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Number of Digit One

#407 Most AskedHard

Number of Digit One #407 Most Asked

Hard

Bloomberg LP

6 views

Topics:

Dynamic ProgrammingRecursionMath

Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.

Example 1:

Input: n = 13
Output: 6

Example 2:

Input: n = 0
Output: 0

Constraints:

0 <= n <= 10⁹

Try coding it in LeetCode

Number of Digit One

#407 Most AskedHard

Number of Digit One #407 Most Asked

Hard

Bloomberg LP

6 views

Topics:

Dynamic ProgrammingRecursionMath

Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.

Example 1:

Input: n = 13
Output: 6

Example 2:

Input: n = 0
Output: 0

Constraints:

0 <= n <= 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 possible values for the input integer n? Are there any specific constraints on its size?
Is n guaranteed to be a non-negative integer, or should I handle negative input values?
For n = 0, should the function return 0?
Could you clarify what is expected when n is a single-digit number, like 1 or 5?
Are we optimizing for space or time complexity, or is there a particular performance target in mind?

Brute Force Solution

Approach

We want to find how many times the digit '1' appears in all the numbers from 1 to a given number. A brute force approach means we'll look at every single number within that range and count the '1's.

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

Start with the number 1.
Check how many times the digit '1' appears in the number. In this case, it appears once.
Move to the next number, which is 2.
Check how many times the digit '1' appears in the number. In this case, it appears zero times.
Keep doing this for every number, one by one, until you reach the given number.
Each time you count the number of '1's in a number, add that count to a running total.
Once you have checked all the numbers, the running total will be the answer: the total number of times the digit '1' appears.

Code Implementation

def count_digit_one_brute_force(number):
    total_ones = 0

    for current_number in range(1, number + 1):
        number_string = str(current_number)
        ones_count = 0

        # Iterate through digits of current number
        for digit in number_string:

            if digit == '1':

                # Increment count if digit is '1'
                ones_count += 1

        # Add the '1' count of current number to running total
        total_ones += ones_count

    return total_ones

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm iterates through each number from 1 to n. For each number, it counts the occurrences of the digit '1'. Counting the digits in a number involves iterating through the digits, and the number of digits in a number is proportional to the logarithm of that number (base 10). Therefore, for each number 'i' from 1 to n, we perform approximately log10(i) operations. Summing this up, we get approximately log10(1) + log10(2) + ... + log10(n), which is roughly proportional to log10(n!). By Stirling's approximation, log(n!) is approximately n log(n), so the overall time complexity is O(n log n).

Space Complexity

O(1) – The provided brute force algorithm iterates from 1 to the input number N, checking each number individually. It maintains a running total of the number of '1's encountered. No auxiliary data structures like arrays, lists, or hash maps are created to store intermediate results or track visited numbers. The space used only involves storing a few integer variables for the loop counter and the running total, which takes constant space regardless of the value of N. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

Instead of counting the ones digit by digit, the optimized solution cleverly calculates how many times the digit '1' appears in each digit place (ones, tens, hundreds, etc.). It analyzes the number's structure and the position of each digit to efficiently determine the count.

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

Look at the given number and consider each digit place (ones, tens, hundreds, and so on) one at a time.
For each digit place, figure out how many complete 'cycles' of that place are contained in the number. For example, for the tens place, find out how many full groups of ten are present.
Multiply the number of full 'cycles' by the value of the digit place. This gives you a base count of how many times '1' appears in that place.
Now, look at the digit in the current place itself. If the digit is greater than 1, add the full value of the digit place to the count (because '1' will appear in every possible number within that place).
If the digit in the current place is exactly 1, add the remainder of the number divided by the digit place, plus 1, to the count (this accounts for the '1's up to the actual number).
If the digit in the current place is 0, then you don't add anything more to the count for this place.
Repeat this process for each digit place in the number, moving from right to left.
Finally, sum up the counts from all the digit places to get the total number of '1's in the entire number.

