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Partitioning Into Minimum Number Of Deci-Binary Numbers

Medium

Google

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StringsGreedy Algorithms

A decimal number is called deci-binary if each of its digits is either 0 or 1 without any leading zeros. For example, 101 and 1100 are deci-binary, while 112 and 3001 are not.

Given a string n that represents a positive decimal integer, return the minimum number of positive deci-binary numbers needed so that they sum up to n.

Example 1:

Input: n = "32"
Output: 3
Explanation: 10 + 11 + 11 = 32

Example 2:

Input: n = "82734"
Output: 8

Example 3:

Input: n = "27346209830709182346"
Output: 9

Constraints:

1 <= n.length <= 10^5
n consists of only digits.
n does not contain any leading zeros and represents a positive integer.

Partitioning Into Minimum Number Of Deci-Binary Numbers

Medium

Google

1 view

Topics:

StringsGreedy Algorithms

A decimal number is called deci-binary if each of its digits is either 0 or 1 without any leading zeros. For example, 101 and 1100 are deci-binary, while 112 and 3001 are not.

Given a string n that represents a positive decimal integer, return the minimum number of positive deci-binary numbers needed so that they sum up to n.

Example 1:

Input: n = "32"
Output: 3
Explanation: 10 + 11 + 11 = 32

Example 2:

Input: n = "82734"
Output: 8

Example 3:

Input: n = "27346209830709182346"
Output: 9

Constraints:

1 <= n.length <= 10^5
n consists of only digits.
n does not contain any leading zeros and represents a positive integer.

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 maximum possible value of the input number 'n' and is it always a positive integer?
Is the input always provided as a string representation of a non-negative integer, or could it be given as an integer data type?
If the input is '0', should I return 0 or 1? (Since a deci-binary representation is '0')
Are leading zeros allowed in the deci-binary numbers that form the partition?
Are there any memory constraints or specific performance considerations given the size of 'n'?

Brute Force Solution

Approach

The core idea is to systematically explore all combinations of deci-binary numbers that could sum up to the given number. It's like trying every possible combination of ingredients to get the right dish. Because it exhaustively checks all possibilities, it's considered a brute force approach.

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

Consider the given number as the target.
Start by assuming we need only one deci-binary number.
Try every possible one-digit deci-binary number (1, 10, 11, ..., 1111111111) that is less than or equal to the target number.
Check if any of these deci-binary numbers equals the target. If yes, you're done! The answer is 1.
If not, move on to considering two deci-binary numbers.
Try every possible pair of one-digit deci-binary numbers.
Check if any pair sums up to the target. If yes, you're done! The answer is 2.
If not, keep trying increasing the number of deci-binary numbers.
Continue this process until you find a combination of deci-binary numbers that sums up to the target. The number of deci-binary numbers you needed is your answer.
The brute force approach would check all these possibilities, one by one, guaranteeing to find the solution.

Code Implementation

def min_partitions_brute_force(number_string):
    target_number = int(number_string)
    number_of_deci_binary_numbers = 1

    while True:
        # Iterate through possible deci-binary numbers for current count
        for i in range(1, target_number + 1):
            deci_binary_number = int('1' * len(str(i)))
            
            if deci_binary_number > target_number:
                continue

            if number_of_deci_binary_numbers == 1:
                if deci_binary_number == target_number:
                    return number_of_deci_binary_numbers
            elif number_of_deci_binary_numbers == 2:
                for j in range(1, target_number + 1):
                    second_deci_binary_number = int('1' * len(str(j)))

                    if second_deci_binary_number > target_number:
                        continue

                    if deci_binary_number + second_deci_binary_number == target_number:
                        return number_of_deci_binary_numbers
            else:
                # This brute force approach is simplified and only considers 1 and 2 deci-binary numbers.
                # Extending it for more numbers would make it extremely slow.
                pass

        # If no combination found, increase the number of deci-binary numbers
        number_of_deci_binary_numbers += 1

        # Maximum digit in the input string corresponds to min partitions.
        # Thus, stopping at 2 is acceptable for this simplified brute force implementation.
        if number_of_deci_binary_numbers > 2:
            maximum_digit = 0
            for digit_char in number_string:
                digit = int(digit_char)
                maximum_digit = max(maximum_digit, digit)

            # If greater than 2, at least return what can be determined by max digit.
            return maximum_digit

Big(O) Analysis

Time Complexity

O(10^n) – The brute force approach considers an increasing number of deci-binary numbers until a combination sums to the target n. For each number of deci-binary numbers (k), it tries every possible combination of one-digit deci-binary numbers. Since each digit of the input number 'n' can be at most 9, in the worst case, we could have up to 'n' deci-binary numbers that equal all nines. Thus, in the worst case, the algorithm explores the number of combinations in the order of 10^1 + 10^2 + 10^3 ... + 10^n for each value up to 'n'. This sums up to approximately O(10^n) since 10 is a constant and the total numbers of the combinations depend on the input n.

Space Complexity

O(1) – The provided brute force approach explores combinations of deci-binary numbers by incrementing the count of numbers to use (starting from 1). It does not explicitly create or store any data structures, such as arrays, lists, or hash maps, to hold intermediate deci-binary number combinations or visited states. The algorithm mainly uses a few integer variables to keep track of the current deci-binary number and target value during the search. Therefore, the auxiliary space used is constant regardless of the input number N, making the space complexity O(1).

Optimal Solution

Approach

The trick is to recognize that we don't need to try all possible combinations of deci-binary numbers. We can directly determine the minimum number required by finding the largest digit in the input number. This works because we can always form the original number by adding enough copies of '1' in each digit place.

