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feat: add solutions to lc problem: No.2531 #3325

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188 changes: 119 additions & 69 deletions solution/2500-2599/2531.Make Number of Distinct Characters Equal/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -70,13 +70,15 @@ tags:

### 方法一：计数 + 枚举

我们先用两个长度为 $26$ 的数组分别统计字符串 $word1$ 和 $word2$ 中每个字母的出现次数，记为 $cnt1$ 和 $cnt2$。
我们先用两个长度为 $26$ 的数组 $\textit{cnt1}$ 和 $\textit{cnt2}$ 分别记录字符串 $\textit{word1}$ 和 $\textit{word2}$ 中每个字符的出现次数。

然后我们枚举 $cnt1$ 中的每个字母，接着枚举 $cnt2$ 中的每个字母，如果 $cnt1[i]$ 和 $cnt2[j]$ 都不为 $0$，那么我们就可以交换 $word1$ 中的第 $i$ 个字母和 $word2$ 中的第 $j$ 个字母。如果交换后 $word1$ 和 $word2$ 中不同字母的数目相等，那么就返回 `true`，否则，我们撤销此次交换，继续枚举。
然后我们分别统计 $\textit{word1}$ 和 $\textit{word2}$ 中不同字符的个数，分别记为 $x$ 和 $y$。

如果枚举完所有的字母对，仍然没有找到一种交换方式，那么就返回 `false`。
接下来我们枚举 $\textit{word1}$ 中的每个字符 $c1$ 和 $\textit{word2}$ 中的每个字符 $c2$，如果 $c1 = c2$，那么我们只需要判断 $x$ 和 $y$ 是否相等；否则，我们需要判断 $x - (\textit{cnt1}[c1] = 1) + (\textit{cnt1}[c2] = 0)$ 和 $y - (\textit{cnt2}[c2] = 1) + (\textit{cnt2}[c1] = 0)$ 是否相等。如果相等，那么我们就找到了一种方案，返回 $\text{true}$。

时间复杂度 $O(n + C^3)$，空间复杂度 $O(C)$，其中 $n$ 是字符串的长度，而 $C$ 是字符集的大小。本题中 $C = 26$。
如果我们枚举完所有的字符都没有找到合适的方案，那么我们就返回 $\text{false}$。

时间复杂度 $O(m + n + |\Sigma|^2)$，其中 $m$ 和 $n$ 分别是字符串 $\textit{word1}$ 和 $\textit{word2}$ 的长度，而 $\Sigma$ 是字符集，本题中字符集为小写字母，所以 $|\Sigma| = 26$。

<!-- tabs:start -->

Expand All @@ -85,21 +87,19 @@ tags:
```python
class Solution:
def isItPossible(self, word1: str, word2: str) -> bool:
cnt1 = [0] * 26
cnt2 = [0] * 26
for c in word1:
cnt1[ord(c) - ord('a')] += 1
for c in word2:
cnt2[ord(c) - ord('a')] += 1
for i, a in enumerate(cnt1):
for j, b in enumerate(cnt2):
if a and b:
cnt1[i], cnt2[j] = cnt1[i] - 1, cnt2[j] - 1
cnt1[j], cnt2[i] = cnt1[j] + 1, cnt2[i] + 1
if sum(v > 0 for v in cnt1) == sum(v > 0 for v in cnt2):
cnt1 = Counter(word1)
cnt2 = Counter(word2)
x, y = len(cnt1), len(cnt2)
for c1, v1 in cnt1.items():
for c2, v2 in cnt2.items():
if c1 == c2:
if x == y:
return True
else:
a = x - (v1 == 1) + (cnt1[c2] == 0)
b = y - (v2 == 1) + (cnt2[c1] == 0)
if a == b:
return True
cnt1[i], cnt2[j] = cnt1[i] + 1, cnt2[j] + 1
cnt1[j], cnt2[i] = cnt1[j] - 1, cnt2[i] - 1
return False
```

