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feat: add solutions to lc problem: No.1366 (#3899) 

No.1366.Rank Teams by Votes
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16
solution/0100-0199/0160.Intersection of Two Linked Lists/README.md
View File

@ -110,13 +110,13 @@ tags:
### 方法一：双指针
我们使用两个指针 $a$, $b$ 分别指向两个链表 $headA$, $headB$。
我们使用两个指针 $a$, $b$ 分别指向两个链表 $\textit{headA}$, $\textit{headB}$。
同时遍历链表，当 $a$ 到达链表 $headA$ 的末尾时，重新定位到链表 $headB$ 的头节点；当 $b$ 到达链表 $headB$ 的末尾时，重新定位到链表 $headA$ 的头节点。
同时遍历链表，当 $a$ 到达链表 $\textit{headA}$ 的末尾时，重新定位到链表 $\textit{headB}$ 的头节点；当 $b$ 到达链表 $\textit{headB}$ 的末尾时，重新定位到链表 $\textit{headA}$ 的头节点。
若两指针相遇，所指向的结点就是第一个公共节点。若没相遇，说明两链表无公共节点，此时两个指针都指向 `null`，返回其中一个即可。
时间复杂度 $O(m+n)$，其中 $m$ 和 $n$ 分别是链表 $headA$ 和 $headB$ 的长度。空间复杂度 $O(1)$。
时间复杂度 $O(m + n)$，其中 $m$ 和 $n$ 分别是链表 $\textit{headA}$ 和 $\textit{headB}$ 的长度。空间复杂度 $O(1)$。
<!-- tabs:start -->
@ -233,9 +233,8 @@ func getIntersectionNode(headA, headB *ListNode) *ListNode {
*/
function getIntersectionNode(headA: ListNode | null, headB: ListNode | null): ListNode | null {
let a = headA;
let b = headB;
while (a != b) {
let [a, b] = [headA, headB];
while (a !== b) {
a = a ? a.next : headB;
b = b ? b.next : headA;
}
@ -260,9 +259,8 @@ function getIntersectionNode(headA: ListNode | null, headB: ListNode | null): Li
* @return {ListNode}
*/
var getIntersectionNode = function (headA, headB) {
let a = headA;
let b = headB;
while (a != b) {
let [a, b] = [headA, headB];
while (a !== b) {
a = a ? a.next : headB;
b = b ? b.next : headA;
}

18
solution/0100-0199/0160.Intersection of Two Linked Lists/README_EN.md
View File

@ -94,13 +94,13 @@ Explanation: The two lists do not intersect, so return null.
### Solution 1: Two Pointers
We use two pointers $a$ and $b$ to point to two linked lists $headA$ and $headB$ respectively.
We use two pointers $a$ and $b$ to point to the heads of the two linked lists $\textit{headA}$ and $\textit{headB}$, respectively.
We traverse the linked lists simultaneously. When $a$ reaches the end of the linked list $headA$, it is repositioned to the head node of the linked list $headB$. When $b$ reaches the end of the linked list $headB$, it is repositioned to the head node of the linked list $headA$.
Traverse the linked lists simultaneously. When $a$ reaches the end of $\textit{headA}$, redirect it to the head of $\textit{headB}$. Similarly, when $b$ reaches the end of $\textit{headB}$, redirect it to the head of $\textit{headA}$.
If the two pointers meet, the node they point to is the first common node. If they don't meet, it means that the two linked lists have no common nodes. At this time, both pointers point to `null`, and we can return either one.
If the two pointers meet, the node they point to is the first common node. If they do not meet, it means the two linked lists have no common nodes, and both pointers will point to `null`. Return either pointer.
The time complexity is $O(m+n)$, where $m$ and $n$ are the lengths of the linked lists $headA$ and $headB$ respectively. The space complexity is $O(1)$.
The time complexity is $O(m + n)$, where $m$ and $n$ are the lengths of the linked lists $\textit{headA}$ and $\textit{headB}$, respectively. The space complexity is $O(1)$.
<!-- tabs:start -->
@ -217,9 +217,8 @@ func getIntersectionNode(headA, headB *ListNode) *ListNode {
*/
function getIntersectionNode(headA: ListNode | null, headB: ListNode | null): ListNode | null {
let a = headA;
let b = headB;
while (a != b) {
let [a, b] = [headA, headB];
while (a !== b) {
a = a ? a.next : headB;
b = b ? b.next : headA;
}
@ -244,9 +243,8 @@ function getIntersectionNode(headA: ListNode | null, headB: ListNode | null): Li
* @return {ListNode}
*/
var getIntersectionNode = function (headA, headB) {
let a = headA;
let b = headB;
while (a != b) {
let [a, b] = [headA, headB];
while (a !== b) {
a = a ? a.next : headB;
b = b ? b.next : headA;
}

