feat: update lc problems (doocs#4106)

Author: yanglbme

solution/0700-0799/0763.Partition Labels/README.md

@@ -60,19 +60,19 @@ tags:
 
 ### 方法一：贪心
 
-我们先用数组或哈希表 $last$ 记录字符串 $s$ 中每个字母最后一次出现的位置。
+我们先用数组或哈希表 $\textit{last}$ 记录字符串 $s$ 中每个字母最后一次出现的位置。
 
 接下来我们使用贪心的方法，将字符串划分为尽可能多的片段。
 
 从左到右遍历字符串 $s$，遍历的同时维护当前片段的开始下标 $j$ 和结束下标 $i$，初始均为 $0$。
 
-对于每个访问到的字母 $c$，获取到最后一次出现的位置 $last[c]$。由于当前片段的结束下标一定不会小于 $last[c]$，因此令 $mx = \max(mx, last[c])$。
+对于每个访问到的字母 $c$，获取到最后一次出现的位置 $\textit{last}[c]$。由于当前片段的结束下标一定不会小于 $\textit{last}[c]$，因此令 $\textit{mx} = \max(\textit{mx}, \textit{last}[c])$。
 
-当访问到下标 $mx$ 时，意味着当前片段访问结束，当前片段的下标范围是 $[j,.. i]$，长度为 $i - j + 1$，我们将其添加到结果数组中。然后令 $j = i + 1$, 继续寻找下一个片段。
+当访问到下标 $\textit{mx}$ 时，意味着当前片段访问结束，当前片段的下标范围是 $[j,.. i]$，长度为 $i - j + 1$，我们将其添加到结果数组中。然后令 $j = i + 1$, 继续寻找下一个片段。
 
 重复上述过程，直至字符串遍历结束，即可得到所有片段的长度。
 
-时间复杂度 $O(n)$，空间复杂度 $O(C)$。其中 $n$ 为字符串 $s$ 的长度，而 $C$ 为字符集的大小。本题中 $C = 26$。
+时间复杂度 $O(n)$，空间复杂度 $O(|\Sigma|)$。其中 $n$ 为字符串 $s$ 的长度，而 $|\Sigma|$ 为字符集的大小。本题中 $|\Sigma| = 26$。
 
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solution/0700-0799/0763.Partition Labels/README_EN.md

@@ -58,7 +58,21 @@ A partition like "ababcbacadefegde", "hijhklij" is incorrect
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Greedy
+
+We first use an array or hash table $\textit{last}$ to record the last occurrence of each letter in the string $s$.
+
+Next, we use a greedy approach to partition the string into as many segments as possible.
+
+Traverse the string $s$ from left to right, while maintaining the start index $j$ and end index $i$ of the current segment, both initially set to $0$.
+
+For each letter $c$ visited, get the last occurrence position $\textit{last}[c]$. Since the end index of the current segment must not be less than $\textit{last}[c]$, let $\textit{mx} = \max(\textit{mx}, \textit{last}[c])$.
+
+When visiting the index $\textit{mx}$, it means the current segment ends. The index range of the current segment is $[j,.. i]$, and the length is $i - j + 1$. We add this length to the result array. Then set $j = i + 1$ and continue to find the next segment.
+
+Repeat the above process until the string traversal is complete to get the lengths of all segments.
+
+Time complexity is $O(n)$, and space complexity is $O(|\Sigma|)$. Where $n$ is the length of the string $s$, and $|\Sigma|$ is the size of the character set. In this problem, $|\Sigma| = 26$.
 
