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feat: add solutions to lc problem: No.1737
No.1737.Change Minimum Characters to Satisfy One of Three Conditions
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‎solution/1700-1799/1737.Change Minimum Characters to Satisfy One of Three Conditions/README.md

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@@ -51,7 +51,17 @@
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<!-- 这里可写通用的实现逻辑 -->
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前缀和
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**方法一：计数 + 枚举**
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我们先统计字符串 $a$ 和 $b$ 中每个字母出现的次数，记为 $cnt_1$ 和 $cnt_2$。
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然后考虑条件 $3$，即 $a$ 和 $b$ 中的每个字母都相同。我们只需要枚举最终的字母 $c$，然后统计 $a$ 和 $b$ 中不是 $c$ 的字母的个数，即为需要改变的字符个数。
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然后考虑条件 $1$ 和 $2$，即 $a$ 中的每个字母都小于 $b$ 中的每个字母，或者 $b$ 中的每个字母都小于 $a$ 中的每个字母。对于条件 $1$，我们令字符串 $a$ 所有字符都小于字符 $c$，字符串 $b$ 所有字符都不小于 $c$，枚举 $c$ 找出最小的答案即可。条件 $2$ 同理。
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最终的答案即为上述三种情况的最小值。
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时间复杂度 $O(m + n + C^2)$，其中 $m$ 和 $n$ 分别为字符串 $a$ 和 $b$ 的长度，而 $C$ 为字符集大小。本题中 $C = 26$。
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<!-- tabs:start -->
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minCharacters(self, a: str, b: str) -> int:
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def f(cnt1, cnt2):
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for i in range(1, 26):
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t = sum(cnt1[i:]) + sum(cnt2[:i])
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nonlocal ans
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ans = min(ans, t)
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m, n = len(a), len(b)
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cnt1 = [0] * 26
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cnt2 = [0] * 26
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for c in a:
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cnt1[ord(c) - ord('a')] += 1
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for c in b:
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cnt2[ord(c) - ord('a')] += 1
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ans = m + n
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for c1, c2 in zip(cnt1, cnt2):
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ans = min(ans, m + n - c1 - c2)
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f(cnt1, cnt2)
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f(cnt2, cnt1)
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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private int ans;
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public int minCharacters(String a, String b) {
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int m = a.length(), n = b.length();
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int[] cnt1 = new int[26];
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int[] cnt2 = new int[26];
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for (int i = 0; i < m; ++i) {
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++cnt1[a.charAt(i) - 'a'];
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}
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for (int i = 0; i < n; ++i) {
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++cnt2[b.charAt(i) - 'a'];
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}
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ans = m + n;
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for (int i = 0; i < 26; ++i) {
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ans = Math.min(ans, m + n - cnt1[i] - cnt2[i]);
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}
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f(cnt1, cnt2);
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f(cnt2, cnt1);
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return ans;
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}
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private void f(int[] cnt1, int[] cnt2) {
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for (int i = 1; i < 26; ++i) {
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int t = 0;
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for (int j = i; j < 26; ++j) {
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t += cnt1[j];
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}
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for (int j = 0; j < i; ++j) {
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t += cnt2[j];
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}
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ans = Math.min(ans, t);
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}
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minCharacters(string a, string b) {
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int m = a.size(), n = b.size();
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vector<int> cnt1(26);
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vector<int> cnt2(26);
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for (char& c : a) ++cnt1[c - 'a'];
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for (char& c : b) ++cnt2[c - 'a'];
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int ans = m + n;
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for (int i = 0; i < 26; ++i) ans = min(ans, m + n - cnt1[i] - cnt2[i]);
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auto f = [&](vector<int>& cnt1, vector<int>& cnt2) {
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for (int i = 1; i < 26; ++i) {
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int t = 0;
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for (int j = i; j < 26; ++j) t += cnt1[j];
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for (int j = 0; j < i; ++j) t += cnt2[j];
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ans = min(ans, t);
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}
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};
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f(cnt1, cnt2);
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f(cnt2, cnt1);
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return ans;
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}
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};
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```
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### **Go**
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```go
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func minCharacters(a string, b string) int {
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cnt1 := [26]int{}
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cnt2 := [26]int{}
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for _, c := range a {
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cnt1[c-'a']++
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}
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for _, c := range b {
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cnt2[c-'a']++
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}
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m, n := len(a), len(b)
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ans := m + n
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for i := 0; i < 26; i++ {
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ans = min(ans, m+n-cnt1[i]-cnt2[i])
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}
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f := func(cnt1, cnt2 [26]int) {
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for i := 1; i < 26; i++ {
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t := 0
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for j := i; j < 26; j++ {
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t += cnt1[j]
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}
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for j := 0; j < i; j++ {
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t += cnt2[j]
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}
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ans = min(ans, t)
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}
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}
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f(cnt1, cnt2)
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f(cnt2, cnt1)
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return ans
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **TypeScript**

‎solution/1700-1799/1737.Change Minimum Characters to Satisfy One of Three Conditions/README_EN.md