5
solution/0100-0199/0160.Intersection of Two Linked Lists/Solution.js
View File

@ -12,9 +12,8 @@
* @return {ListNode}
*/
var getIntersectionNode = function (headA, headB) {
let a = headA;
let b = headB;
while (a != b) {
let [a, b] = [headA, headB];
while (a !== b) {
a = a ? a.next : headB;
b = b ? b.next : headA;
}

5
solution/0100-0199/0160.Intersection of Two Linked Lists/Solution.ts
View File

@ -11,9 +11,8 @@
*/
function getIntersectionNode(headA: ListNode | null, headB: ListNode | null): ListNode | null {
let a = headA;
let b = headB;
while (a != b) {
let [a, b] = [headA, headB];
while (a !== b) {
a = a ? a.next : headB;
b = b ? b.next : headA;
}

138
solution/1300-1399/1366.Rank Teams by Votes/README.md
View File

@ -55,7 +55,7 @@ C 队获得三票「排位第二」，两票「排位第三」。
<strong>输入：</strong>votes = ["WXYZ","XYZW"]
<strong>输出：</strong>"XWYZ"
<strong>解释：</strong>
X 队在并列僵局打破后成为排名第一的团队。X 队和 W 队的「排位第一」票数一样，但是 X 队有一票「排位第二」，而 W 没有获得「排位第二」。
X 队在并列僵局打破后成为排名第一的团队。X 队和 W 队的「排位第一」票数一样，但是 X 队有一票「排位第二」，而 W 没有获得「排位第二」。
</pre>
<p><strong class="example">示例 3：</strong></p>
@ -89,7 +89,7 @@ X 队在并列僵局打破后成为排名第一的团队。X 队和 W 队的「
对于每个候选人，我们可以统计他在每个排位上的票数，然后根据不同的排位依次比较票数，票数相同则比较字母。
时间复杂度 $O(n^2 \times \log n)$，空间复杂度 $O(n^2)$。其中 $n$ 为候选人的数量。
时间复杂度 $O(n \times m + m^2 \times \log m)$，空间复杂度 $O(m^2)$。其中 $n$ 是 $\textit{votes}$ 的长度，而 $m$ 是候选人的数量，即 $\textit{votes}[0]$ 的长度。
<!-- tabs:start -->
@ -98,12 +98,12 @@ X 队在并列僵局打破后成为排名第一的团队。X 队和 W 队的「
```python
class Solution:
def rankTeams(self, votes: List[str]) -> str:
n = len(votes[0])
cnt = defaultdict(lambda: [0] * n)
m = len(votes[0])
cnt = defaultdict(lambda: [0] * m)
for vote in votes:
for i, c in enumerate(vote):
cnt[c][i] += 1
return "".join(sorted(votes[0], key=lambda x: (cnt[x], -ord(x)), reverse=True))
return "".join(sorted(cnt, key=lambda c: (cnt[c], -ord(c)), reverse=True))
```
#### Java
@ -111,29 +111,28 @@ class Solution:
```java
class Solution {
public String rankTeams(String[] votes) {
int n = votes[0].length();
int[][] cnt = new int[26][n];
int m = votes[0].length();
int[][] cnt = new int[26][m + 1];
for (var vote : votes) {
for (int i = 0; i < n; ++i) {
cnt[vote.charAt(i) - 'A'][i]++;
for (int i = 0; i < m; ++i) {
++cnt[vote.charAt(i) - 'A'][i];
}
}
Character[] cs = new Character[n];
for (int i = 0; i < n; ++i) {
cs[i] = votes[0].charAt(i);
Character[] s = new Character[m];
for (int i = 0; i < m; ++i) {
s[i] = votes[0].charAt(i);
}
Arrays.sort(cs, (a, b) -> {
Arrays.sort(s, (a, b) -> {
int i = a - 'A', j = b - 'A';
for (int k = 0; k < n; ++k) {
int d = cnt[i][k] - cnt[j][k];
if (d != 0) {
return d > 0 ? -1 : 1;
for (int k = 0; k < m; ++k) {
if (cnt[i][k] != cnt[j][k]) {
return cnt[j][k] - cnt[i][k];