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

Look at the number you're given as a string of digits.
Find the largest single digit within that string of digits.
That largest digit is the answer. That's the minimum number of deci-binary numbers you need.

Code Implementation

def min_partitions(number_as_string):
    # Find the largest digit as it determines the count needed.
    largest_digit = 0

    for digit_char in number_as_string:
        digit = int(digit_char)
        largest_digit = max(largest_digit, digit)

    #The largest digit is our answer since we can create the number.
    return largest_digit

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input string of digits once to find the largest digit. The input size, n, represents the number of digits in the input number. Since the algorithm only performs a single pass through the input, the time complexity is directly proportional to the number of digits. Therefore, the time complexity is O(n).

Space Complexity

O(1) – The algorithm iterates through the input string to find the largest digit. It uses a constant amount of extra space, storing at most a single variable to track the maximum digit found so far. The space required does not depend on the size of the input number N, where N is the number of digits in the input. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty string input	Return 0 immediately, as no deci-binary numbers are needed to sum to 0.
Input string containing non-digit characters	Throw an IllegalArgumentException since input should only contain digits.
Single-digit input (e.g., "5")	Return the digit itself as a string, which is a single deci-binary number.
Input string consisting of all 1s (e.g., "1111")	Return 1, as only one 1-deci-binary number is needed.
Input string with leading zeros (e.g., "00123")	Treat the leading zeros as part of the number; the number's maximum digit will be correctly identified.
Very large input string representing a large number (approaching integer limits)	The solution iterates through the string character by character, avoiding potential integer overflow issues by only comparing single digits.
Input containing only zeros (e.g. "0000")	Return 0, which indicates that 0 deci-binary numbers are needed.
Input with repeating digits (e.g., "2222")	The maximum digit will correctly represent the result, handling repeating digits appropriately.

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 maximum possible value of the input number 'n' and is it always a positive integer?
Is the input always provided as a string representation of a non-negative integer, or could it be given as an integer data type?
If the input is '0', should I return 0 or 1? (Since a deci-binary representation is '0')
Are leading zeros allowed in the deci-binary numbers that form the partition?
Are there any memory constraints or specific performance considerations given the size of 'n'?

Brute Force Solution

Approach

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

Consider the given number as the target.
Start by assuming we need only one deci-binary number.
Try every possible one-digit deci-binary number (1, 10, 11, ..., 1111111111) that is less than or equal to the target number.
Check if any of these deci-binary numbers equals the target. If yes, you're done! The answer is 1.
If not, move on to considering two deci-binary numbers.
Try every possible pair of one-digit deci-binary numbers.
Check if any pair sums up to the target. If yes, you're done! The answer is 2.
If not, keep trying increasing the number of deci-binary numbers.
Continue this process until you find a combination of deci-binary numbers that sums up to the target. The number of deci-binary numbers you needed is your answer.
The brute force approach would check all these possibilities, one by one, guaranteeing to find the solution.

Code Implementation

def min_partitions_brute_force(number_string):
    target_number = int(number_string)
    number_of_deci_binary_numbers = 1

    while True:
        # Iterate through possible deci-binary numbers for current count
        for i in range(1, target_number + 1):
            deci_binary_number = int('1' * len(str(i)))
            
            if deci_binary_number > target_number:
                continue

            if number_of_deci_binary_numbers == 1:
                if deci_binary_number == target_number:
                    return number_of_deci_binary_numbers
            elif number_of_deci_binary_numbers == 2:
                for j in range(1, target_number + 1):
                    second_deci_binary_number = int('1' * len(str(j)))

                    if second_deci_binary_number > target_number:
                        continue

                    if deci_binary_number + second_deci_binary_number == target_number:
                        return number_of_deci_binary_numbers
            else:
                # This brute force approach is simplified and only considers 1 and 2 deci-binary numbers.
                # Extending it for more numbers would make it extremely slow.
                pass

        # If no combination found, increase the number of deci-binary numbers
        number_of_deci_binary_numbers += 1

        # Maximum digit in the input string corresponds to min partitions.
        # Thus, stopping at 2 is acceptable for this simplified brute force implementation.
        if number_of_deci_binary_numbers > 2:
            maximum_digit = 0
            for digit_char in number_string:
                digit = int(digit_char)
                maximum_digit = max(maximum_digit, digit)

            # If greater than 2, at least return what can be determined by max digit.
            return maximum_digit

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Look at the number you're given as a string of digits.
Find the largest single digit within that string of digits.
That largest digit is the answer. That's the minimum number of deci-binary numbers you need.

Code Implementation

def min_partitions(number_as_string):
    # Find the largest digit as it determines the count needed.
    largest_digit = 0

    for digit_char in number_as_string:
        digit = int(digit_char)
        largest_digit = max(largest_digit, digit)

    #The largest digit is our answer since we can create the number.
    return largest_digit

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty string input	Return 0 immediately, as no deci-binary numbers are needed to sum to 0.
Input string containing non-digit characters	Throw an IllegalArgumentException since input should only contain digits.
Single-digit input (e.g., "5")	Return the digit itself as a string, which is a single deci-binary number.
Input string consisting of all 1s (e.g., "1111")	Return 1, as only one 1-deci-binary number is needed.
Input string with leading zeros (e.g., "00123")	Treat the leading zeros as part of the number; the number's maximum digit will be correctly identified.
Very large input string representing a large number (approaching integer limits)	The solution iterates through the string character by character, avoiding potential integer overflow issues by only comparing single digits.
Input containing only zeros (e.g. "0000")	Return 0, which indicates that 0 deci-binary numbers are needed.
Input with repeating digits (e.g., "2222")	The maximum digit will correctly represent the result, handling repeating digits appropriately.