Maximum Deletions on a String

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ArraysStringsDynamic Programming

You are given a string s consisting of only lowercase English letters. In one operation, you can:

Delete the entire string s, or
Delete the first i letters of s if the first i letters of s are equal to the following i letters in s, for any i in the range 1 <= i <= s.length / 2.

For example, if s = "ababc", then in one operation, you could delete the first two letters of s to get "abc", since the first two letters of s and the following two letters of s are both equal to "ab".

Return the maximum number of operations needed to delete all of s.

Example 1:

Input: s = "abcabcdabc"
Output: 2
Explanation:
- Delete the first 3 letters ("abc") since the next 3 letters are equal. Now, s = "abcdabc".
- Delete all the letters.
We used 2 operations so return 2. It can be proven that 2 is the maximum number of operations needed.
Note that in the second operation we cannot delete "abc" again because the next occurrence of "abc" does not happen in the next 3 letters.

Example 2:

Input: s = "aaabaab"
Output: 4
Explanation:
- Delete the first letter ("a") since the next letter is equal. Now, s = "aabaab".
- Delete the first 3 letters ("aab") since the next 3 letters are equal. Now, s = "aab".
- Delete the first letter ("a") since the next letter is equal. Now, s = "ab".
- Delete all the letters.
We used 4 operations so return 4. It can be proven that 4 is the maximum number of operations needed.

Example 3:

Input: s = "aaaaa"
Output: 5
Explanation: In each operation, we can delete the first letter of s.

Constraints:

1 <= s.length <= 4000
s consists only of lowercase English letters.

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 maximum possible length of the input string s?
If no deletion operations can be performed, what should the function return? Should I return 0, or is there a specific error code or null value expected?
Can the input string s be empty or null? If so, what should be the expected behavior?
If there are multiple valid series of deletion operations that result in the maximum number of deletions, is any valid series acceptable, or is there a specific deletion order that should be prioritized?
Does case sensitivity matter when comparing the prefix and suffix? For instance, should 'abc' be considered equal to 'ABC'?

Brute Force Solution

Approach

The brute force method for this problem involves checking every possible way to delete parts of the string. We want to find the maximum number of deletions we can make while still having a remaining string that satisfies the condition: it must be made up of repeated, identical substrings.

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

First, consider deleting nothing at all. Check if the entire original string can be made by repeating a smaller string.
Next, consider deleting one character from the original string. For each character you delete, check if the remaining string can be made by repeating a smaller string.
Now, consider deleting two characters from the original string. Try every possible combination of two character deletions, and for each combination, check if the remaining string can be made by repeating a smaller string.
Keep doing this, deleting three characters, then four, and so on, until you have tried deleting almost all the characters (leaving at least one character remaining).
For each of these 'remaining strings' which you have after some deletions, see if it can be formed by repeating some smaller string. For example, 'ABABAB' can be formed by repeating 'AB' three times.
Keep track of the maximum number of deletions you made and still have a 'remaining string' that meets the repeating substring condition.
The highest number of deletions you recorded is your answer.

Code Implementation

def maximum_deletions_on_a_string(input_string):
    max_deletions = -1
    string_length = len(input_string)

    for i in range(1 << string_length):
        current_string = ""
        deletions = 0

        for j in range(string_length):
            if (i >> j) & 1:
                current_string += input_string[j]
            else:
                deletions += 1

        if not current_string:
            continue

        # Check if the current string can be formed by repeating a substring
        for substring_length in range(1, len(current_string) + 1):
            if len(current_string) % substring_length == 0:
                substring = current_string[:substring_length]
                repetitions = len(current_string) // substring_length
                repeated_string = substring * repetitions

                # Ensure formed by repeating the substring
                if repeated_string == current_string:
                    max_deletions = max(max_deletions, deletions)
                    break

    return max_deletions

Big(O) Analysis

Time Complexity

O(2^n * n) – The algorithm explores all possible subsets of the string by deleting characters. There are 2^n possible subsets (each character can either be present or deleted). For each subset, we check if it can be formed by repeating a smaller string, which requires iterating through the subset (at most n operations). Therefore, the time complexity is approximately O(2^n * n), dominated by generating and checking each subset.

Space Complexity

O(N!) – The brute force method explores all possible subsets of the input string by considering deletions. Generating all combinations of character deletions requires storing intermediate strings. In the worst-case scenario, where almost all characters are deleted, the number of combinations grows factorially with N, the length of the original string. Therefore, the space complexity is dominated by the need to store these string combinations, leading to a space complexity of O(N!).

Optimal Solution

Approach

The best way to solve this is to use something called dynamic programming. Instead of trying every single combination of deletions, we'll build up the solution step by step, remembering the best results along the way to avoid repeating work.