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@@ -54,13 +54,137 @@ Prefix Sum
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### **Python3**
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```python
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class Solution:
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def minCharacters(self, a: str, b: str) -> int:
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def f(cnt1, cnt2):
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for i in range(1, 26):
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t = sum(cnt1[i:]) + sum(cnt2[:i])
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nonlocal ans
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ans = min(ans, t)
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m, n = len(a), len(b)
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cnt1 = [0] * 26
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cnt2 = [0] * 26
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for c in a:
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cnt1[ord(c) - ord('a')] += 1
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for c in b:
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cnt2[ord(c) - ord('a')] += 1
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ans = m + n
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for c1, c2 in zip(cnt1, cnt2):
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ans = min(ans, m + n - c1 - c2)
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f(cnt1, cnt2)
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f(cnt2, cnt1)
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return ans
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```
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### **Java**
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```java
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class Solution {
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private int ans;
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public int minCharacters(String a, String b) {
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int m = a.length(), n = b.length();
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int[] cnt1 = new int[26];
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int[] cnt2 = new int[26];
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for (int i = 0; i < m; ++i) {
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++cnt1[a.charAt(i) - 'a'];
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}
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for (int i = 0; i < n; ++i) {
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++cnt2[b.charAt(i) - 'a'];
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}
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ans = m + n;
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for (int i = 0; i < 26; ++i) {
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ans = Math.min(ans, m + n - cnt1[i] - cnt2[i]);
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}
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f(cnt1, cnt2);
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f(cnt2, cnt1);
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return ans;
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}
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private void f(int[] cnt1, int[] cnt2) {
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for (int i = 1; i < 26; ++i) {
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int t = 0;
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for (int j = i; j < 26; ++j) {
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t += cnt1[j];
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}
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for (int j = 0; j < i; ++j) {
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t += cnt2[j];
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}
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ans = Math.min(ans, t);
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}
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minCharacters(string a, string b) {
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int m = a.size(), n = b.size();
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vector<int> cnt1(26);
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vector<int> cnt2(26);
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for (char& c : a) ++cnt1[c - 'a'];
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for (char& c : b) ++cnt2[c - 'a'];
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int ans = m + n;
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for (int i = 0; i < 26; ++i) ans = min(ans, m + n - cnt1[i] - cnt2[i]);
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auto f = [&](vector<int>& cnt1, vector<int>& cnt2) {
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for (int i = 1; i < 26; ++i) {
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int t = 0;
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for (int j = i; j < 26; ++j) t += cnt1[j];
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for (int j = 0; j < i; ++j) t += cnt2[j];
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ans = min(ans, t);
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}
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};
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f(cnt1, cnt2);
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f(cnt2, cnt1);
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return ans;
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}
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};
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```
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### **Go**
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```go
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func minCharacters(a string, b string) int {
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cnt1 := [26]int{}
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cnt2 := [26]int{}
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for _, c := range a {
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cnt1[c-'a']++
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}
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for _, c := range b {
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cnt2[c-'a']++
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}
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m, n := len(a), len(b)
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ans := m + n
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for i := 0; i < 26; i++ {
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ans = min(ans, m+n-cnt1[i]-cnt2[i])
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}
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f := func(cnt1, cnt2 [26]int) {
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for i := 1; i < 26; i++ {
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t := 0
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for j := i; j < 26; j++ {
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t += cnt1[j]
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}
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for j := 0; j < i; j++ {
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t += cnt2[j]
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}
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ans = min(ans, t)
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}
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}
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f(cnt1, cnt2)
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f(cnt2, cnt1)
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return ans
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **TypeScript**

‎solution/1700-1799/1737.Change Minimum Characters to Satisfy One of Three Conditions/Solution.cpp

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

‎solution/1700-1799/1737.Change Minimum Characters to Satisfy One of Three Conditions/Solution.go

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

‎solution/1700-1799/1737.Change Minimum Characters to Satisfy One of Three Conditions/Solution.java

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class Solution {
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private int ans;
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public int minCharacters(String a, String b) {
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int m = a.length(), n = b.length();
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int[] cnt1 = new int[26];
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int[] cnt2 = new int[26];
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for (int i = 0; i < m; ++i) {
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++cnt1[a.charAt(i) - 'a'];
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}
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for (int i = 0; i < n; ++i) {
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++cnt2[b.charAt(i) - 'a'];
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}
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ans = m + n;
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for (int i = 0; i < 26; ++i) {
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ans = Math.min(ans, m + n - cnt1[i] - cnt2[i]);
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}
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f(cnt1, cnt2);
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f(cnt2, cnt1);
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return ans;
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}
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private void f(int[] cnt1, int[] cnt2) {
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for (int i = 1; i < 26; ++i) {
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int t = 0;
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for (int j = i; j < 26; ++j) {
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t += cnt1[j];
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}
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for (int j = 0; j < i; ++j) {
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t += cnt2[j];
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}
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ans = Math.min(ans, t);
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}
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}
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}

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