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Find the Count of Numbers Which Are Not Special

Medium

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Topics:

ArraysDynamic Programming

You are given 2 positive integers l and r. For any number x, all positive divisors of x except x are called the proper divisors of x.

A number is called special if it has exactly 2 proper divisors. For example:

The number 4 is special because it has proper divisors 1 and 2.
The number 6 is not special because it has proper divisors 1, 2, and 3.

Return the count of numbers in the range [l, r] that are not special.

Example 1:

Input: l = 5, r = 7

Output: 3

Explanation:

There are no special numbers in the range [5, 7].

Example 2:

Input: l = 4, r = 16

Output: 11

Explanation:

The special numbers in the range [4, 16] are 4 and 9.

Constraints:

1 <= l <= r <= 10⁹

Try coding it in LeetCode

Find the Count of Numbers Which Are Not Special

Medium

Asked by:

17 views

Topics:

ArraysDynamic Programming

You are given 2 positive integers l and r. For any number x, all positive divisors of x except x are called the proper divisors of x.

A number is called special if it has exactly 2 proper divisors. For example:

The number 4 is special because it has proper divisors 1 and 2.
The number 6 is not special because it has proper divisors 1, 2, and 3.

Return the count of numbers in the range [l, r] that are not special.

Example 1:

Input: l = 5, r = 7

Output: 3

Explanation:

There are no special numbers in the range [5, 7].

Example 2:

Input: l = 4, r = 16

Output: 11

Explanation:

The special numbers in the range [4, 16] are 4 and 9.

Constraints:

1 <= l <= r <= 10⁹

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of the input numbers? Can I expect negative numbers, zeros, or very large positive numbers?
What defines a 'special' number? Is there a mathematical relationship, a specific property, or a provided function to determine if a number is 'special'?
Can the input array be empty or null? If so, what should the return value be?
Are duplicate numbers in the input significant? Should they be counted multiple times if they are not 'special'?
What data type should the return value be? Should I return an integer representing the count, or something else?

Brute Force Solution

Approach

We want to find how many numbers are *not* special. The brute force strategy involves checking every single number in the given range to see if it meets the criteria of being *not special*. By individually examining each number, we can count how many fit the description.

Here's how the algorithm would work step-by-step:

Start with the first number in the range.
Check if the number is special according to the definition of special (e.g. its digits are not strictly increasing).
If the number IS special, ignore it.
If the number is NOT special, then we have found a number we want to count, so don't count it.
Move to the next number in the range.
Repeat the checking process for this new number.
Keep doing this for every number in the range.
The final answer will be how many numbers were *not* special after you've checked every single number.

Code Implementation

def count_non_special_numbers_brute_force(lower_bound, upper_bound):
    count_of_non_special_numbers = 0

    for current_number in range(lower_bound, upper_bound + 1):
        number_as_string = str(current_number)
        is_special = True # Assume special until proven otherwise.

        # Check each digit to determine if the number is strictly increasing.
        for digit_index in range(len(number_as_string) - 1):
            if int(number_as_string[digit_index]) >= int(number_as_string[digit_index + 1]):
                is_special = False
                break # No need to continue checking, its not special

        if not is_special:
            # This is a number which is NOT special, so we count it.
            count_of_non_special_numbers += 1

    return count_of_non_special_numbers

Big(O) Analysis

Time Complexity

O(n*m) – The brute force approach iterates through each number from 1 to n, where n is the upper bound of the range. For each number, it checks if the number is 'special' which requires examining the digits of the number. Let 'm' be the number of digits in the largest number (n). In the worst case, checking if a number is special might require examining all its digits. Thus, the overall time complexity is O(n*m) where n is the range and m is the maximum number of digits of a number in the range. Since m is related to the log of n, the complexity could be viewed as O(n log n) depending on the specific 'special' check but we assume 'm' here, which gives a simplified bound that captures the core idea.

Space Complexity

O(1) – The provided solution iterates through a range of numbers and checks each one individually. It does not use any auxiliary data structures like arrays, lists, or hash maps to store intermediate results or visited numbers. The algorithm only requires a few constant space variables to keep track of the current number being checked and possibly a counter, so the space used does not depend on the input range size, N. Thus, the space complexity is constant.

Optimal Solution

Approach

The problem asks us to count numbers that don't have repeated digits. The efficient approach figures out how many numbers *do* have repeated digits, then subtracts that count from the total number of possible numbers to find the answer more quickly.

Here's how the algorithm would work step-by-step:

First, calculate the total number of numbers possible within the given range. This is usually a straightforward subtraction and addition.
Next, figure out how many numbers have at least one repeating digit. A clever way to do this is to determine how many numbers don't have any repeating digits.
To count numbers *without* repeating digits, think about filling each digit place one at a time, choosing from the available digits. When you choose a digit, remove it from being available for subsequent choices.
For example, if you're building a 3-digit number without repeating digits, you have 9 choices for the first digit (1-9, can't be 0), then 9 choices for the second digit (0 and the digits that weren't first), and finally 8 choices for the third digit.
Multiply these choices together for each length of number to get the total count of numbers *without* any repeating digits.
Subtract the count of numbers without any repeating digits from the total number of numbers within the range to determine the count of numbers with at least one repeating digit (which are considered 'special').
Finally, subtract the count of 'special' numbers from the total count of numbers to determine the count of numbers that aren't special. That's your answer.

