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3085. Minimum Deletions to Make String K-Special

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, $|\Sigma| = 26$.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  };
  

• 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))
  
  

• 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)));
  }
  
  

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