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solution/0700-0799/0765.Couples Holding Hands/README_EN.md

@@ -60,7 +60,17 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Union-Find
+
+We can assign a number to each pair of couples. Person with number $0$ and $1$ corresponds to couple $0$, person with number $2$ and $3$ corresponds to couple $1$, and so on. In other words, the person corresponding to $row[i]$ has a couple number of $\lfloor \frac{row[i]}{2} \rfloor$.
+
+If there are $k$ pairs of couples who are seated incorrectly with respect to each other, i.e., if $k$ pairs of couples are in the same permutation cycle, it will take $k-1$ swaps for all of them to be seated correctly.
+
+Why? Consider the following: we first adjust the positions of a couple to their correct seats. After this, the problem reduces from $k$ couples to $k-1$ couples. This process continues, and when $k = 1$, the number of swaps required is $0$. Therefore, if $k$ pairs of couples are in the wrong positions, we need $k-1$ swaps.
+
+Thus, we only need to traverse the array once, use union-find to determine how many permutation cycles there are. Suppose there are $x$ cycles, and the size of each cycle (in terms of couple pairs) is $y_1, y_2, \cdots, y_x$. The number of swaps required is $y_1-1 + y_2-1 + \cdots + y_x-1 = y_1 + y_2 + \cdots + y_x - x = n - x$.
+
+The time complexity is $O(n \times \alpha(n))$, and the space complexity is $O(n)$, where $\alpha(n)$ is the inverse Ackermann function, which can be considered a very small constant.
 
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solution/0700-0799/0768.Max Chunks To Make Sorted II/README.md

@@ -33,8 +33,8 @@ tags:
 <strong>输入：</strong>arr = [5,4,3,2,1]
 <strong>输出：</strong>1
 <strong>解释：</strong>
-将数组分成2块或者更多块，都无法得到所需的结果。 
-例如，分成 [5, 4], [3, 2, 1] 的结果是 [4, 5, 1, 2, 3]，这不是有序的数组。 
+将数组分成2块或者更多块，都无法得到所需的结果。
+例如，分成 [5, 4], [3, 2, 1] 的结果是 [4, 5, 1, 2, 3]，这不是有序的数组。
 </pre>
 
 <p><strong class="example">示例 2：</strong></p>
@@ -43,8 +43,8 @@ tags:
 <strong>输入：</strong>arr = [2,1,3,4,4]
 <strong>输出：</strong>4
 <strong>解释：</strong>
-可以把它分成两块，例如 [2, 1], [3, 4, 4]。 
-然而，分成 [2, 1], [3], [4], [4] 可以得到最多的块数。 
+可以把它分成两块，例如 [2, 1], [3, 4, 4]。
+然而，分成 [2, 1], [3], [4], [4] 可以得到最多的块数。
 </pre>
 
 <p>&nbsp;</p>
@@ -66,7 +66,7 @@ tags:
 
 根据题目，我们可以发现，从左到右，每个分块都有一个最大值，并且这些分块的最大值呈单调递增（非严格递增）。我们可以用一个栈来存储这些分块的最大值。最后得到的栈的大小，也就是题目所求的最多能完成排序的块。
 
-时间复杂度 $O(n)$，其中 $n$ 表示 $arr$ 的长度。
+时间复杂度 $O(n)$，其中 $n$ 表示 $\textit{arr}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -156,19 +156,19 @@ func maxChunksToSorted(arr []int) int {
 
 ```ts
 function maxChunksToSorted(arr: number[]): number {
-    const stack = [];
-    for (const num of arr) {
-        if (stack.length !== 0 && num < stack[stack.length - 1]) {
-            const max = stack.pop();
-            while (stack.length !== 0 && num < stack[stack.length - 1]) {
-                stack.pop();
-            }
-            stack.push(max);
+    const stk: number[] = [];
+    for (let v of arr) {
+        if (stk.length === 0 || v >= stk[stk.length - 1]) {
+            stk.push(v);
         } else {
-            stack.push(num);
+            let mx = stk.pop()!;
+            while (stk.length > 0 && stk[stk.length - 1] > v) {
+                stk.pop();
+            }
+            stk.push(mx);
         }
     }
-    return stack.length;
+    return stk.length;
 }
 ```
 