Code Implementation

def count_digit_one(number):
    digit_one_count = 0
    factor = 1

    while factor <= number:
        # Calculate the divisor which is ten times current factor
        divisor = factor * 10

        # Count how many times '1' appears at the current digit place.
        digit_one_count += (number // divisor) * factor

        # Check the current digit to determine additional '1's
        current_digit = (number % divisor) // factor

        if current_digit == 1:
            # Add the remainder plus 1 to the count
            digit_one_count += (number % factor) + 1

        elif current_digit > 1:
            # Add the full value of the current digit place
            digit_one_count += factor

        factor *= 10

    return digit_one_count

Big(O) Analysis

Time Complexity

O(log n) – The algorithm iterates through each digit place of the input number n. The number of digit places is logarithmic with respect to n (specifically, it's the base-10 logarithm, but the base doesn't affect Big O notation). Therefore, the loop runs a number of times proportional to log n, and each step within the loop performs constant time operations. Thus the time complexity is O(log n).

Space Complexity

O(1) – The algorithm calculates the number of '1's by iterating through each digit place of the input number. It uses a few constant space variables to store intermediate calculations such as the current digit place, counts, and temporary values for determining cycles and remainders. The space required does not depend on the input number N itself, but on the number of digits in N, which is logarithmic to N. Since the number of digits impacts the number of iterations, but the variables used within each iteration is constant, the auxiliary space complexity is constant O(1).

Edge Cases

Case	How to Handle
n is 0	Return 0 since there are no ones in numbers from 0 to 0.
n is a single-digit number less than 1	Return 0 because the only valid number (0) does not contain the digit 1.
n is a single-digit number equal to 1	Return 1 because the range [0,1] contains 1 one.
n is a single-digit number greater than 1	Return 1 because the only '1' appears as the number 1 itself within the [0, n] range.
n is a large number (e.g., close to INT_MAX) to test integer overflow	Use 'long' data type for intermediate calculations to prevent potential integer overflow during the counting process.
n contains leading zeros (should be handled like any valid integer)	Leading zeros should be ignored by the input processing; '0123' is treated as '123'.
n is a number with many ones	The algorithm must accurately count the ones in all place values in the given number.
n consists only of nines	This number requires correctly handling the carry-over when incrementing digit places.

n is 0

How to Handle:

Return 0 since there are no ones in numbers from 0 to 0.

n is a single-digit number less than 1

How to Handle:

Return 0 because the only valid number (0) does not contain the digit 1.

n is a single-digit number equal to 1

How to Handle:

Return 1 because the range [0,1] contains 1 one.

n is a single-digit number greater than 1

How to Handle:

Return 1 because the only '1' appears as the number 1 itself within the [0, n] range.

n is a large number (e.g., close to INT_MAX) to test integer overflow

How to Handle:

Use 'long' data type for intermediate calculations to prevent potential integer overflow during the counting process.

n contains leading zeros (should be handled like any valid integer)

How to Handle:

Leading zeros should be ignored by the input processing; '0123' is treated as '123'.

n is a number with many ones

How to Handle:

The algorithm must accurately count the ones in all place values in the given number.

n consists only of nines

How to Handle:

This number requires correctly handling the carry-over when incrementing digit places.

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 possible values for the input integer n? Are there any specific constraints on its size?
Is n guaranteed to be a non-negative integer, or should I handle negative input values?
For n = 0, should the function return 0?
Could you clarify what is expected when n is a single-digit number, like 1 or 5?
Are we optimizing for space or time complexity, or is there a particular performance target in mind?

Brute Force Solution

Approach

We want to find how many times the digit '1' appears in all the numbers from 1 to a given number. A brute force approach means we'll look at every single number within that range and count the '1's.