}
}
return a - b;
});
StringBuilder ans = new StringBuilder();
for (char c : cs) {
for (var c : s) {
ans.append(c);
}
return ans.toString();
@ -147,25 +146,25 @@ class Solution {
class Solution {
public:
string rankTeams(vector<string>& votes) {
int n = votes[0].size();
int cnt[26][n];
memset(cnt, 0, sizeof cnt);
for (auto& vote : votes) {
for (int i = 0; i < n; ++i) {
cnt[vote[i] - 'A'][i]++;
int m = votes[0].size();
array<array<int, 27>, 26> cnt{};
for (const auto& vote : votes) {
for (int i = 0; i < m; ++i) {
++cnt[vote[i] - 'A'][i];
}
}
string ans = votes[0];
sort(ans.begin(), ans.end(), [&](auto& a, auto& b) {
string s = votes[0];
ranges::sort(s, [&](char a, char b) {
int i = a - 'A', j = b - 'A';
for (int k = 0; k < n; ++k) {
for (int k = 0; k < m; ++k) {
if (cnt[i][k] != cnt[j][k]) {
return cnt[i][k] > cnt[j][k];
}
}
return a < b;
});
return ans;
return string(s.begin(), s.end());
}
};
```
@ -174,24 +173,83 @@ public:
```go
func rankTeams(votes []string) string {
cnt := [26][26]int{}
m := len(votes[0])
cnt := [26][27]int{}
for _, vote := range votes {
for i, c := range vote {
cnt[c-'A'][i]++
for i, ch := range vote {
cnt[ch-'A'][i]++
}
}
ans := []byte(votes[0])
sort.Slice(ans, func(i, j int) bool {
cnt1, cnt2 := cnt[ans[i]-'A'], cnt[ans[j]-'A']
for k, a := range cnt1 {
b := cnt2[k]
if a != b {
return a > b
s := []rune(votes[0])
sort.Slice(s, func(i, j int) bool {
a, b := s[i]-'A', s[j]-'A'
for k := 0; k < m; k++ {
if cnt[a][k] != cnt[b][k] {
return cnt[a][k] > cnt[b][k]
}
}
return ans[i] < ans[j]
return s[i] < s[j]
})
return string(ans)
return string(s)
}
```
#### TypeScript
```ts
function rankTeams(votes: string[]): string {
const m = votes[0].length;
const cnt: number[][] = Array.from({ length: 26 }, () => Array(m + 1).fill(0));
for (const vote of votes) {
for (let i = 0; i < m; i++) {
cnt[vote.charCodeAt(i) - 65][i]++;
}
}
const s: string[] = votes[0].split('');
s.sort((a, b) => {
const i = a.charCodeAt(0) - 65;
const j = b.charCodeAt(0) - 65;
for (let k = 0; k < m; k++) {
if (cnt[i][k] !== cnt[j][k]) {
return cnt[j][k] - cnt[i][k];
}
}
return a.localeCompare(b);
});
return s.join('');
}
```
#### Rust
```rust
impl Solution {
pub fn rank_teams(votes: Vec<String>) -> String {
let m = votes[0].len();
let mut cnt = vec![vec![0; m + 1]; 26];
for vote in &votes {
for (i, ch) in vote.chars().enumerate() {
cnt[(ch as u8 - b'A') as usize][i] += 1;
}
}
let mut s: Vec<char> = votes[0].chars().collect();
s.sort_by(|&a, &b| {
let i = (a as u8 - b'A') as usize;
let j = (b as u8 - b'A') as usize;
for k in 0..m {
if cnt[i][k] != cnt[j][k] {
return cnt[j][k].cmp(&cnt[i][k]);
}
}
a.cmp(&b)
});
s.into_iter().collect()
}
}
```

