Problem Description

You are given a string word containing only lowercase letters and an integer k.

A string is considered k-special if the difference between the frequencies of any two characters in the string is at most k. In other words, for every pair of characters in the string, the absolute difference between their occurrence counts should not exceed k.

Mathematically, word is k-special if |freq(word[i]) - freq(word[j])| <= k for all indices i and j, where freq(x) represents how many times character x appears in the string.

Your task is to find the minimum number of characters you need to delete from word to make it k-special.

For example, if you have a string where one character appears 10 times and another appears 2 times, and k = 3, the difference is |10 - 2| = 8, which is greater than 3. So you would need to delete some characters to reduce this gap to at most 3.

The solution works by:

1. Counting the frequency of each character in the string
2. Trying different possible minimum frequencies v from 0 to the length of the string
3. For each minimum frequency v, calculating how many deletions are needed:
   • Characters with frequency less than v must be completely removed
   • Characters with frequency greater than v + k must be reduced to v + k
4. Returning the minimum number of deletions across all possible values of v

Intuition

The key insight is that in a k-special string, all character frequencies must fall within a range of size k. Think of it as finding a "window" of width k on a number line where we want to fit all our character frequencies.

Since we can only delete characters (not add them), we can only reduce frequencies. This means we need to find an optimal range [v, v+k] where v is some minimum frequency value, and adjust all character frequencies to fit within this range.

For any character with frequency less than v, we have no choice but to delete all occurrences since we can't increase its frequency. For characters with frequency greater than v+k, we need to delete just enough to bring them down to v+k.

The challenge is finding the best value of v. If we set v too high, we'll need to delete many characters that have low frequencies. If we set v too low, we might need to delete many occurrences from high-frequency characters to bring them down to v+k.

The solution elegantly handles this by trying all possible values of v from 0 to n (the length of the string). For each candidate value of v, we calculate the total deletions needed using a simple formula:

• For frequencies below v: delete all (cost = x)
• For frequencies above v+k: delete excess (cost = x - v - k)
• For frequencies in range [v, v+k]: no deletions needed

By checking all possible values of v and taking the minimum, we guarantee finding the optimal solution. This brute force approach works efficiently because we only have at most 26 distinct character frequencies (one for each lowercase letter), making the computation fast regardless of the string length.

Learn more about Greedy and Sorting patterns.

Solution Approach

The implementation follows a counting and enumeration strategy:

Step 1: Count Character Frequencies

First, we use Python's Counter to count the frequency of each character in the string and extract just the frequency values into nums:

nums = Counter(word).values()

This gives us a collection of frequencies, one for each unique character in the string. For example, if word = "aabbbc", then nums would contain [2, 3, 1] (frequencies of 'a', 'b', and 'c').

Step 2: Define the Cost Function

The helper function f(v) calculates the minimum deletions needed when we choose v as the minimum frequency in our target range [v, v+k]:

def f(v: int) -> int:
    ans = 0
    for x in nums:
        if x < v:
            ans += x  # Delete all occurrences
        elif x > v + k:
            ans += x - v - k  # Delete excess to bring down to v+k
    return ans

The logic is straightforward:

• If a character's frequency x is less than v, we must delete all x occurrences
• If x is greater than v + k, we delete x - (v + k) characters to bring it down to exactly v + k
• If x is already in the range [v, v+k], no deletions are needed

Step 3: Find the Minimum

We enumerate all possible values of v from 0 to the length of the string:

return min(f(v) for v in range(len(word) + 1))

Why this range? The minimum frequency v can be at most n (if we keep all occurrences of a single character). Values beyond n don't make sense since no character can appear more than n times in a string of length n.

Time Complexity: O(n × m) where n is the length of the string and m is the number of unique characters (at most 26). For each value of v (n+1 iterations), we check all character frequencies (at most 26).

Space Complexity: O(m) for storing the frequency counts, where m is at most 26.

The elegance of this solution lies in its simplicity - by trying all possible minimum frequencies and calculating the cost for each, we guarantee finding the optimal solution without complex optimization techniques.

Example Walkthrough

Let's walk through the solution with a concrete example:

Input: word = "aaabbbcc", k = 1

Step 1: Count Character Frequencies

• 'a' appears 3 times
• 'b' appears 3 times
• 'c' appears 2 times
• So nums = [3, 3, 2]

Step 2: Check if Already k-special

• Maximum difference: |3 - 2| = 1 ≤ k, so the string is already k-special!
• But let's see how the algorithm confirms this.

Step 3: Try All Possible Minimum Frequencies

For each value v from 0 to 8 (length of string), calculate deletions needed:

• v = 0: Range is [0, 1]

  • 'a' (freq=3): 3 > 1, delete 3-1 = 2 characters
  • 'b' (freq=3): 3 > 1, delete 3-1 = 2 characters
  • 'c' (freq=2): 2 > 1, delete 2-1 = 1 character
  • Total deletions: 2 + 2 + 1 = 5

• v = 1: Range is [1, 2]

  • 'a' (freq=3): 3 > 2, delete 3-2 = 1 character
  • 'b' (freq=3): 3 > 2, delete 3-2 = 1 character
  • 'c' (freq=2): 2 is in range, delete 0
  • Total deletions: 1 + 1 + 0 = 2

• v = 2: Range is [2, 3]

  • 'a' (freq=3): 3 is in range, delete 0
  • 'b' (freq=3): 3 is in range, delete 0
  • 'c' (freq=2): 2 is in range, delete 0
  • Total deletions: 0 + 0 + 0 = 0 ✓

• v = 3: Range is [3, 4]

  • 'a' (freq=3): 3 is in range, delete 0
  • 'b' (freq=3): 3 is in range, delete 0
  • 'c' (freq=2): 2 < 3, delete all 2 characters
  • Total deletions: 0 + 0 + 2 = 2

• v ≥ 4: All frequencies are below v, so we'd delete everything (8 deletions)

Step 4: Return Minimum The minimum deletions across all values is 0 (at v=2), confirming the string is already k-special.

