you can take all integers a up to sqrt(n – 1) instead of all integers up to n. For each a, count the number of b >= a such that a * b < n. Then multiply the result by two to count the pair (b, a) for each pair (a, b) you saw. After that, subtract the integer part of sqrt(n – 1) to ensure the pairs (a, a) were counted exactly once.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return count of Ordered pairs
// whose product are less than N
int countOrderedPairs(int N)
{
    // Initialize count to 0
    int count_pairs = 0;
  
    // count total pairs
    for (int i = 1; i <= sqrt(N - 1); ++i) {
        for (int j = i; j * i < N; ++j)
            ++count_pairs;
    }
  
    // multiply by 2 to get ordered_pairs
    count_pairs *= 2;
  
    // subtract redundant pairs (a, b) where a==b.
    count_pairs -= int(sqrt(N - 1));
  
    // return answer
    return count_pairs;
}
  
// Driver code
int main()
{
    int N = 5;
  
    // function call to print required answer
    cout << countOrderedPairs(N);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of above approach
  
class GFG{
// Function to return count of Ordered pairs
// whose product are less than N
static int countOrderedPairs(int N)
{
    // Initialize count to 0
    int count_pairs = 0;
  
    // count total pairs
    for (int i = 1; i <= (int)Math.sqrt(N - 1); ++i) {
        for (int j = i; j * i < N; ++j)
            ++count_pairs;
    }
  
    // multiply by 2 to get ordered_pairs
    count_pairs *= 2;
  
    // subtract redundant pairs (a, b) where a==b.
    count_pairs -= (int)(Math.sqrt(N - 1));
  
    // return answer
    return count_pairs;
}
  
// Driver code
public static void main(String[] args)
{
    int N = 5;
  
    // function call to print required answer
    System.out.println(countOrderedPairs(N));
}
}
// This code is contributed by mits

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Pyhton 3 implementation of above approach
  
from math import sqrt
# Function to return count of Ordered pairs
# whose product are less than N
def countOrderedPairs(N):
    # Initialize count to 0
    count_pairs = 0
  
    # count total pairs
    p = int(sqrt(N-1)) + 1
    q = int(sqrt(N))+2
    for i in range(1,p,1):
        for j in range(i,q,1):
            count_pairs += 1
      
    # multiply by 2 to get ordered_pairs
    count_pairs *= 2
  
    # subtract redundant pairs (a, b) where a==b.
    count_pairs -= int(sqrt(N - 1))
  
    # return answer
    return count_pairs
  
# Driver code
if __name__ == '__main__':
    N = 5
  
    # function call to print required answer
    print(countOrderedPairs(N))
  
# This code is contributed by
# Surendra_Gangwar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

//C# implementation of above approach 
  
using System;
  
public class GFG{
    // Function to return count of Ordered pairs 
// whose product are less than N 
static int countOrderedPairs(int N) 
    // Initialize count to 0 
    int count_pairs = 0; 
  
    // count total pairs 
    for (int i = 1; i <= (int)Math.Sqrt(N - 1); ++i) { 
        for (int j = i; j * i < N; ++j) 
            ++count_pairs; 
    
  
    // multiply by 2 to get ordered_pairs 
    count_pairs *= 2; 
  
    // subtract redundant pairs (a, b) where a==b. 
    count_pairs -= (int)(Math.Sqrt(N - 1)); 
  
    // return answer 
    return count_pairs; 
  
// Driver code 
    static public void Main (){
      
    int N = 5; 
    // function call to print required answer 
    Console.WriteLine(countOrderedPairs(N)); 
// This code is contributed by Sachin.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of above approach 
  
// Function to return count of Ordered 
// pairs whose products are less than N 
function countOrderedPairs($N
    // Initialize count to 0 
    $count_pairs = 0; 
  
    // count total pairs 
    for ($i = 1; $i <= sqrt($N - 1); ++$i
    
        for ( $j = $i; $j * $i < $N; ++$j
            ++$count_pairs
    
  
    // multiply by 2 to get ordered_pairs 
    $count_pairs *= 2; 
  
    // subtract redundant pairs 
    // (a, b) where a==b. 
    $count_pairs -= (sqrt($N - 1)); 
  
    // return answer 
    return $count_pairs
  
// Driver code 
$N = 5; 
  
// function call to print
// required answer 
echo countOrderedPairs($N); 
  
// This code is contributed
// by Sach_Code
?>

chevron_right


Output:

8

Time Complexity: O(N*sqrt(N))



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.