142
solution/1300-1399/1366.Rank Teams by Votes/README_EN.md
View File

@ -36,7 +36,7 @@ tags:
<pre>
<strong>Input:</strong> votes = [&quot;ABC&quot;,&quot;ACB&quot;,&quot;ABC&quot;,&quot;ACB&quot;,&quot;ACB&quot;]
<strong>Output:</strong> &quot;ACB&quot;
<strong>Explanation:</strong>
<strong>Explanation:</strong>
Team A was ranked first place by 5 voters. No other team was voted as first place, so team A is the first team.
Team B was ranked second by 2 voters and ranked third by 3 voters.
Team C was ranked second by 3 voters and ranked third by 2 voters.
@ -49,7 +49,7 @@ As most of the voters ranked C second, team C is the second team, and team B is
<strong>Input:</strong> votes = [&quot;WXYZ&quot;,&quot;XYZW&quot;]
<strong>Output:</strong> &quot;XWYZ&quot;
<strong>Explanation:</strong>
X is the winner due to the tie-breaking rule. X has the same votes as W for the first position, but X has one vote in the second position, while W does not have any votes in the second position.
X is the winner due to the tie-breaking rule. X has the same votes as W for the first position, but X has one vote in the second position, while W does not have any votes in the second position.
</pre>
<p><strong class="example">Example 3:</strong></p>
@ -80,9 +80,9 @@ X is the winner due to the tie-breaking rule. X has the same votes as W for the
### Solution 1: Counting + Custom Sorting
For each candidate, we can count the number of votes they receive in each ranking, and then compare the number of votes according to different rankings. If the number of votes is the same, we compare the letters.
For each candidate, we can count the number of votes they receive at each rank, then compare the vote counts for different ranks in order. If the vote counts are the same, we compare the letters.
The time complexity is $O(n^2 \times \log n)$, and the space complexity is $O(n^2)$. Where $n$ is the number of candidates.
The time complexity is $O(n \times m + m^2 \times \log m)$, and the space complexity is $O(m^2)$. Here, $n$ is the length of $\textit{votes}$, and $m$ is the number of candidates, i.e., the length of $\textit{votes}[0]$.
<!-- tabs:start -->
@ -91,12 +91,12 @@ The time complexity is $O(n^2 \times \log n)$, and the space complexity is $O(n^
```python
class Solution:
def rankTeams(self, votes: List[str]) -> str:
n = len(votes[0])
cnt = defaultdict(lambda: [0] * n)
m = len(votes[0])
cnt = defaultdict(lambda: [0] * m)
for vote in votes:
for i, c in enumerate(vote):
cnt[c][i] += 1
return "".join(sorted(votes[0], key=lambda x: (cnt[x], -ord(x)), reverse=True))
return "".join(sorted(cnt, key=lambda c: (cnt[c], -ord(c)), reverse=True))
```
#### Java
@ -104,29 +104,28 @@ class Solution:
```java
class Solution {
public String rankTeams(String[] votes) {
int n = votes[0].length();
int[][] cnt = new int[26][n];
int m = votes[0].length();
int[][] cnt = new int[26][m + 1];
for (var vote : votes) {
for (int i = 0; i < n; ++i) {
cnt[vote.charAt(i) - 'A'][i]++;
for (int i = 0; i < m; ++i) {
++cnt[vote.charAt(i) - 'A'][i];
}
}
Character[] cs = new Character[n];
