Problem Description

You are given a string s that contains only lowercase English letters and an integer k.

The goal is to delete some characters (you can delete zero or more characters) from the string such that the resulting string has at most k distinct characters.

For example:

• If s = "aabbcc" and k = 2, you could delete all 'c' characters to get "aabb", which has only 2 distinct characters ('a' and 'b'). The minimum deletions would be 2.
• If s = "abcdef" and k = 3, you need to remove 3 characters to keep only 3 distinct characters. You could delete 'd', 'e', 'f' to get "abc".

The task is to find the minimum number of character deletions needed to ensure the string has no more than k distinct characters.

The solution approach counts the frequency of each character using Counter(s).values(), sorts these frequencies in ascending order, and then sums up the frequencies of the characters that need to be removed (the first 26 - k characters when sorted, since there are at most 26 lowercase English letters). By removing characters with the lowest frequencies first, we minimize the total number of deletions.

Intuition

To minimize the number of deletions while keeping at most k distinct characters, we need to decide which characters to keep and which to remove completely.

The key insight is that if we're going to remove a character, we should remove all occurrences of that character. Why? Because keeping even one occurrence of a character still counts as one distinct character, so partial removal doesn't help us reduce the distinct character count.

Given this, the problem becomes: which characters should we completely remove to minimize total deletions?

The greedy approach is to remove the characters that appear least frequently first. This makes intuitive sense - if we have to remove entire characters, we want to remove the ones that contribute the fewest characters to our total count.

For example, if we have character frequencies: a:10, b:8, c:2, d:1 and we need to keep only 2 distinct characters, we should remove 'd' (1 deletion) and 'c' (2 deletions) for a total of 3 deletions, rather than removing 'a' or 'b' which would require many more deletions.

This leads us to the solution strategy:

1. Count the frequency of each character
2. Sort these frequencies in ascending order
3. Remove characters starting from the lowest frequency until we have at most k distinct characters remaining
4. The sum of frequencies of removed characters is our answer

The expression sorted(Counter(s).values())[:-k] elegantly captures this - it takes all frequency values except the k largest ones (which we keep), and sums the rest (which we delete).

Learn more about Greedy and Sorting patterns.

Solution Approach

The implementation uses a counting and greedy strategy to solve the problem efficiently.

Step 1: Count Character Frequencies

We use Python's Counter from the collections module to count the frequency of each character in the string s. The expression Counter(s) creates a dictionary-like object where keys are characters and values are their frequencies.

Step 2: Extract and Sort Frequencies

Counter(s).values() extracts just the frequency values (we don't need to know which character has which frequency). These values are then sorted in ascending order using sorted().

For example, if s = "aabbbbccc", we get:

• Counter: {'a': 2, 'b': 4, 'c': 3}
• Values: [2, 4, 3]
• Sorted: [2, 3, 4]

Step 3: Select Characters to Remove

The slice notation [:-k] selects all elements except the last k elements. Since our frequencies are sorted in ascending order, the last k elements represent the k most frequent characters - these are the ones we want to keep.

The characters corresponding to the frequencies in [:-k] are the ones we need to remove entirely.

Step 4: Calculate Total Deletions

sum() adds up all the frequencies of characters we're removing. This gives us the total number of deletions needed.

Example Walkthrough: If s = "aaabbbccd" and k = 2:

1. Counter gives us: {'a': 3, 'b': 3, 'c': 2, 'd': 1}
2. Sorted frequencies: [1, 2, 3, 3]
3. With k = 2, we take [:-2] which is [1, 2]
4. Sum is 1 + 2 = 3, meaning we delete 3 characters total

The time complexity is O(n + m log m) where n is the length of the string and m is the number of distinct characters (at most 26). The space complexity is O(m) for storing the character counts.

Example Walkthrough

Let's walk through the solution with s = "aabbccddde" and k = 2.

Step 1: Count Character Frequencies

• 'a' appears 2 times
• 'b' appears 2 times
• 'c' appears 2 times
• 'd' appears 3 times
• 'e' appears 1 time

So we have 5 distinct characters with frequencies: {a:2, b:2, c:2, d:3, e:1}

Step 2: Sort Frequencies in Ascending Order Extract just the frequency values: [2, 2, 2, 3, 1] Sort them: [1, 2, 2, 2, 3]

Step 3: Determine Which Characters to Remove We want to keep at most k = 2 distinct characters. Since we have 5 distinct characters total, we need to remove 3 characters completely.