Let's trace through a second example where deletions are needed:

Input: word = "aaaaabbc", k = 1

• Frequencies: 'a'=5, 'b'=2, 'c'=1, so nums = [5, 2, 1]
• Maximum difference: |5 - 1| = 4 > 1, so we need deletions

Testing key values of v:

• v = 1: Range [1, 2] → Delete 3 'a's (5→2), keep 'b's, keep 'c' → Total: 3
• v = 2: Range [2, 3] → Delete 2 'a's (5→3), keep 'b's, delete all 'c' → Total: 3

The minimum is 3 deletions, resulting in a string like "aabbc" where all frequencies are within range [1, 2].

Solution Implementation

Time and Space Complexity

The time complexity is O(n × |Σ|), where n is the length of the string word and |Σ| represents the size of the character set (which is 26 for lowercase English letters).

Breaking down the analysis:

• Counter(word).values() takes O(n) time to count character frequencies
• The outer loop for v in range(len(word) + 1) iterates up to n + 1 times
• For each value of v, the function f(v) iterates through all unique character frequencies in nums, which is at most |Σ| (since there can be at most 26 unique characters)
• Therefore, the total time complexity is O(n) + O(n × |Σ|) = O(n × |Σ|)

The space complexity is O(|Σ|):

• The Counter object and nums store at most |Σ| frequency values (one for each unique character in the string)
• No other significant auxiliary space is used
• Thus, the space complexity is O(|Σ|), which equals O(26) = O(1) for this specific problem with lowercase English letters

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

Common Pitfalls

1. Incorrectly Assuming We Need to Match a Specific Frequency

A common misconception is thinking we need to make all characters have the same frequency or match some existing frequency in the string. This leads to solutions that only consider existing frequencies as targets.

Wrong Approach:

def minimumDeletions(self, word: str, k: int) -> int:
    frequencies = list(Counter(word).values())
    min_deletions = float('inf')
  
    # WRONG: Only considering existing frequencies as targets
    for target in frequencies:
        deletions = 0
        for freq in frequencies:
            if abs(freq - target) > k:
                # This doesn't correctly handle the range [v, v+k]
                deletions += abs(freq - target) - k
        min_deletions = min(min_deletions, deletions)
  
    return min_deletions

Why it's wrong: The problem allows frequencies to be in a range [v, v+k], not centered around a specific value. We need to consider all possible minimum values v, not just existing frequencies.

2. Misunderstanding the Frequency Range Constraint

Another pitfall is misinterpreting how the k-special condition works, thinking characters must be within k distance of some central frequency rather than all being within a range of size k.

Wrong Approach:

def minimumDeletions(self, word: str, k: int) -> int:
    frequencies = list(Counter(word).values())
    avg_freq = sum(frequencies) // len(frequencies)
  
    deletions = 0
    for freq in frequencies:
        # WRONG: Trying to bring everything close to average
        if freq < avg_freq - k:
            deletions += avg_freq - k - freq
        elif freq > avg_freq + k:
            deletions += freq - avg_freq - k
  
    return deletions

Why it's wrong: The solution doesn't require centering around an average or median. We need to find an optimal range [v, v+k] that minimizes deletions.

3. Off-by-One Errors in Range Iteration

A subtle but critical error is using the wrong upper bound when iterating through possible minimum frequencies.

Wrong Approach:

def minimumDeletions(self, word: str, k: int) -> int:
    frequencies = list(Counter(word).values())
  
    # WRONG: Should be range(len(word) + 1), not range(max(frequencies) + 1)
    return min(calculate_cost(v) for v in range(max(frequencies) + 1))

Why it's wrong: While iterating only up to max(frequencies) might seem logical, we could miss optimal solutions where v is higher than the current maximum frequency. The correct upper bound is len(word) + 1 because that's the theoretical maximum any character could appear.

4. Incorrect Deletion Calculation for Out-of-Range Frequencies

When a frequency exceeds v + k, some might incorrectly calculate the deletions needed.

Wrong Approach:

def calculate_deletions(v: int) -> int:
    deletions = 0
    for freq in frequencies:
        if freq < v:
            deletions += freq
        elif freq > v + k:
            # WRONG: Should delete (freq - v - k), not just k
            deletions += k
    return deletions

Correct Approach: When frequency > v + k, we need to delete exactly freq - (v + k) characters to bring it down to v + k, not some fixed amount.

Solution to Avoid These Pitfalls:

1. Always enumerate all possible minimum frequencies from 0 to the string length
2. Understand the range constraint means all frequencies must fit within a window of size k
3. Carefully calculate deletions:
   • Below range: delete all
   • Above range: delete excess to reach upper bound
   • Within range: keep as is
4. Test with edge cases like single character strings, k=0, and strings where all characters have the same frequency