for (int i = 0; i < n; ++i) {
cs[i] = votes[0].charAt(i);
Character[] s = new Character[m];
for (int i = 0; i < m; ++i) {
s[i] = votes[0].charAt(i);
}
Arrays.sort(cs, (a, b) -> {
Arrays.sort(s, (a, b) -> {
int i = a - 'A', j = b - 'A';
for (int k = 0; k < n; ++k) {
int d = cnt[i][k] - cnt[j][k];
if (d != 0) {
return d > 0 ? -1 : 1;
for (int k = 0; k < m; ++k) {
if (cnt[i][k] != cnt[j][k]) {
return cnt[j][k] - cnt[i][k];
}
}
return a - b;
});
StringBuilder ans = new StringBuilder();
for (char c : cs) {
for (var c : s) {
ans.append(c);
}
return ans.toString();
@ -140,25 +139,25 @@ class Solution {
class Solution {
public:
string rankTeams(vector<string>& votes) {
int n = votes[0].size();
int cnt[26][n];
memset(cnt, 0, sizeof cnt);
for (auto& vote : votes) {
for (int i = 0; i < n; ++i) {
cnt[vote[i] - 'A'][i]++;
int m = votes[0].size();
array<array<int, 27>, 26> cnt{};
for (const auto& vote : votes) {
for (int i = 0; i < m; ++i) {
++cnt[vote[i] - 'A'][i];
}
}
string ans = votes[0];
sort(ans.begin(), ans.end(), [&](auto& a, auto& b) {
string s = votes[0];
ranges::sort(s, [&](char a, char b) {
int i = a - 'A', j = b - 'A';
for (int k = 0; k < n; ++k) {
for (int k = 0; k < m; ++k) {
if (cnt[i][k] != cnt[j][k]) {
return cnt[i][k] > cnt[j][k];
}
}
return a < b;
});
return ans;
return string(s.begin(), s.end());
}
};
```
@ -167,24 +166,83 @@ public:
```go
func rankTeams(votes []string) string {
cnt := [26][26]int{}
m := len(votes[0])
cnt := [26][27]int{}
for _, vote := range votes {
for i, c := range vote {
cnt[c-'A'][i]++
for i, ch := range vote {
cnt[ch-'A'][i]++
}
}
ans := []byte(votes[0])
sort.Slice(ans, func(i, j int) bool {
cnt1, cnt2 := cnt[ans[i]-'A'], cnt[ans[j]-'A']
for k, a := range cnt1 {
b := cnt2[k]
if a != b {
return a > b
s := []rune(votes[0])
sort.Slice(s, func(i, j int) bool {
a, b := s[i]-'A', s[j]-'A'
for k := 0; k < m; k++ {
if cnt[a][k] != cnt[b][k] {
return cnt[a][k] > cnt[b][k]
}
}
return ans[i] < ans[j]
return s[i] < s[j]
})
return string(ans)
return string(s)
}
```
#### TypeScript
```ts
function rankTeams(votes: string[]): string {
const m = votes[0].length;
const cnt: number[][] = Array.from({ length: 26 }, () => Array(m + 1).fill(0));
for (const vote of votes) {
for (let i = 0; i < m; i++) {
cnt[vote.charCodeAt(i) - 65][i]++;
}
}
const s: string[] = votes[0].split('');
s.sort((a, b) => {
const i = a.charCodeAt(0) - 65;
const j = b.charCodeAt(0) - 65;
for (let k = 0; k < m; k++) {
if (cnt[i][k] !== cnt[j][k]) {
return cnt[j][k] - cnt[i][k];
}
}
return a.localeCompare(b);
});
return s.join('');
}
```
#### Rust
```rust
impl Solution {
pub fn rank_teams(votes: Vec<String>) -> String {
let m = votes[0].len();
let mut cnt = vec![vec![0; m + 1]; 26];
for vote in &votes {
for (i, ch) in vote.chars().enumerate() {
cnt[(ch as u8 - b'A') as usize][i] += 1;
}
}
let mut s: Vec<char> = votes[0].chars().collect();
s.sort_by(|&a, &b| {
let i = (a as u8 - b'A') as usize;
let j = (b as u8 - b'A') as usize;
for k in 0..m {
if cnt[i][k] != cnt[j][k] {
return cnt[j][k].cmp(&cnt[i][k]);
}
}
a.cmp(&b)
});
s.into_iter().collect()
}
}
```