Expand All @@ -110,35 +110,31 @@ class Solution {
public boolean isItPossible(String word1, String word2) {
int[] cnt1 = new int[26];
int[] cnt2 = new int[26];
int x = 0, y = 0;
for (int i = 0; i < word1.length(); ++i) {
++cnt1[word1.charAt(i) - 'a'];
if (++cnt1[word1.charAt(i) - 'a'] == 1) {
++x;
}
}
for (int i = 0; i < word2.length(); ++i) {
++cnt2[word2.charAt(i) - 'a'];
if (++cnt2[word2.charAt(i) - 'a'] == 1) {
++y;
}
}
for (int i = 0; i < 26; ++i) {
for (int j = 0; j < 26; ++j) {
if (cnt1[i] > 0 && cnt2[j] > 0) {
--cnt1[i];
--cnt2[j];
++cnt1[j];
++cnt2[i];
int d = 0;
for (int k = 0; k < 26; ++k) {
if (cnt1[k] > 0) {
++d;
if (i == j) {
if (x == y) {
return true;
}
if (cnt2[k] > 0) {
--d;
} else {
int a = x - (cnt1[i] == 1 ? 1 : 0) + (cnt1[j] == 0 ? 1 : 0);
int b = y - (cnt2[j] == 1 ? 1 : 0) + (cnt2[i] == 0 ? 1 : 0);
if (a == b) {
return true;
}
}
if (d == 0) {
return true;
}
++cnt1[i];
++cnt2[j];
--cnt1[j];
--cnt2[i];
}
}
}
Expand All @@ -155,35 +151,31 @@ public:
bool isItPossible(string word1, string word2) {
int cnt1[26]{};
int cnt2[26]{};
int x = 0, y = 0;
for (char& c : word1) {
++cnt1[c - 'a'];
if (++cnt1[c - 'a'] == 1) {
++x;
}
}
for (char& c : word2) {
++cnt2[c - 'a'];
if (++cnt2[c - 'a'] == 1) {
++y;
}
}
for (int i = 0; i < 26; ++i) {
for (int j = 0; j < 26; ++j) {
if (cnt1[i] > 0 && cnt2[j] > 0) {
--cnt1[i];
--cnt2[j];
++cnt1[j];
++cnt2[i];
int d = 0;
for (int k = 0; k < 26; ++k) {
if (cnt1[k] > 0) {
++d;
if (i == j) {
if (x == y) {
return true;
}
if (cnt2[k] > 0) {
--d;
} else {
int a = x - (cnt1[i] == 1 ? 1 : 0) + (cnt1[j] == 0 ? 1 : 0);
int b = y - (cnt2[j] == 1 ? 1 : 0) + (cnt2[i] == 0 ? 1 : 0);
if (a == b) {
return true;
}
}
if (d == 0) {
return true;
}
++cnt1[i];
++cnt2[j];
--cnt1[j];
--cnt2[i];
}
}
}
Expand All @@ -198,38 +190,96 @@ public:
func isItPossible(word1 string, word2 string) bool {
cnt1 := [26]int{}
cnt2 := [26]int{}
x, y := 0, 0
for _, c := range word1 {
cnt1[c-'a']++
if cnt1[c-'a'] == 1 {
x++
}
}
for _, c := range word2 {
cnt2[c-'a']++
if cnt2[c-'a'] == 1 {
y++
}
}
for i := range cnt1 {
for j := range cnt2 {
if cnt1[i] > 0 && cnt2[j] > 0 {
cnt1[i], cnt2[j] = cnt1[i]-1, cnt2[j]-1
cnt1[j], cnt2[i] = cnt1[j]+1, cnt2[i]+1
d := 0
for k, a := range cnt1 {
if a > 0 {
d++
if i == j {
if x == y {
return true
}
if cnt2[k] > 0 {
d--
} else {
a := x
if cnt1[i] == 1 {
a--
}
if cnt1[j] == 0 {
a++
}

b := y
if cnt2[j] == 1 {
b--
}
if cnt2[i] == 0 {
b++
}

if a == b {
return true
}
}
if d == 0 {
return true
}
cnt1[i], cnt2[j] = cnt1[i]+1, cnt2[j]+1
cnt1[j], cnt2[i] = cnt1[j]-1, cnt2[i]-1
}
}
}
return false
}
```

#### TypeScript

```ts
function isItPossible(word1: string, word2: string): boolean {
const cnt1: number[] = Array(26).fill(0);
const cnt2: number[] = Array(26).fill(0);
let [x, y] = [0, 0];

for (const c of word1) {
if (++cnt1[c.charCodeAt(0) - 'a'.charCodeAt(0)] === 1) {
++x;
}
}

for (const c of word2) {
if (++cnt2[c.charCodeAt(0) - 'a'.charCodeAt(0)] === 1) {
++y;
}
}

for (let i = 0; i < 26; ++i) {
for (let j = 0; j < 26; ++j) {
if (cnt1[i] > 0 && cnt2[j] > 0) {
if (i === j) {
if (x === y) {
return true;
}
} else {
const a = x - (cnt1[i] === 1 ? 1 : 0) + (cnt1[j] === 0 ? 1 : 0);
const b = y - (cnt2[j] === 1 ? 1 : 0) + (cnt2[i] === 0 ? 1 : 0);
if (a === b) {
return true;
}
}
}
}
}

return false;
}
```

<!-- tabs:end -->

<!-- solution:end -->
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