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Number of Ways to Separate Numbers

Hard

Walmart

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Topics:

ArraysDynamic Programming

You wrote down many positive integers in a string called num. However, you realized that you forgot to add commas to seperate the different numbers. You remember that the list of integers was non-decreasing and that no integer had leading zeros.

Return the number of possible lists of integers that you could have written down to get the string num. Since the answer may be large, return it modulo 10⁹ + 7.

Example 1:

Input: num = "327"
Output: 2
Explanation: You could have written down the numbers:
3, 27
327

Example 2:

Input: num = "094"
Output: 0
Explanation: No numbers can have leading zeros and all numbers must be positive.

Example 3:

Input: num = "0"
Output: 0
Explanation: No numbers can have leading zeros and all numbers must be positive.

Constraints:

1 <= num.length <= 3500
num consists of digits '0' through '9'.

Number of Ways to Separate Numbers

Hard

Walmart

0 views

Topics:

ArraysDynamic Programming

Return the number of possible lists of integers that you could have written down to get the string num. Since the answer may be large, return it modulo 10⁹ + 7.

Example 1:

Input: num = "327"
Output: 2
Explanation: You could have written down the numbers:
3, 27
327

Example 2:

Input: num = "094"
Output: 0
Explanation: No numbers can have leading zeros and all numbers must be positive.

Example 3:

Input: num = "0"
Output: 0
Explanation: No numbers can have leading zeros and all numbers must be positive.

Constraints:

1 <= num.length <= 3500
num consists of digits '0' through '9'.

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 length of the input string `s`? Are there any constraints on the digits within the string (e.g., leading zeros for individual numbers after separation)?
If there are no valid ways to separate the string into increasing numbers, what should the function return?
Can the input string `s` be empty or null? If so, what is the expected output?
Does each separated number need to be strictly greater than the previous one, or is equality allowed?
Could you provide an example of a string where the numbers separated could have leading zeros (after the initial separation) and clarify whether those leading zeros should be considered valid?

Brute Force Solution

Approach

The brute force approach to this problem involves trying every single possible way to break up the given number string into smaller number strings. We want to find all valid ways to do this, checking certain conditions each time. The idea is simple: exhaustively check all possibilities.

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

Imagine you're given a long string of digits with no spaces.
Start by considering the first digit as the first number.
Then, consider the first two digits as the first number, and so on, until you consider all digits as one big number.
For each of these possible 'first numbers', check if it's a valid number according to the problem's rules.
If a 'first number' is valid, take the remaining digits and repeat the process, finding all possible 'second numbers'.
Continue this process of splitting the digits and validating the numbers until there are no digits left.
Every time you successfully break down the entire string into valid numbers, you've found one possible way.
Keep track of all the valid ways you find.
Once you've tried every possible way to split the string, you'll have the total number of valid ways.

Code Implementation

def number_of_ways_to_separate_numbers_brute_force(number_string):
    number_of_ways = 0
    length_of_number_string = len(number_string)

    def find_all_valid_separations(current_index, previous_number):
        nonlocal number_of_ways

        if current_index == length_of_number_string:
            # Base case: we've reached the end of the string
            number_of_ways += 1
            return

        for next_index in range(current_index + 1, length_of_number_string + 1):
            substring = number_string[current_index:next_index]

            # Ensure the substring is not empty
            if not substring:
                continue

            current_number = int(substring)

            #Leading zero check
            if len(substring) > 1 and substring[0] == '0':
                continue

            #Check if the current number is greater or equal
            if previous_number is None or current_number >= previous_number:
                find_all_valid_separations(next_index, current_number)

    # Start the recursion with no previous number
    find_all_valid_separations(0, None)

    return number_of_ways

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores all possible ways to split the input string of length n. For each position in the string, we have two choices: either split the string at that position or don't. This branching creates a binary tree of possibilities. Therefore, the number of possible splits grows exponentially with the input size n. Since we explore each of these possibilities, the time complexity is O(2^n).

Space Complexity

O(N) – The described brute force approach implicitly uses recursion. In the worst-case scenario, where each 'first number' consists of only the first digit, the recursion can go as deep as N, where N is the length of the input string. Each recursive call adds a new frame to the call stack, requiring space to store function arguments, local variables, and the return address. Therefore, the auxiliary space used by the recursion stack grows linearly with the input size N, resulting in O(N) space complexity.

