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 $f(v)$.

Python3JavaC++GoTypeScript

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)));
}

3085. Minimum Deletions to Make String K-Special

Description

Solutions

Solution 1: Counting + Enumeration

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