Question

3085. Minimum Deletions to Make String K-Special

中文文档

Description

You are given a string word and an integer k.

We consider word to be k-special if |freq(word[i]) - freq(word[j])| <= k for all indices i and j in the string.

Here, freq(x) denotes the frequency of the character x in word, and |y| denotes the absolute value of y.

Return _the minimum number of characters you need to delete to make_ word _k-special_.

 

Example 1:

Input: word = "aabcaba", k = 0

Output: 3

Explanation: We can make word 0-special by deleting 2 occurrences of "a" and 1 occurrence of "c". Therefore, word becomes equal to "baba" where freq('a') == freq('b') == 2.

Example 2:

Input: word = "dabdcbdcdcd", k = 2

Output: 2

Explanation: We can make word 2-special by deleting 1 occurrence of "a" and 1 occurrence of "d". Therefore, word becomes equal to "bdcbdcdcd" where freq('b') == 2, freq('c') == 3, and freq('d') == 4.

Example 3:

Input: word = "aaabaaa", k = 2

Output: 1

Explanation: We can make word 2-special by deleting 1 occurrence of "b". Therefore, word becomes equal to "aaaaaa" where each letter's frequency is now uniformly 6.

 

Constraints:

• 1 <= word.length <= 105
• 0 <= k <= 105
• word consists only of lowercase English letters.

Solutions

Solution 1: Counting + Enumeration

First, we can count the occurrence of each character in the string and put all the counts into an array $nums$. Since the string only contains lowercase letters, the length of the array $nums$ will not exceed $26$.

Next, we can enumerate the minimum frequency $v$ of characters in the $K$ special strings within the range $[0,..n]$, and then use a function $f(v)$ to calculate the minimum number of deletions to adjust the frequency of all characters to $v$. The minimum value of all $f(v)$ is the answer.

The calculation method of function $f(v)$ is as follows:

Traverse each element $x$ in the array $nums$. If $x < v$, it means that we need to delete all characters with a frequency of $x$, and the number of deletions is $x$. If $x > v + k$, it means that we need to adjust all characters with a frequency of $x$ to $v + k$, and the number of deletions is $x - v - k$. The sum of all deletion counts is the value of $f(v)$.

The time complexity is $O(n \times |\Sigma|)$, and the space complexity is $O(|\Sigma|)$. Here, $n$ is the length of the string, and $|\Sigma|$ is the size of the character set. In this problem, $|\Sigma| = 26$.

class Solution:
  def minimumDeletions(self, word: str, k: int) -> int:
    def f(v: int) -> int:
      ans = 0
      for x in nums:
        if x < v:
          ans += x
        elif x > v + k:
          ans += x - v - k
      return ans

    nums = Counter(word).values()
    return min(f(v) for v in range(len(word) + 1))

class Solution {
  private List<Integer> nums = new ArrayList<>();

  public int minimumDeletions(String word, int k) {
    int[] freq = new int[26];
    int n = word.length();
    for (int i = 0; i < n; ++i) {
      ++freq[word.charAt(i) - 'a'];
    }
    for (int v : freq) {
      if (v > 0) {
        nums.add(v);
      }
    }
    int ans = n;
    for (int i = 0; i <= n; ++i) {
      ans = Math.min(ans, f(i, k));
    }
    return ans;
  }

  private int f(int v, int k) {
    int ans = 0;
    for (int x : nums) {
      if (x < v) {
        ans += x;
      } else if (x > v + k) {
        ans += x - v - k;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int minimumDeletions(string word, int k) {
    int freq[26]{};
    for (char& c : word) {
      ++freq[c - 'a'];
    }
    vector<int> nums;
    for (int v : freq) {
      if (v) {
        nums.push_back(v);
      }
    }
    int n = word.size();
    int ans = n;
    auto f = [&](int v) {
      int ans = 0;
      for (int x : nums) {
        if (x < v) {
          ans += x;
        } else if (x > v + k) {
          ans += x - v - k;
        }
      }
      return ans;
    };
    for (int i = 0; i <= n; ++i) {
      ans = min(ans, f(i));
    }
    return ans;
  }
};

func minimumDeletions(word string, k int) int {
  freq := [26]int{}
  for _, c := range word {
    freq[c-'a']++
  }
  nums := []int{}
  for _, v := range freq {
    if v > 0 {
      nums = append(nums, v)
    }
  }
  f := func(v int) int {
    ans := 0
    for _, x := range nums {
      if x < v {
        ans += x
      } else if x > v+k {
        ans += x - v - k
      }
    }
    return ans
  }
  ans := len(word)
  for i := 0; i <= len(word); i++ {
    ans = min(ans, f(i))
  }
  return ans
}

function minimumDeletions(word: string, k: number): number {
  const freq: number[] = Array(26).fill(0);
  for (const ch of word) {
    ++freq[ch.charCodeAt(0) - 97];
  }
  const nums = freq.filter(x => x > 0);
  const f = (v: number): number => {
    let ans = 0;
    for (const x of nums) {
      if (x < v) {
        ans += x;
      } else if (x > v + k) {
        ans += x - v - k;
      }
    }
    return ans;
  };
  return Math.min(...Array.from({ length: word.length + 1 }, (_, i) => f(i)));
}