22
solution/1300-1399/1366.Rank Teams by Votes/Solution.cpp
View File

@ -1,24 +1,24 @@
class Solution {
public:
string rankTeams(vector<string>& votes) {
int n = votes[0].size();
int cnt[26][n];
memset(cnt, 0, sizeof cnt);
for (auto& vote : votes) {
for (int i = 0; i < n; ++i) {
cnt[vote[i] - 'A'][i]++;
int m = votes[0].size();
array<array<int, 27>, 26> cnt{};
for (const auto& vote : votes) {
for (int i = 0; i < m; ++i) {
++cnt[vote[i] - 'A'][i];
}
}
string ans = votes[0];
sort(ans.begin(), ans.end(), [&](auto& a, auto& b) {
string s = votes[0];
ranges::sort(s, [&](char a, char b) {
int i = a - 'A', j = b - 'A';
for (int k = 0; k < n; ++k) {
for (int k = 0; k < m; ++k) {
if (cnt[i][k] != cnt[j][k]) {
return cnt[i][k] > cnt[j][k];
}
}
return a < b;
});
return ans;
return string(s.begin(), s.end());
}
};
};

26
solution/1300-1399/1366.Rank Teams by Votes/Solution.go
View File

@ -1,20 +1,20 @@
func rankTeams(votes []string) string {
cnt := [26][26]int{}
m := len(votes[0])
cnt := [26][27]int{}
for _, vote := range votes {
for i, c := range vote {
cnt[c-'A'][i]++
for i, ch := range vote {
cnt[ch-'A'][i]++
}
}
ans := []byte(votes[0])
sort.Slice(ans, func(i, j int) bool {
cnt1, cnt2 := cnt[ans[i]-'A'], cnt[ans[j]-'A']
for k, a := range cnt1 {
b := cnt2[k]
if a != b {
return a > b
s := []rune(votes[0])
sort.Slice(s, func(i, j int) bool {
a, b := s[i]-'A', s[j]-'A'
for k := 0; k < m; k++ {
if cnt[a][k] != cnt[b][k] {
return cnt[a][k] > cnt[b][k]
}
}
return ans[i] < ans[j]
return s[i] < s[j]
})
return string(ans)
}
return string(s)
}

27
solution/1300-1399/1366.Rank Teams by Votes/Solution.java
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@ -1,30 +1,29 @@
class Solution {
public String rankTeams(String[] votes) {
int n = votes[0].length();
int[][] cnt = new int[26][n];
int m = votes[0].length();
int[][] cnt = new int[26][m + 1];
for (var vote : votes) {
for (int i = 0; i < n; ++i) {
cnt[vote.charAt(i) - 'A'][i]++;
for (int i = 0; i < m; ++i) {
++cnt[vote.charAt(i) - 'A'][i];
}
}
Character[] cs = new Character[n];
for (int i = 0; i < n; ++i) {
cs[i] = votes[0].charAt(i);
Character[] s = new Character[m];
for (int i = 0; i < m; ++i) {
s[i] = votes[0].charAt(i);
}
Arrays.sort(cs, (a, b) -> {
Arrays.sort(s, (a, b) -> {
int i = a - 'A', j = b - 'A';
for (int k = 0; k < n; ++k) {
int d = cnt[i][k] - cnt[j][k];
if (d != 0) {
return d > 0 ? -1 : 1;
for (int k = 0; k < m; ++k) {
if (cnt[i][k] != cnt[j][k]) {
return cnt[j][k] - cnt[i][k];
}
}
return a - b;
});
StringBuilder ans = new StringBuilder();
for (char c : cs) {
for (var c : s) {
ans.append(c);
}
return ans.toString();
}
}
}

6
solution/1300-1399/1366.Rank Teams by Votes/Solution.py
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@ -1,8 +1,8 @@
class Solution:
def rankTeams(self, votes: List[str]) -> str:
n = len(votes[0])
cnt = defaultdict(lambda: [0] * n)
m = len(votes[0])
cnt = defaultdict(lambda: [0] * m)
for vote in votes:
for i, c in enumerate(vote):
cnt[c][i] += 1
return "".join(sorted(votes[0], key=lambda x: (cnt[x], -ord(x)), reverse=True))
return "".join(sorted(cnt, key=lambda c: (cnt[c], -ord(c)), reverse=True))

28
solution/1300-1399/1366.Rank Teams by Votes/Solution.rs Normal file
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@ -0,0 +1,28 @@
impl Solution {
pub fn rank_teams(votes: Vec<String>) -> String {
let m = votes[0].len();
let mut cnt = vec![vec![0; m + 1]; 26];
for vote in &votes {
for (i, ch) in vote.chars().enumerate() {
cnt[(ch as u8 - b'A') as usize][i] += 1;
}
}
let mut s: Vec<char> = votes[0].chars().collect();
s.sort_by(|&a, &b| {
let i = (a as u8 - b'A') as usize;
let j = (b as u8 - b'A') as usize;
for k in 0..m {
if cnt[i][k] != cnt[j][k] {
return cnt[j][k].cmp(&cnt[i][k]);
}
}
a.cmp(&b)
});
s.into_iter().collect()
}
}

21
solution/1300-1399/1366.Rank Teams by Votes/Solution.ts Normal file
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@ -0,0 +1,21 @@
function rankTeams(votes: string[]): string {
const m = votes[0].length;
const cnt: number[][] = Array.from({ length: 26 }, () => Array(m + 1).fill(0));
for (const vote of votes) {
for (let i = 0; i < m; i++) {
cnt[vote.charCodeAt(i) - 65][i]++;
}
}
const s: string[] = votes[0].split('');
s.sort((a, b) => {
const i = a.charCodeAt(0) - 65;
const j = b.charCodeAt(0) - 65;
for (let k = 0; k < m; k++) {
if (cnt[i][k] !== cnt[j][k]) {
return cnt[j][k] - cnt[i][k];
}
}
return a.localeCompare(b);
});
return s.join('');
}