Number of Beautiful Partitions

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StringsDynamic Programming

You are given a string s that consists of the digits '1' to '9' and two integers k and minLength.

A partition of s is called beautiful if:

s is partitioned into k non-intersecting substrings.
Each substring has a length of at least minLength.
Each substring starts with a prime digit and ends with a non-prime digit. Prime digits are '2', '3', '5', and '7', and the rest of the digits are non-prime.

Return the number of beautiful partitions of s. Since the answer may be very large, return it modulo 10⁹ + 7.

A substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = "23542185131", k = 3, minLength = 2
Output: 3
Explanation: There exists three ways to create a beautiful partition:
"2354 | 218 | 5131"
"2354 | 21851 | 31"
"2354218 | 51 | 31"

Example 2:

Input: s = "23542185131", k = 3, minLength = 3
Output: 1
Explanation: There exists one way to create a beautiful partition: "2354 | 218 | 5131".

Example 3:

Input: s = "3312958", k = 3, minLength = 1
Output: 1
Explanation: There exists one way to create a beautiful partition: "331 | 29 | 58".

Constraints:

1 <= k, minLength <= s.length <= 1000
s consists of the digits '1' to '9'.

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 are the constraints on the integer values within the input array, and can they be negative or zero?
What is the maximum size of the input array, and should I be concerned about integer overflow during intermediate calculations?
Can you clarify the definition of a 'beautiful partition' more precisely, specifically how the sum of each partition affects the definition?
If there are multiple possible numbers of beautiful partitions, is any valid count acceptable, or is there a specific criteria for choosing which to return?
If no beautiful partitions exist, what value should I return? Is it 0, -1, or should I throw an exception?

Brute Force Solution

Approach

The brute force approach to finding beautiful partitions involves trying every single possible way to split a sequence into smaller groups. We exhaustively check if each grouping satisfies the 'beautiful' condition as we go. If it does, we count it; otherwise, we discard it.

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

Consider all possible ways to divide the sequence into the smallest possible number of groups, ensuring each group meets the minimum size requirement to be considered 'beautiful'.
Next, explore ways to divide the sequence into one more group than before, again checking if each grouping is 'beautiful'.
Continue increasing the number of groups, repeating the process of checking each possible division for the 'beautiful' condition.
Keep going until you've considered dividing the sequence into the maximum possible number of groups, still validating for the 'beautiful' requirement at each stage.
Tally up the number of ways you found to create 'beautiful' partitions across all the group sizes you considered.
The final count represents the total number of 'beautiful' ways you could divide the original sequence.

Code Implementation

def number_of_beautiful_partitions_brute_force(sequence, minimum_group_size):
    sequence_length = len(sequence)
    number_of_beautiful_partitions = 0

    # Iterate through the possible number of groups
    for number_of_groups in range(1, sequence_length // minimum_group_size + 1):
        # Helper function to generate all possible partitions
        def generate_partitions(index, current_partition):
            nonlocal number_of_beautiful_partitions

            if index == sequence_length:
                # Base case: We've reached the end of the sequence
                if len(current_partition) == number_of_groups:
                    is_beautiful = True
                    for group in current_partition:
                        if len(group) < minimum_group_size:
                            is_beautiful = False
                            break
                        if sum(group) < minimum_group_size:
                            is_beautiful = False
                            break

                    if is_beautiful:
                        number_of_beautiful_partitions += 1
                return

            # Explore adding the current element to the last group
            if current_partition and sum(current_partition[-1]) < minimum_group_size * len(current_partition[-1]):
                generate_partitions(index + 1, current_partition[:-1] + [current_partition[-1] + [sequence[index]]])
            else:
                # Create a new group
                if not current_partition or len(current_partition) < number_of_groups:
                    # Create a new partition
                    generate_partitions(index + 1, current_partition + [[sequence[index]]])

        # Start generating partitions from the beginning of the sequence
        generate_partitions(0, [])

    return number_of_beautiful_partitions

Big(O) Analysis

Time Complexity

O(2^n) – The described brute force approach explores all possible partitions of the input sequence of size n. Generating all possible partitions inherently requires considering each element and deciding whether to include it in the current group or start a new one, leading to 2 choices for each element. This results in approximately 2^n possible partitions to evaluate. Since each partition needs to be checked for the 'beautiful' condition, and the number of partitions dominates the checking cost, the overall time complexity is O(2^n).