Optimal Solution

Approach

The problem asks us to count how many ways a given string of digits can be split into valid numbers. Instead of checking every possible split, we can use a clever counting technique to avoid repeated work by leveraging the fact that valid splits must follow specific ordering rules.

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

First, realize the numbers in the split need to be non-decreasing. If you see a number that's smaller than the one before, that split is invalid and we should skip it.
Consider how the length of a valid next number compares to the previous number. If the next number is shorter or longer, we can easily determine if it is a valid choice.
If the next number has the same length as the previous one, we need to compare the numbers themselves to make sure the sequence is non-decreasing.
Use a technique that builds up the number of ways to split the string. Keep track of how many valid splits you've found up to a certain point in the string.
Instead of recalculating splits, use the information you already have to figure out how many new splits you can make by adding the next number.
By systematically checking validity and building upon prior calculations, we efficiently count all valid splits without exhaustively trying every possibility.

Code Implementation

def number_of_ways_to_separate_numbers(number_string):
    string_length = len(number_string)
    dp_table = [0] * (string_length + 1)
    dp_table[0] = 1
    modulo = 10**9 + 7

    for index in range(1, string_length + 1):
        for previous_index in range(index):
            current_number_string = number_string[previous_index:index]

            # Avoid leading zeros.
            if current_number_string[0] == '0':
                continue

            if previous_index == 0:
                dp_table[index] = (dp_table[index] + dp_table[previous_index]) % modulo
                continue

            previous_number_string = number_string[find_start_of_previous(dp_table, previous_index):previous_index]

            # Check if length is shorter.
            if len(current_number_string) < len(previous_number_string):
                dp_table[index] = (dp_table[index] + dp_table[previous_index]) % modulo

            # Compare number if length is the same.
            elif len(current_number_string) == len(previous_number_string):
                if current_number_string >= previous_number_string:

                    # This is a valid split so add it to ways
                    dp_table[index] = (dp_table[index] + dp_table[previous_index]) % modulo

            # Check if length is longer than prev
            elif len(current_number_string) > len(previous_number_string):
                dp_table[index] = (dp_table[index] + dp_table[previous_index]) % modulo

    return dp_table[string_length]

def find_start_of_previous(dp_table, previous_index):
    current_length = previous_index
    while current_length > 0 and dp_table[current_length] == 0:
        current_length -= 1
    return current_length

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the string of length n. For each position i, it considers possible splits from i onwards. The comparison of the current substring (representing the next number in the split) with the previous one might involve comparing each character in the substring, which is at most n in length. Because the algorithm calculates how many ways there are to split up to a certain point, it may go through the string multiple times based on previous splits. Therefore the algorithm effectively makes comparisons that approximate a nested loop structure, leading to approximately n * n operations. This simplifies to O(n²).

Space Complexity

O(N) – The algorithm uses a dynamic programming approach, storing the number of valid splits up to each index of the input string. This requires an auxiliary array, typically named 'dp', whose size is proportional to the length of the input string, N. Consequently, the space complexity scales linearly with the input size as we store intermediate results for each prefix of the string. Therefore the auxiliary space used is O(N).

Edge Cases

Case	How to Handle
Empty input string	Return 0 immediately since there are no valid partitions of an empty string.
Input string with a single '0'	Return 0, because leading zeros are invalid and any partition will contain one.
Input string consisting of all zeros	Return 0 as no valid partition is possible due to leading zero constraints.
Input string with leading zeros in substrings	Explicitly check for leading zeros after each partition to determine if it's valid.
Very large input string that can lead to integer overflow	Use dynamic programming with modulo operator to prevent integer overflow.
Input string that requires recursion exceeding stack limit	Implement the solution using iterative dynamic programming instead of recursion.
Input string already in non-decreasing order and long length	Check if the current number is strictly greater than previous number to eliminate redundant computations.
Input string with very large numbers such that string comparison is more efficient than numeric conversion.	Use string comparison instead of converting substrings to numbers.

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 length of the input string `s`? Are there any constraints on the digits within the string (e.g., leading zeros for individual numbers after separation)?
If there are no valid ways to separate the string into increasing numbers, what should the function return?
Can the input string `s` be empty or null? If so, what is the expected output?
Does each separated number need to be strictly greater than the previous one, or is equality allowed?
Could you provide an example of a string where the numbers separated could have leading zeros (after the initial separation) and clarify whether those leading zeros should be considered valid?

