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Jan 24, 2025
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feat: add solutions to lc problem: No.1941 #3991

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119 changes: 48 additions & 71 deletions ...00-1999/1941.Check if All Characters Have Equal Number of Occurrences/README.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -58,11 +58,11 @@ tags:

### 方法一：计数

我们用一个哈希表或一个长度为 $26$ 的数组 $cnt$ 记录字符串 $s$ 中每个字符出现的次数。
我们用一个哈希表或者一个长度为 $26$ 的数组 $\textit{cnt}$ 记录字符串 $s$ 中每个字符出现的次数。

接下来遍历 $cnt$ 中的每个值，判断所有非零值是否相等即可。
接下来遍历 $\textit{cnt}$ 中的每个值，判断所有非零值是否相等即可。

时间复杂度 $O(n)$，空间复杂度 $O(C)$。其中 $n$ 是字符串 $s$ 的长度；而 $C$ 是字符集大小，本题中字符集为小写英文字母，因此 $C=26$。
时间复杂度 $O(n)$，空间复杂度 $O(|\Sigma|)$。其中 $n$ 是字符串 $s$ 的长度；而 $\Sigma$ 是字符集大小，本题中字符集为小写英文字母，因此 $|\Sigma|=26$。

<!-- tabs:start -->

Expand All @@ -71,8 +71,7 @@ tags:
```python
class Solution:
def areOccurrencesEqual(self, s: str) -> bool:
cnt = Counter(s)
return len(set(cnt.values())) == 1
return len(set(Counter(s).values())) == 1
```

#### Java
Expand All @@ -81,18 +80,18 @@ class Solution:
class Solution {
public boolean areOccurrencesEqual(String s) {
int[] cnt = new int[26];
for (int i = 0; i < s.length(); ++i) {
++cnt[s.charAt(i) - 'a'];
for (char c : s.toCharArray()) {
++cnt[c - 'a'];
}
int x = 0;
for (int v : cnt) {
if (v > 0) {
if (x == 0) {
x = v;
} else if (x != v) {
return false;
}
int v = 0;
for (int x : cnt) {
if (x == 0) {
continue;
}
if (v > 0 && v != x) {
return false;
}
v = x;
}
return true;
}
Expand All @@ -105,19 +104,19 @@ class Solution {
class Solution {
public:
bool areOccurrencesEqual(string s) {
int cnt[26]{};
for (char& c : s) {
vector<int> cnt(26);
for (char c : s) {
++cnt[c - 'a'];
}
int x = 0;
for (int& v : cnt) {
if (v) {
if (!x) {
x = v;
} else if (x != v) {
return false;
}
int v = 0;
for (int x : cnt) {
if (x == 0) {
continue;
}
if (v && v != x) {
return false;
}
v = x;
}
return true;
}
Expand All @@ -132,15 +131,15 @@ func areOccurrencesEqual(s string) bool {
for _, c := range s {
cnt[c-'a']++
}
x := 0
for _, v := range cnt {
if v > 0 {
if x == 0 {
x = v
} else if x != v {
return false
}
v := 0
for _, x := range cnt {
if x == 0 {
continue
}
if v > 0 && v != x {
return false
}
v = x
}
return true
}
Expand All @@ -150,21 +149,12 @@ func areOccurrencesEqual(s string) bool {

```ts
function areOccurrencesEqual(s: string): boolean {
const cnt: number[] = new Array(26).fill(0);
const cnt: number[] = Array(26).fill(0);
for (const c of s) {
++cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)];
}
let x = 0;
for (const v of cnt) {
if (v) {
if (!x) {
x = v;
} else if (x !== v) {
return false;
}
}
}
return true;
const v = cnt.find(v => v);
return cnt.every(x => !x || v === x);
}
```

Expand All @@ -177,35 +167,22 @@ class Solution {
* @return Boolean
*/
function areOccurrencesEqual($s) {
$cnt = array_fill(0, 26, 0);
for ($i = 0; $i < strlen($s); $i++) {
$hashtable[$s[$i]] += 1;
$cnt[ord($s[$i]) - ord('a')]++;
}
$rs = array_unique($hashtable);
return count($rs) === 1;
}
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- solution:start -->

### 方法二

<!-- tabs:start -->

#### TypeScript

```ts
function areOccurrencesEqual(s: string): boolean {
const cnt: number[] = new Array(26).fill(0);
for (const c of s) {
++cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)];
$v = 0;
foreach ($cnt as $x) {
if ($x == 0) {
continue;
}
if ($v && $v != $x) {
return false;
}
$v = $x;
}
return true;
}
const x = cnt.find(v => v);
return cnt.every(v => !v || v === x);
}
```

Expand Down
121 changes: 52 additions & 69 deletions ...1999/1941.Check if All Characters Have Equal Number of Occurrences/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -56,7 +56,13 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Counting

We use a hash table or an array of length $26$ called $\textit{cnt}$ to record the number of occurrences of each character in the string $s$.