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

Start with the number 1.
Check how many times the digit '1' appears in the number. In this case, it appears once.
Move to the next number, which is 2.
Check how many times the digit '1' appears in the number. In this case, it appears zero times.
Keep doing this for every number, one by one, until you reach the given number.
Each time you count the number of '1's in a number, add that count to a running total.
Once you have checked all the numbers, the running total will be the answer: the total number of times the digit '1' appears.

Code Implementation

def count_digit_one_brute_force(number):
    total_ones = 0

    for current_number in range(1, number + 1):
        number_string = str(current_number)
        ones_count = 0

        # Iterate through digits of current number
        for digit in number_string:

            if digit == '1':

                # Increment count if digit is '1'
                ones_count += 1

        # Add the '1' count of current number to running total
        total_ones += ones_count

    return total_ones

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Look at the given number and consider each digit place (ones, tens, hundreds, and so on) one at a time.
For each digit place, figure out how many complete 'cycles' of that place are contained in the number. For example, for the tens place, find out how many full groups of ten are present.
Multiply the number of full 'cycles' by the value of the digit place. This gives you a base count of how many times '1' appears in that place.
Now, look at the digit in the current place itself. If the digit is greater than 1, add the full value of the digit place to the count (because '1' will appear in every possible number within that place).
If the digit in the current place is exactly 1, add the remainder of the number divided by the digit place, plus 1, to the count (this accounts for the '1's up to the actual number).
If the digit in the current place is 0, then you don't add anything more to the count for this place.
Repeat this process for each digit place in the number, moving from right to left.
Finally, sum up the counts from all the digit places to get the total number of '1's in the entire number.

Code Implementation

def count_digit_one(number):
    digit_one_count = 0
    factor = 1

    while factor <= number:
        # Calculate the divisor which is ten times current factor
        divisor = factor * 10

        # Count how many times '1' appears at the current digit place.
        digit_one_count += (number // divisor) * factor

        # Check the current digit to determine additional '1's
        current_digit = (number % divisor) // factor

        if current_digit == 1:
            # Add the remainder plus 1 to the count
            digit_one_count += (number % factor) + 1

        elif current_digit > 1:
            # Add the full value of the current digit place
            digit_one_count += factor

        factor *= 10

    return digit_one_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n is 0	Return 0 since there are no ones in numbers from 0 to 0.
n is a single-digit number less than 1	Return 0 because the only valid number (0) does not contain the digit 1.
n is a single-digit number equal to 1	Return 1 because the range [0,1] contains 1 one.
n is a single-digit number greater than 1	Return 1 because the only '1' appears as the number 1 itself within the [0, n] range.
n is a large number (e.g., close to INT_MAX) to test integer overflow	Use 'long' data type for intermediate calculations to prevent potential integer overflow during the counting process.
n contains leading zeros (should be handled like any valid integer)	Leading zeros should be ignored by the input processing; '0123' is treated as '123'.
n is a number with many ones	The algorithm must accurately count the ones in all place values in the given number.
n consists only of nines	This number requires correctly handling the carry-over when incrementing digit places.

n is 0

How to Handle:

Return 0 since there are no ones in numbers from 0 to 0.

n is a single-digit number less than 1

How to Handle:

Return 0 because the only valid number (0) does not contain the digit 1.

n is a single-digit number equal to 1

How to Handle:

Return 1 because the range [0,1] contains 1 one.

n is a single-digit number greater than 1

How to Handle:

Return 1 because the only '1' appears as the number 1 itself within the [0, n] range.

n is a large number (e.g., close to INT_MAX) to test integer overflow

How to Handle:

Use 'long' data type for intermediate calculations to prevent potential integer overflow during the counting process.

n contains leading zeros (should be handled like any valid integer)

How to Handle:

Leading zeros should be ignored by the input processing; '0123' is treated as '123'.

n is a number with many ones

How to Handle:

The algorithm must accurately count the ones in all place values in the given number.

n consists only of nines

How to Handle:

This number requires correctly handling the carry-over when incrementing digit places.

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