Number of co-prime pairs from 1 to N which consists of given two digits

Last Updated : 23 Feb, 2021

Given an integer Nand two integers D1 and D2 ( < 10), the task is to find the number of co-prime pairs less than or equal to N consisting of only digits D1 and D2.

Examples:

Input: N = 30, D1 = 2, D2 = 3
Output: 5
Explanation:
All possible pairs of numbers upto 30 having digits 2 and 3 which are co-prime to each other are (2, 3), (2, 23), (3, 22), (3, 23), (22, 23) .

Input: N = 100, D1 = 5, D2 = 6
Output: 8
Explanation:
All possible pairs of numbers upto 100 having digits 5 and 6 which are co-prime to each other are (5, 6), (5, 56), (5, 66), (6, 55), (6, 65), (55, 56), (56, 65), (65, 66).

Approach: The idea to solve this problem is based on the following observations:

Observation:

Find and append every number consisting only given two digits which are less than or equal to N into a list.

Now it’s easy to find the number of unordered pairs which are co-primes.

Note that there can be maximum 1 + 2 + 2² + 2³ + 2⁴ + ………2¹⁰= 2047 numbers in the list i.e. Overall Time Complexity cannot exceed O(2047 * 2047), as the maximum no. of digits possible is 9.

Follow the steps below to solve the problem:

Initialize an empty list, say l, and append the given two digits as a string into the list.
Sort the list.
Initialize two lists, say total and temp2 for further use.
Iterate until l[0] < 10:
- Append the given two digits as a string to all the elements present in the list l.
- Keep updating l in the way shown below:
  - [2, 3] -> [‘2’ + ‘2’, ‘2’ + ‘3’, ‘3’ + ‘2’, ‘3’ + ‘3’]
  - [22, 23, 32, 33] – > [’22’ + ‘2’, ’22’ + ‘3’, ’23’ + ‘2’, ’23’ + ‘3’, ’32’ + ‘2’, ’32’ + ‘3’, ’33’ + ‘2’, ’33’ + ‘3’] and so on.
Define a function numOfPairs() which returns the count of unordered co-prime pairs.
Print the count returned by the above function as the answer.

Below is the implementation of the above approach:

Python3

# Python3 program for the above approach 
  
from copy import deepcopy 
import math 
  
# Function to check whether given 
# integers are co-prime or not 
def coprime(a, b): 
    return (math.gcd(a, b)) == 1
  
# Utility function to count 
# number of co-prime pairs 
def numOfPairs(arr, N): 
    count = 0
  
    # Traverse the array 
    for i in range(0, N-1): 
        for j in range(i+1, N): 
  
            # If co-prime 
            if (coprime(int(arr[i]), int(arr[j]))): 
  
                # Increment count 
                count = count + 1
  
    # Return count 
    return count 
  
# Function to count number 
# of co-prime pairs 
def noOfCoPrimePairs(N, d1, d2): 
  
    # Stores digits in string form 
    l = [] 
    l.append(str(d1)) 
    l.append(str(d2)) 
  
    # Sort the list 
    l.sort() 
  
    if int(N) < int(l[1]): 
        return 0
  
    # Keep two copies of list l 
    total = temp2 = deepcopy(l) 
    flag = 0
    temp3 = [] 
  
    # Generate 2 digit numbers 
    # using d1 and d2 
    while len(l[0]) < 10: 
        for i in range(len(l)): 
            for j in range(2): 
  
                # If current number 
                # does not exceed N 
                if int(l[i]+temp2[j]) > int(N): 
                    flag = 1
                    break
  
                total.append(l[i]+temp2[j]) 
                temp3.append(l[i]+temp2[j]) 
  
            if flag == 1: 
                break
        if flag == 1: 
            break
        l = deepcopy(temp3) 
        temp3 = [] 
  
    # Stores length of list 
    lenOfTotal = len(total) 
  
    # Stores number of co-prime pairs 
    ans = numOfPairs(total, lenOfTotal) 
  
    # Print number of co-prime pairs 
    print(ans) 
  
  
# Driver Code 
if __name__ == "__main__": 
  
    # Given value of N, d1, d2 
    N = 30
    d1 = 2
    d2 = 3
  
    # Function call to count 
    # number of co-prime pairs 
    noOfCoPrimePairs(N, d1, d2) 

Output:

Time Complexity: O(2^logN)
Auxiliary Space: O(2^logN)

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