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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 inb
in the alphabet.  Every letter in
b
is strictly less than every letter ina
in the alphabet.  Both
a
andb
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 <= 10^{5}
a
andb
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+ncnt1[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; }