Using [:-k] on our sorted list [1, 2, 2, 2, 3]:

• [:-2] gives us [1, 2, 2]
• These are the 3 smallest frequencies, representing the characters we'll remove

Step 4: Calculate Minimum Deletions Sum the frequencies of characters to remove: 1 + 2 + 2 = 5

This means we need to delete 5 characters total. Looking back at our original frequencies:

• We keep the 2 characters with highest frequencies: 'd' (3 occurrences) and one of the characters with 2 occurrences
• We remove: 'e' (1 deletion), and two of the characters that appear twice (4 deletions)
• Total: 5 deletions

The resulting string would have exactly 2 distinct characters remaining, achieved with the minimum possible deletions.

Solution Implementation

Time and Space Complexity

Time Complexity: O(|Σ| × log |Σ|) where |Σ| is the size of the character set. In this problem, |Σ| = 26 since we're dealing with lowercase English letters.

The time complexity breaks down as follows:

• Counter(s) takes O(n) time where n is the length of string s
• Counter(s).values() returns at most |Σ| frequency values
• sorted() on these values takes O(|Σ| × log |Σ|) time
• Slicing [:-k] takes O(|Σ|) time
• sum() takes O(|Σ|) time

Since n ≥ |Σ| (the string length is at least as large as the number of unique characters), and |Σ| is constant (26), the dominant operation is the sorting step, giving us O(|Σ| × log |Σ|).

Space Complexity: O(|Σ|) where |Σ| = 26

The space complexity comes from:

• Counter(s) stores at most |Σ| key-value pairs: O(|Σ|)
• Counter(s).values() creates a view/list of at most |Σ| values: O(|Σ|)
• sorted() creates a new list of at most |Σ| elements: O(|Σ|)

The overall space complexity is O(|Σ|).

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

Common Pitfalls

1. Edge Case: When k = 0

Pitfall: The slice [:-k] with k = 0 results in [:-0], which returns an empty list instead of the full list. This would incorrectly return 0 deletions when we actually need to delete all characters.

Example:

• Input: s = "abc", k = 0
• Expected: Delete all 3 characters
• Buggy result: sorted_frequencies[:-0] returns [], sum is 0

Solution:

def minDeletion(self, s: str, k: int) -> int:
    char_frequencies = Counter(s)
    sorted_frequencies = sorted(char_frequencies.values())
  
    # Handle k = 0 case explicitly
    if k == 0:
        return len(s)
  
    # For k > 0, use the slicing approach
    return sum(sorted_frequencies[:-k])

2. When k Exceeds Number of Distinct Characters

Pitfall: If k is greater than or equal to the number of distinct characters, the slice [:-k] might behave unexpectedly. For instance, if there are 3 distinct characters and k = 5, [:-5] would return an empty list, which is correct but not immediately obvious.

Better Approach for Clarity:

def minDeletion(self, s: str, k: int) -> int:
    char_frequencies = Counter(s)
    sorted_frequencies = sorted(char_frequencies.values())
  
    # No deletions needed if k covers all distinct characters
    if k >= len(sorted_frequencies):
        return 0
  
    # Sum frequencies of characters to remove (smallest frequencies first)
    return sum(sorted_frequencies[:len(sorted_frequencies) - k])

3. Misunderstanding the Slice Notation

Pitfall: Confusing [:-k] with [:k]. The notation [:-k] means "all elements except the last k", while [:k] means "the first k elements".

Incorrect Implementation:

# WRONG: This would keep characters with lowest frequencies
return sum(sorted_frequencies[:k])

Correct Implementation:

# RIGHT: Remove characters with lowest frequencies
return sum(sorted_frequencies[:-k])

4. Not Handling Empty String

Pitfall: While Counter("") handles empty strings correctly (returns empty Counter), it's good practice to explicitly handle this edge case for clarity.

Robust Solution:

def minDeletion(self, s: str, k: int) -> int:
    if not s or k >= 26:  # Empty string or k covers all possible lowercase letters
        return 0
  
    char_frequencies = Counter(s)
    sorted_frequencies = sorted(char_frequencies.values())
  
    if k == 0:
        return len(s)
  
    if k >= len(sorted_frequencies):
        return 0
  
    return sum(sorted_frequencies[:-k])