Space Complexity

O(N) – The brute force approach, as described, explores all possible partitions. In the worst-case scenario, recursive calls (if implicit in checking each grouping) could stack up to a depth proportional to N, where N is the length of the sequence, to explore all possible splits. Each recursive call would maintain its own state, leading to a call stack of size N. Therefore, the auxiliary space used is proportional to the depth of the recursion, resulting in O(N) space complexity. The algorithm also keeps track of intermediate results to determine whether a given partition is 'beautiful', potentially using data structures of size at most N.

Optimal Solution

Approach

The problem asks us to count the number of ways to split a sequence into valid groups. The optimal approach uses dynamic programming to avoid redundant calculations and efficiently build the solution from smaller subproblems. We make a table to remember the number of ways to split the sequence up to a certain point, ensuring we only calculate each part once.

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

Create a table to store the number of beautiful partitions we've found for the sequence up to each point.
Start from the beginning of the sequence, and consider each possible group you can form that meets the minimum length requirement.
For each valid group, look up how many beautiful partitions you can make of the remaining sequence using the table we created.
Add the number of beautiful partitions of the remaining sequence to the number of beautiful partitions up to the end of the current group.
Store the total count of beautiful partitions in the table for that specific position in the sequence.
Repeat this process for each position in the sequence, building up the table as you go.
At the end, the table entry corresponding to the end of the entire sequence will contain the total number of beautiful partitions.

Code Implementation

def number_of_beautiful_partitions(numbers, minimum_length):    sequence_length = len(numbers)
    modulo = 10**9 + 7

    # Stores the number of beautiful partitions up to each index.
    beautiful_partitions_count = [0] * (sequence_length + 1)
    beautiful_partitions_count[0] = 1

    for i in range(1, sequence_length + 1):
        current_sum = 0
        for j in range(i, 0, -1):
            current_sum += numbers[j - 1]

            # Check if the current group is valid.
            if i - j + 1 >= minimum_length and current_sum % 2 == 0:

                # Add the partitions of the remaining sequence.
                beautiful_partitions_count[i] = (beautiful_partitions_count[i] + beautiful_partitions_count[j - 1]) % modulo

    # The last element contains the total number of partitions.
    return beautiful_partitions_count[sequence_length]

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the input sequence of size n to build a DP table. For each element, it checks potential group sizes, up to n elements. Therefore, in the worst-case scenario, the algorithm performs a nested loop-like operation where for each of the n elements, it potentially iterates through the rest of the sequence to find a valid partition point. This results in roughly n * n/2 operations. Thus, the overall time complexity is O(n²).

Space Complexity

O(N) – The algorithm creates a table (presumably an array or list) to store the number of beautiful partitions found for the sequence up to each point. Since this table stores values for each position in the sequence, the space required grows linearly with the input size N, where N is the length of the sequence. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 since there are no partitions possible.
k is zero or negative	If k <= 0, return 0 as no partition is considered beautiful.
Array size is smaller than k	Return 0 as it is impossible to have at least k beautiful partitions.
Array contains only 0s	The sum of any subarray will be 0, and we need to ensure valid partitions are counted depending on the condition whether a partition's sum can be 0 or not.
All elements in the array are the same and non-zero	We can potentially form many partitions as the sums will be multiples of that number, so calculate and return the possible beautiful partitions efficiently.
The sum of the entire array is not divisible by k	If the total sum of the array is not divisible by k, and the condition requires divisibility by k, return 0 since there will be no valid partitions.
Large array size, potential for integer overflow during calculations.	Use modulo operation during sum calculations to prevent integer overflow and potentially use long data types.
k is a very large number, potentially exceeding the total sum of elements in the array	Handle such cases by ensuring k is a valid divisor and if not return 0.

Null or empty input array

How to Handle:

Return 0 since there are no partitions possible.

k is zero or negative

How to Handle:

If k <= 0, return 0 as no partition is considered beautiful.

Array size is smaller than k

How to Handle:

Return 0 as it is impossible to have at least k beautiful partitions.

Array contains only 0s

How to Handle:

The sum of any subarray will be 0, and we need to ensure valid partitions are counted depending on the condition whether a partition's sum can be 0 or not.

All elements in the array are the same and non-zero

How to Handle:

We can potentially form many partitions as the sums will be multiples of that number, so calculate and return the possible beautiful partitions efficiently.

The sum of the entire array is not divisible by k

How to Handle:

If the total sum of the array is not divisible by k, and the condition requires divisibility by k, return 0 since there will be no valid partitions.

Large array size, potential for integer overflow during calculations.

How to Handle:

Use modulo operation during sum calculations to prevent integer overflow and potentially use long data types.

k is a very large number, potentially exceeding the total sum of elements in the array

How to Handle:

Handle such cases by ensuring k is a valid divisor and if not return 0.