Brute Force Solution

Approach

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

Imagine you're given a long string of digits with no spaces.
Start by considering the first digit as the first number.
Then, consider the first two digits as the first number, and so on, until you consider all digits as one big number.
For each of these possible 'first numbers', check if it's a valid number according to the problem's rules.
If a 'first number' is valid, take the remaining digits and repeat the process, finding all possible 'second numbers'.
Continue this process of splitting the digits and validating the numbers until there are no digits left.
Every time you successfully break down the entire string into valid numbers, you've found one possible way.
Keep track of all the valid ways you find.
Once you've tried every possible way to split the string, you'll have the total number of valid ways.

Code Implementation

def number_of_ways_to_separate_numbers_brute_force(number_string):
    number_of_ways = 0
    length_of_number_string = len(number_string)

    def find_all_valid_separations(current_index, previous_number):
        nonlocal number_of_ways

        if current_index == length_of_number_string:
            # Base case: we've reached the end of the string
            number_of_ways += 1
            return

        for next_index in range(current_index + 1, length_of_number_string + 1):
            substring = number_string[current_index:next_index]

            # Ensure the substring is not empty
            if not substring:
                continue

            current_number = int(substring)

            #Leading zero check
            if len(substring) > 1 and substring[0] == '0':
                continue

            #Check if the current number is greater or equal
            if previous_number is None or current_number >= previous_number:
                find_all_valid_separations(next_index, current_number)

    # Start the recursion with no previous number
    find_all_valid_separations(0, None)

    return number_of_ways

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, realize the numbers in the split need to be non-decreasing. If you see a number that's smaller than the one before, that split is invalid and we should skip it.
Consider how the length of a valid next number compares to the previous number. If the next number is shorter or longer, we can easily determine if it is a valid choice.
If the next number has the same length as the previous one, we need to compare the numbers themselves to make sure the sequence is non-decreasing.
Use a technique that builds up the number of ways to split the string. Keep track of how many valid splits you've found up to a certain point in the string.
Instead of recalculating splits, use the information you already have to figure out how many new splits you can make by adding the next number.
By systematically checking validity and building upon prior calculations, we efficiently count all valid splits without exhaustively trying every possibility.

Code Implementation

def number_of_ways_to_separate_numbers(number_string):
    string_length = len(number_string)
    dp_table = [0] * (string_length + 1)
    dp_table[0] = 1
    modulo = 10**9 + 7

    for index in range(1, string_length + 1):
        for previous_index in range(index):
            current_number_string = number_string[previous_index:index]

            # Avoid leading zeros.
            if current_number_string[0] == '0':
                continue

            if previous_index == 0:
                dp_table[index] = (dp_table[index] + dp_table[previous_index]) % modulo
                continue

            previous_number_string = number_string[find_start_of_previous(dp_table, previous_index):previous_index]

            # Check if length is shorter.
            if len(current_number_string) < len(previous_number_string):
                dp_table[index] = (dp_table[index] + dp_table[previous_index]) % modulo

            # Compare number if length is the same.
            elif len(current_number_string) == len(previous_number_string):
                if current_number_string >= previous_number_string:

                    # This is a valid split so add it to ways
                    dp_table[index] = (dp_table[index] + dp_table[previous_index]) % modulo

            # Check if length is longer than prev
            elif len(current_number_string) > len(previous_number_string):
                dp_table[index] = (dp_table[index] + dp_table[previous_index]) % modulo

    return dp_table[string_length]

def find_start_of_previous(dp_table, previous_index):
    current_length = previous_index
    while current_length > 0 and dp_table[current_length] == 0:
        current_length -= 1
    return current_length

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input string	Return 0 immediately since there are no valid partitions of an empty string.
Input string with a single '0'	Return 0, because leading zeros are invalid and any partition will contain one.
Input string consisting of all zeros	Return 0 as no valid partition is possible due to leading zero constraints.
Input string with leading zeros in substrings	Explicitly check for leading zeros after each partition to determine if it's valid.
Very large input string that can lead to integer overflow	Use dynamic programming with modulo operator to prevent integer overflow.
Input string that requires recursion exceeding stack limit	Implement the solution using iterative dynamic programming instead of recursion.
Input string already in non-decreasing order and long length	Check if the current number is strictly greater than previous number to eliminate redundant computations.
Input string with very large numbers such that string comparison is more efficient than numeric conversion.	Use string comparison instead of converting substrings to numbers.