Code Implementation

def find_count_of_numbers_which_are_not_special(number):
    number_string = str(number)
    number_length = len(number_string)
    total_numbers = number + 1

    def count_numbers_with_unique_digits(digits):
        if digits > 10:
            return 0

        unique_digit_count = 0
        # Start with one-digit numbers
        for number_of_digits in range(1, digits + 1):
            if number_of_digits == 1:
                unique_digit_count += 9
            else:
                available_digits = 9
                unique_numbers = 9

                for _ in range(number_of_digits - 1):
                    unique_numbers *= available_digits
                    available_digits -= 1

                unique_digit_count += unique_numbers

        return unique_digit_count

    count_unique = count_numbers_with_unique_digits(number_length - 1)
    
    seen = set()
    for index, digit_char in enumerate(number_string):
        digit = int(digit_char)
        for j in range(0 if index > 0 else 1, digit):
            if j not in seen:
                available_digits = 9 - index
                unique_numbers = 1

                for _ in range(number_length - index - 1):
                    unique_numbers *= available_digits
                    available_digits -= 1

                count_unique += unique_numbers

        if digit in seen:
            break
        seen.add(digit)
    else:
        count_unique += 1
    
    # Subtract the count of unique digit numbers.
    return total_numbers - count_unique

Big(O) Analysis

Time Complexity

O(log n) – The algorithm's runtime is dominated by the process of determining the number of digits in the input number 'n' and iterating from 1 to the number of digits. The number of digits in 'n' is logarithmic with respect to n (base 10). Therefore, the outer loop iterates a number of times proportional to log n. Operations within the loop involve calculations with a fixed number of digits, so their time complexity is constant, O(1). Thus, the overall time complexity is O(log n).

Space Complexity

O(1) – The described algorithm primarily involves calculations and does not explicitly mention creating auxiliary data structures that scale with the input number (N). Although intermediate calculations are performed, these are typically stored in a fixed number of variables. Therefore, the space complexity remains constant, independent of the input range, resulting in O(1) auxiliary space.

Edge Cases

Case	How to Handle
Null or undefined input list	Return 0 or throw an IllegalArgumentException/TypeError if null/undefined inputs are not allowed.
Empty input list (length 0)	Return 0 since there are no numbers to evaluate.
List contains only one element	Return 1 as the single element cannot form a 'special' number by itself.
List contains all identical numbers	If all numbers are identical, they will not be special; therefore, return the list's length.
List contains very large numbers (potential for overflow)	Use appropriate data types (e.g., long in Java, 64-bit integers) or modulo operations to prevent integer overflow during calculations.
List contains negative numbers	The definition of 'special' should apply to both positive and negative numbers, as long as the calculation doesn't cause errors.
List with a very large number of elements, approaching system memory limits	Ensure the algorithm scales efficiently (O(n) or O(n log n)) and doesn't consume excessive memory.
List contains zero(s)	Handle zero values appropriately based on the definition of a special number; division by zero must be avoided.

Null or undefined input list

How to Handle:

Return 0 or throw an IllegalArgumentException/TypeError if null/undefined inputs are not allowed.

Empty input list (length 0)

How to Handle:

Return 0 since there are no numbers to evaluate.

List contains only one element

How to Handle:

Return 1 as the single element cannot form a 'special' number by itself.

List contains all identical numbers

How to Handle:

If all numbers are identical, they will not be special; therefore, return the list's length.

List contains very large numbers (potential for overflow)

How to Handle:

Use appropriate data types (e.g., long in Java, 64-bit integers) or modulo operations to prevent integer overflow during calculations.

List contains negative numbers

How to Handle:

The definition of 'special' should apply to both positive and negative numbers, as long as the calculation doesn't cause errors.

List with a very large number of elements, approaching system memory limits

How to Handle:

Ensure the algorithm scales efficiently (O(n) or O(n log n)) and doesn't consume excessive memory.

List contains zero(s)

How to Handle:

Handle zero values appropriately based on the definition of a special number; division by zero must be avoided.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of the input numbers? Can I expect negative numbers, zeros, or very large positive numbers?
What defines a 'special' number? Is there a mathematical relationship, a specific property, or a provided function to determine if a number is 'special'?
Can the input array be empty or null? If so, what should the return value be?
Are duplicate numbers in the input significant? Should they be counted multiple times if they are not 'special'?
What data type should the return value be? Should I return an integer representing the count, or something else?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start with the first number in the range.
Check if the number is special according to the definition of special (e.g. its digits are not strictly increasing).
If the number IS special, ignore it.
If the number is NOT special, then we have found a number we want to count, so don't count it.
Move to the next number in the range.
Repeat the checking process for this new number.
Keep doing this for every number in the range.
The final answer will be how many numbers were *not* special after you've checked every single number.

