Question

1737. Change Minimum Characters to Satisfy One of Three Conditions

中文文档

Description

You are given two strings a and b that consist of lowercase letters. In one operation, you can change any character in a or b to any lowercase letter.

Your goal is to satisfy one of the following three conditions:

• Every letter in a is strictly less than every letter in b in the alphabet.
• Every letter in b is strictly less than every letter in a in the alphabet.
• Both a and b consist of only one distinct letter.

Return _the minimum number of operations needed to achieve your goal._

 

Example 1:

Input: a = "aba", b = "caa"
Output: 2
Explanation: Consider the best way to make each condition true:
1) Change b to "ccc" in 2 operations, then every letter in a is less than every letter in b.
2) Change a to "bbb" and b to "aaa" in 3 operations, then every letter in b is less than every letter in a.
3) Change a to "aaa" and b to "aaa" in 2 operations, then a and b consist of one distinct letter.
The best way was done in 2 operations (either condition 1 or condition 3).

Example 2:

Input: a = "dabadd", b = "cda"
Output: 3
Explanation: The best way is to make condition 1 true by changing b to "eee".

 

Constraints:

• 1 <= a.length, b.length <= 105
• a and b consist only of lowercase letters.

Solutions

Solution 1: Counting + Enumeration

First, we count the number of occurrences of each letter in strings $a$ and $b$, denoted as $cnt_1$ and $cnt_2$.

Then, we consider condition $3$, i.e., every letter in $a$ and $b$ is the same. We just need to enumerate the final letter $c$, and then count the number of letters in $a$ and $b$ that are not $c$. This is the number of characters that need to be changed.

Next, we consider conditions $1$ and $2$, i.e., every letter in $a$ is less than every letter in $b$, or every letter in $b$ is less than every letter in $a$. For condition $1$, we make all characters in string $a$ less than character $c$, and all characters in string $b$ not less than $c$. We enumerate $c$ to find the smallest answer. Condition $2$ is similar.

The final answer is the minimum of the above three cases.

The time complexity is $O(m + n + C^2)$, where $m$ and $n$ are the lengths of strings $a$ and $b$ respectively, and $C$ is the size of the character set. In this problem, $C = 26$.

class Solution:
  def minCharacters(self, a: str, b: str) -> int:
    def f(cnt1, cnt2):
      for i in range(1, 26):
        t = sum(cnt1[i:]) + sum(cnt2[:i])
        nonlocal ans
        ans = min(ans, t)

    m, n = len(a), len(b)
    cnt1 = [0] * 26
    cnt2 = [0] * 26
    for c in a:
      cnt1[ord(c) - ord('a')] += 1
    for c in b:
      cnt2[ord(c) - ord('a')] += 1
    ans = m + n
    for c1, c2 in zip(cnt1, cnt2):
      ans = min(ans, m + n - c1 - c2)
    f(cnt1, cnt2)
    f(cnt2, cnt1)
    return ans

class Solution {
  private int ans;

  public int minCharacters(String a, String b) {
    int m = a.length(), n = b.length();
    int[] cnt1 = new int[26];
    int[] cnt2 = new int[26];
    for (int i = 0; i < m; ++i) {
      ++cnt1[a.charAt(i) - 'a'];
    }
    for (int i = 0; i < n; ++i) {
      ++cnt2[b.charAt(i) - 'a'];
    }
    ans = m + n;
    for (int i = 0; i < 26; ++i) {
      ans = Math.min(ans, m + n - cnt1[i] - cnt2[i]);
    }
    f(cnt1, cnt2);
    f(cnt2, cnt1);
    return ans;
  }

  private void f(int[] cnt1, int[] cnt2) {
    for (int i = 1; i < 26; ++i) {
      int t = 0;
      for (int j = i; j < 26; ++j) {
        t += cnt1[j];
      }
      for (int j = 0; j < i; ++j) {
        t += cnt2[j];
      }
      ans = Math.min(ans, t);
    }
  }
}

class Solution {
public:
  int minCharacters(string a, string b) {
    int m = a.size(), n = b.size();
    vector<int> cnt1(26);
    vector<int> cnt2(26);
    for (char& c : a) ++cnt1[c - 'a'];
    for (char& c : b) ++cnt2[c - 'a'];
    int ans = m + n;
    for (int i = 0; i < 26; ++i) ans = min(ans, m + n - cnt1[i] - cnt2[i]);
    auto f = [&](vector<int>& cnt1, vector<int>& cnt2) {
      for (int i = 1; i < 26; ++i) {
        int t = 0;
        for (int j = i; j < 26; ++j) t += cnt1[j];
        for (int j = 0; j < i; ++j) t += cnt2[j];
        ans = min(ans, t);
      }
    };
    f(cnt1, cnt2);
    f(cnt2, cnt1);
    return ans;
  }
};

func minCharacters(a string, b string) int {
  cnt1 := [26]int{}
  cnt2 := [26]int{}
  for _, c := range a {
    cnt1[c-'a']++
  }
  for _, c := range b {
    cnt2[c-'a']++
  }
  m, n := len(a), len(b)
  ans := m + n
  for i := 0; i < 26; i++ {
    ans = min(ans, m+n-cnt1[i]-cnt2[i])
  }
  f := func(cnt1, cnt2 [26]int) {
    for i := 1; i < 26; i++ {
      t := 0
      for j := i; j < 26; j++ {
        t += cnt1[j]
      }
      for j := 0; j < i; j++ {
        t += cnt2[j]
      }
      ans = min(ans, t)
    }
  }
  f(cnt1, cnt2)
  f(cnt2, cnt1)
  return ans
}

function minCharacters(a: string, b: string): number {
  const m = a.length,
    n = b.length;
  let count1 = new Array(26).fill(0);
  let count2 = new Array(26).fill(0);
  const base = 'a'.charCodeAt(0);

  for (let char of a) {
    count1[char.charCodeAt(0) - base]++;
  }
  for (let char of b) {
    count2[char.charCodeAt(0) - base]++;
  }

  let pre1 = 0,
    pre2 = 0;
  let ans = m + n;
  for (let i = 0; i < 25; i++) {
    pre1 += count1[i];
    pre2 += count2[i];
    // case1， case2， case3
    ans = Math.min(ans, m - pre1 + pre2, pre1 + n - pre2, m + n - count1[i] - count2[i]);
  }
  ans = Math.min(ans, m + n - count1[25] - count2[25]);

  return ans;
}