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1737. Change Minimum Characters to Satisfy One of Three Conditions

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 {
    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;
    }
};

  • 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


  • 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;
}

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