Code Implementation

def count_non_special_numbers_brute_force(lower_bound, upper_bound):
    count_of_non_special_numbers = 0

    for current_number in range(lower_bound, upper_bound + 1):
        number_as_string = str(current_number)
        is_special = True # Assume special until proven otherwise.

        # Check each digit to determine if the number is strictly increasing.
        for digit_index in range(len(number_as_string) - 1):
            if int(number_as_string[digit_index]) >= int(number_as_string[digit_index + 1]):
                is_special = False
                break # No need to continue checking, its not special

        if not is_special:
            # This is a number which is NOT special, so we count it.
            count_of_non_special_numbers += 1

    return count_of_non_special_numbers

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, calculate the total number of numbers possible within the given range. This is usually a straightforward subtraction and addition.
Next, figure out how many numbers have at least one repeating digit. A clever way to do this is to determine how many numbers don't have any repeating digits.
To count numbers *without* repeating digits, think about filling each digit place one at a time, choosing from the available digits. When you choose a digit, remove it from being available for subsequent choices.
For example, if you're building a 3-digit number without repeating digits, you have 9 choices for the first digit (1-9, can't be 0), then 9 choices for the second digit (0 and the digits that weren't first), and finally 8 choices for the third digit.
Multiply these choices together for each length of number to get the total count of numbers *without* any repeating digits.
Subtract the count of numbers without any repeating digits from the total number of numbers within the range to determine the count of numbers with at least one repeating digit (which are considered 'special').
Finally, subtract the count of 'special' numbers from the total count of numbers to determine the count of numbers that aren't special. That's your answer.

Code Implementation

def find_count_of_numbers_which_are_not_special(number):
    number_string = str(number)
    number_length = len(number_string)
    total_numbers = number + 1

    def count_numbers_with_unique_digits(digits):
        if digits > 10:
            return 0

        unique_digit_count = 0
        # Start with one-digit numbers
        for number_of_digits in range(1, digits + 1):
            if number_of_digits == 1:
                unique_digit_count += 9
            else:
                available_digits = 9
                unique_numbers = 9

                for _ in range(number_of_digits - 1):
                    unique_numbers *= available_digits
                    available_digits -= 1

                unique_digit_count += unique_numbers

        return unique_digit_count

    count_unique = count_numbers_with_unique_digits(number_length - 1)
    
    seen = set()
    for index, digit_char in enumerate(number_string):
        digit = int(digit_char)
        for j in range(0 if index > 0 else 1, digit):
            if j not in seen:
                available_digits = 9 - index
                unique_numbers = 1

                for _ in range(number_length - index - 1):
                    unique_numbers *= available_digits
                    available_digits -= 1

                count_unique += unique_numbers

        if digit in seen:
            break
        seen.add(digit)
    else:
        count_unique += 1
    
    # Subtract the count of unique digit numbers.
    return total_numbers - count_unique

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or undefined input list	Return 0 or throw an IllegalArgumentException/TypeError if null/undefined inputs are not allowed.
Empty input list (length 0)	Return 0 since there are no numbers to evaluate.
List contains only one element	Return 1 as the single element cannot form a 'special' number by itself.
List contains all identical numbers	If all numbers are identical, they will not be special; therefore, return the list's length.
List contains very large numbers (potential for overflow)	Use appropriate data types (e.g., long in Java, 64-bit integers) or modulo operations to prevent integer overflow during calculations.
List contains negative numbers	The definition of 'special' should apply to both positive and negative numbers, as long as the calculation doesn't cause errors.
List with a very large number of elements, approaching system memory limits	Ensure the algorithm scales efficiently (O(n) or O(n log n)) and doesn't consume excessive memory.
List contains zero(s)	Handle zero values appropriately based on the definition of a special number; division by zero must be avoided.

Null or undefined input list

How to Handle:

Return 0 or throw an IllegalArgumentException/TypeError if null/undefined inputs are not allowed.

Empty input list (length 0)

How to Handle:

Return 0 since there are no numbers to evaluate.

List contains only one element

How to Handle:

Return 1 as the single element cannot form a 'special' number by itself.

List contains all identical numbers

How to Handle:

If all numbers are identical, they will not be special; therefore, return the list's length.

List contains very large numbers (potential for overflow)

How to Handle:

Use appropriate data types (e.g., long in Java, 64-bit integers) or modulo operations to prevent integer overflow during calculations.

List contains negative numbers

How to Handle:

The definition of 'special' should apply to both positive and negative numbers, as long as the calculation doesn't cause errors.

List with a very large number of elements, approaching system memory limits

How to Handle:

Ensure the algorithm scales efficiently (O(n) or O(n log n)) and doesn't consume excessive memory.

List contains zero(s)

How to Handle:

Handle zero values appropriately based on the definition of a special number; division by zero must be avoided.