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

Imagine starting at the end of the string. Think about how many characters we need to delete to make the remaining string a palindrome (reads the same forwards and backwards).
We'll work backwards, one character at a time. For each character, we'll consider two choices: either keep it or delete it.
If we keep the character, we'll check if it matches the corresponding character at the other end of the remaining string. If they match, great! We've made progress towards a longer palindrome. If they don't match, we might need to delete it eventually.
If we delete the character, we move on to the next character in the string and see what happens.
We'll store the smallest number of deletions needed to make the string a palindrome, starting from each position in the string. This stored information will help us make the best choices without recalculating things.
By working backwards and remembering the best deletion counts, we can find the fewest number of deletions needed for the whole string very efficiently.

Code Implementation

def maximum_deletions_on_a_string(input_string):
    string_length = len(input_string)
    
    # dp_table[i][j] is min deletions to make
    # input_string[i:j+1] a palindrome
    dp_table = [[0] * string_length for _ in range(string_length)]

    for sub_string_length in range(2, string_length + 1):
        for start_index in range(string_length - sub_string_length + 1):
            end_index = start_index + sub_string_length - 1

            if input_string[start_index] == input_string[end_index]:
                # Inherit optimal deletions from inner substring
                dp_table[start_index][end_index] = dp_table[start_index + 1][end_index - 1]

            else:
                # Delete either start or end and take min deletions
                dp_table[start_index][end_index] = min(
                    dp_table[start_index + 1][end_index] + 1,
                    dp_table[start_index][end_index - 1] + 1
                )

    # Return the minimum number of deletions for the entire string
    return dp_table[0][string_length - 1]

Big(O) Analysis

Time Complexity

O(n²) – The described dynamic programming approach involves iterating through the string to populate a table that stores the minimum number of deletions needed to make a substring a palindrome. The outer loop iterates backwards through the string of length n. For each character in the outer loop, the inner operation (checking if the characters at both ends match and then updating the dynamic programming table) takes O(1) time. Therefore, the number of inner loop executions is also proportional to n. So, the time complexity is approximately proportional to n * n, which is O(n²).

Space Complexity

O(N^2) – The dynamic programming approach, as described, involves storing the minimum number of deletions needed to make a substring a palindrome, starting from each position. This suggests the use of a table or matrix, likely of size N x N, where N is the length of the input string. Each cell in this table stores an intermediate result (the minimum deletions), contributing to auxiliary space. Therefore, the space complexity is proportional to the square of the input string's length, or O(N^2).

Edge Cases

Case	How to Handle
Empty string	Return 0 immediately as no deletion operations are possible.
Single character string	Return 0 immediately, since no prefix can equal the remaining suffix after its removal.
String with length 2 where characters are the same	Return 1, as the single character prefix matches the single character suffix after removing the prefix.
String with length 2 where characters are different	Return 0, as no prefix can equal the remaining suffix after its removal.
String of all the same characters (e.g., 'aaaa')	The optimal solution involves repeatedly removing single-character prefixes, resulting in a deletion count equal to the string length minus 1.
String where no prefix equals the corresponding suffix after removal.	Return 0 because no valid deletion operations can be performed.
Maximum size string to test efficiency of algorithm (e.g., length 10^5)	The algorithm should use a string comparison strategy to achieve O(n^2) time complexity in the worst case, or optimize it further potentially using dynamic programming or rolling hashes if it is not fast enough.
Very long string with repetitive patterns that cause many deletions	The algorithm needs to correctly handle multiple consecutive valid deletion operations without incorrectly short-circuiting or infinite looping.

Empty string

How to Handle:

Return 0 immediately as no deletion operations are possible.

Single character string

How to Handle:

Return 0 immediately, since no prefix can equal the remaining suffix after its removal.

String with length 2 where characters are the same

How to Handle:

Return 1, as the single character prefix matches the single character suffix after removing the prefix.

String with length 2 where characters are different

How to Handle:

Return 0, as no prefix can equal the remaining suffix after its removal.

String of all the same characters (e.g., 'aaaa')

How to Handle:

The optimal solution involves repeatedly removing single-character prefixes, resulting in a deletion count equal to the string length minus 1.

String where no prefix equals the corresponding suffix after removal.

How to Handle:

Return 0 because no valid deletion operations can be performed.

Maximum size string to test efficiency of algorithm (e.g., length 10^5)

How to Handle:

The algorithm should use a string comparison strategy to achieve O(n^2) time complexity in the worst case, or optimize it further potentially using dynamic programming or rolling hashes if it is not fast enough.

Very long string with repetitive patterns that cause many deletions

How to Handle:

The algorithm needs to correctly handle multiple consecutive valid deletion operations without incorrectly short-circuiting or infinite looping.