Next, we traverse each value in $\textit{cnt}$ and check if all non-zero values are equal.

The time complexity is $O(n)$, and the space complexity is $O(|\Sigma|)$. Here, $n$ is the length of the string $s$, and $\Sigma$ is the size of the character set. In this problem, the character set consists of lowercase English letters, so $|\Sigma|=26$.

<!-- tabs:start -->

Expand All @@ -65,8 +71,7 @@ tags:
```python
class Solution:
def areOccurrencesEqual(self, s: str) -> bool:
cnt = Counter(s)
return len(set(cnt.values())) == 1
return len(set(Counter(s).values())) == 1
```

#### Java
Expand All @@ -75,18 +80,18 @@ class Solution:
class Solution {
public boolean areOccurrencesEqual(String s) {
int[] cnt = new int[26];
for (int i = 0; i < s.length(); ++i) {
++cnt[s.charAt(i) - 'a'];
for (char c : s.toCharArray()) {
++cnt[c - 'a'];
}
int x = 0;
for (int v : cnt) {
if (v > 0) {
if (x == 0) {
x = v;
} else if (x != v) {
return false;
}
int v = 0;
for (int x : cnt) {
if (x == 0) {
continue;
}
if (v > 0 && v != x) {
return false;
}
v = x;
}
return true;
}
Expand All @@ -99,19 +104,19 @@ class Solution {
class Solution {
public:
bool areOccurrencesEqual(string s) {
int cnt[26]{};
for (char& c : s) {
vector<int> cnt(26);
for (char c : s) {
++cnt[c - 'a'];
}
int x = 0;
for (int& v : cnt) {
if (v) {
if (!x) {
x = v;
} else if (x != v) {
return false;
}
int v = 0;
for (int x : cnt) {
if (x == 0) {
continue;
}
if (v && v != x) {
return false;
}
v = x;
}
return true;
}
Expand All @@ -126,15 +131,15 @@ func areOccurrencesEqual(s string) bool {
for _, c := range s {
cnt[c-'a']++
}
x := 0
for _, v := range cnt {
if v > 0 {
if x == 0 {
x = v
} else if x != v {
return false
}
v := 0
for _, x := range cnt {
if x == 0 {
continue
}
if v > 0 && v != x {
return false
}
v = x
}
return true
}
Expand All @@ -144,21 +149,12 @@ func areOccurrencesEqual(s string) bool {

```ts
function areOccurrencesEqual(s: string): boolean {
const cnt: number[] = new Array(26).fill(0);
const cnt: number[] = Array(26).fill(0);
for (const c of s) {
++cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)];
}
let x = 0;
for (const v of cnt) {
if (v) {
if (!x) {
x = v;
} else if (x !== v) {
return false;
}
}
}
return true;
const v = cnt.find(v => v);
return cnt.every(x => !x || v === x);
}
```

Expand All @@ -171,35 +167,22 @@ class Solution {
* @return Boolean
*/
function areOccurrencesEqual($s) {
$cnt = array_fill(0, 26, 0);
for ($i = 0; $i < strlen($s); $i++) {
$hashtable[$s[$i]] += 1;
$cnt[ord($s[$i]) - ord('a')]++;
}
$rs = array_unique($hashtable);
return count($rs) === 1;
}
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- solution:start -->

### Solution 2

<!-- tabs:start -->

#### TypeScript

```ts
function areOccurrencesEqual(s: string): boolean {
const cnt: number[] = new Array(26).fill(0);
for (const c of s) {
++cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)];
$v = 0;
foreach ($cnt as $x) {
if ($x == 0) {
continue;
}
if ($v && $v != $x) {
return false;
}
$v = $x;
}
return true;
}
const x = cnt.find(v => v);
return cnt.every(v => !v || v === x);
}
```

Expand Down
22 changes: 11 additions & 11 deletions ...tion/1900-1999/1941.Check if All Characters Have Equal Number of Occurrences/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,20 +1,20 @@
class Solution {
public:
bool areOccurrencesEqual(string s) {
int cnt[26]{};
for (char& c : s) {
vector<int> cnt(26);
for (char c : s) {
++cnt[c - 'a'];
}
int x = 0;
for (int& v : cnt) {
if (v) {
if (!x) {
x = v;
} else if (x != v) {
return false;
}
int v = 0;
for (int x : cnt) {
if (x == 0) {
continue;
}
if (v && v != x) {
return false;
}
v = x;
}
return true;
}
};
};
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