3039. Apply Operations to Make String Empty

Problem Description

The task is to repeatedly delete the first occurrence of each alphabet character in a string s from 'a' to 'z', until the string becomes empty. At each stage, only one occurrence of each letter is removed, if that letter is present in the string. This operation is carried out repeatedly. The goal is to find the string s right before the last round of the operation is applied. In simpler terms, we want to know what s looks like when it contains just enough characters for one last complete pass (removing 'a' to 'z' sequence). For example, if we begin with s = "aabcbbca", after performing these operations consecutively, prior to the final operation that empties it, the string is reduced to "ba".

Intuition

To compute the string right before the last operation, the core idea relies on the observation that only the characters that have the highest frequency will be potentially left before the final removal (because all others will get eliminated in earlier rounds). So our approach involves two main steps:

1. Count the frequency of each character in the string. The character(s) with the maximum frequency will determine the number of operations needed before the string becomes empty, as they define the 'pace' at which the string is reduced through each operation set.

2. For each character that has the maximum frequency, we need to check if its current position in the string corresponds to the last occurrence of that character. If both conditions hold true for a character (maximum frequency and the position is the last occurrence), it will survive until right before the final operation.

To implement this idea:

• Use a data structure like a hash table or Counter (in Python) to record the number of occurrences of each character in the string. Determine the maximum occurrence count mx.

• Create another hash table to record the last occurrence index of each character in the string.

• Iterate through the string and for each character, check if the number of occurrences is equal to mx and its index is the last occurrence. If so, this character is part of the string right before the last deletion operation.

• Combine all such characters that meet the criteria to form the desired result.

By following this solution approach, we can ensure that only the characters that could possibly remain till the penultimate operation are included in the final string.

Learn more about Sorting patterns.

Solution Approach

The provided solution uses the Counter class from the Python collections module to efficiently count the occurrences of each character in the string. An additional dictionary, last, is created to store the index of the last occurrence of each character in the string. This approach leverages hash tables (dictionaries in Python), which offer efficient O(1) average time complexity for lookup, insert, and update operations. This is vital for keeping the overall solution efficient.

The implementation steps are as follows:

1. Count Occurrences: A Counter object, cnt, captures the frequency of each character by iterating over the string once (O(n) time complexity, where n is the length of s).

2. Find Maximum Frequency: The most_common method of the Counter object is then used to find character frequency, mx, that occurs most often (O(k) time complexity, where k is the number of distinct characters in the string; typically k <= 26).

3. Record Last Index: A dictionary, last, maps each character to the last index at which it appears in the string. This also involves iterating over the string once.

4. Construct Result String: Finally, the program iterates through s again and includes a character c in the result only if c meets both of the following conditions:

• cnt[c] == mx: The occurrence count of c matches the maximum frequency. This ensures we only consider characters that can last until just before the final operation.
• last[c] == i: The current index i is the last occurrence index of c. This condition ensures that for a character to be included in the final string, it must be the last one of its kind within s.

These combined conditions ensure we only append to the resulting string those characters that will be removed in the last operation. This guarantees the string constructed will be the exact string available right before the last operation is performed.

By iterating through the characters and checking these conditions, the solution constructs the answer in a single pass, i.e., in O(n) time complexity, which makes the overall algorithm run in linear time with respect to the length of the input string.

Example Walkthrough

Let's illustrate the solution approach using the example string s = "aabcddcbba".

1. Count Occurrences: We first count the occurrences of each character using a Counter object.

   • {'a': 3, 'b': 3, 'c': 2, 'd': 2}

2. Find Maximum Frequency: Using the most_common method of the Counter object, we find the maximum occurrence frequency, mx.

   • mx = 3 (both 'a' and 'b' occur 3 times)

3. Record Last Index: We record the last index where each character appears in the string.

   • {'a': 8, 'b': 9, 'c': 7, 'd': 4}

4. Construct Result String: We iterate through s from left to right, checking if each character meets the conditions to be appended to the result.

   • First 'a' at index 0: cnt['a'] == mx is True, but last['a'] == 0 is False. We do not include this 'a'.
   • Second 'a' at index 1: cnt['a'] == mx is True, but last['a'] == 1 is False. We do not include this 'a'.
   • Third 'a' at index 8: cnt['a'] == mx is True and last['a'] == 8 is also True. We include this 'a'.
   • Apply similar logic to other characters.

Result would include the third 'a' and the second 'b' because those are the last occurrences and they also have the maximum frequency (mx). No other character satisfies both conditions, so they will all have been removed before the penultimate operation.

Hence, right before the last round that removes 'a' to 'z', string s looks like "ab".

Solution Implementation

Time and Space Complexity

The time complexity of the provided code is O(n) where n is the length of the input string s. This is because the code iterates over the string multiple times independently: first, to count the occurrences of each character using Counter, and then to create a dictionary with the index of the last occurrence of each character. Finally, it iterates over the string again to build the result string.

The space complexity is O(|Σ|) where |Σ| is the size of the character set which, in the case of lowercase English letters, is 26. The space used by the Counter object and the last occurrence dictionary both depend on the number of different characters in the string, not the size of s itself.

Learn more about how to find time and space complexity quickly using problem constraints.