@@ -177,19 +177,23 @@ function maxChunksToSorted(arr: number[]): number {
 ```rust
 impl Solution {
     pub fn max_chunks_to_sorted(arr: Vec<i32>) -> i32 {
-        let mut stack = vec![];
-        for num in arr.iter() {
-            if !stack.is_empty() && num < stack.last().unwrap() {
-                let max = stack.pop().unwrap();
-                while !stack.is_empty() && num < stack.last().unwrap() {
-                    stack.pop();
-                }
-                stack.push(max);
+        let mut stk = Vec::new();
+        for &v in arr.iter() {
+            if stk.is_empty() || v >= *stk.last().unwrap() {
+                stk.push(v);
             } else {
-                stack.push(*num);
+                let mut mx = stk.pop().unwrap();
+                while let Some(&top) = stk.last() {
+                    if top > v {
+                        stk.pop();
+                    } else {
+                        break;
+                    }
+                }
+                stk.push(mx);
             }
         }
-        stack.len() as i32
+        stk.len() as i32
     }
 }
 ```

solution/0700-0799/0768.Max Chunks To Make Sorted II/README_EN.md

@@ -61,7 +61,11 @@ However, splitting into [2, 1], [3], [4], [4] is the highest number of chunks po
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Monotonic Stack
+
+According to the problem, we can find that from left to right, each chunk has a maximum value, and these maximum values are monotonically increasing (non-strictly increasing). We can use a stack to store these maximum values of the chunks. The size of the final stack is the maximum number of chunks that can be sorted.
+
+Time complexity is $O(n)$, where $n$ represents the length of $\textit{arr}$.
 
 <!-- tabs:start -->
 
@@ -151,19 +155,19 @@ func maxChunksToSorted(arr []int) int {
 
 ```ts
 function maxChunksToSorted(arr: number[]): number {
-    const stack = [];
-    for (const num of arr) {
-        if (stack.length !== 0 && num < stack[stack.length - 1]) {
-            const max = stack.pop();
-            while (stack.length !== 0 && num < stack[stack.length - 1]) {
-                stack.pop();
-            }
-            stack.push(max);
+    const stk: number[] = [];
+    for (let v of arr) {
+        if (stk.length === 0 || v >= stk[stk.length - 1]) {
+            stk.push(v);
         } else {
-            stack.push(num);
+            let mx = stk.pop()!;
+            while (stk.length > 0 && stk[stk.length - 1] > v) {
+                stk.pop();
+            }
+            stk.push(mx);
         }
     }
-    return stack.length;
+    return stk.length;
 }
 ```
 
@@ -172,19 +176,23 @@ function maxChunksToSorted(arr: number[]): number {
 ```rust
 impl Solution {
     pub fn max_chunks_to_sorted(arr: Vec<i32>) -> i32 {
-        let mut stack = vec![];
-        for num in arr.iter() {
-            if !stack.is_empty() && num < stack.last().unwrap() {
-                let max = stack.pop().unwrap();
-                while !stack.is_empty() && num < stack.last().unwrap() {
-                    stack.pop();
-                }
-                stack.push(max);
+        let mut stk = Vec::new();
+        for &v in arr.iter() {
+            if stk.is_empty() || v >= *stk.last().unwrap() {
+                stk.push(v);
             } else {
-                stack.push(*num);
+                let mut mx = stk.pop().unwrap();
+                while let Some(&top) = stk.last() {
+                    if top > v {
+                        stk.pop();
+                    } else {
+                        break;
+                    }
+                }
+                stk.push(mx);
             }
         }
-        stack.len() as i32
+        stk.len() as i32
     }
 }
 ```

solution/0700-0799/0768.Max Chunks To Make Sorted II/Solution.rs

@@ -1,17 +1,21 @@
 impl Solution {
     pub fn max_chunks_to_sorted(arr: Vec<i32>) -> i32 {
-        let mut stack = vec![];
-        for num in arr.iter() {
-            if !stack.is_empty() && num < stack.last().unwrap() {
-                let max = stack.pop().unwrap();
-                while !stack.is_empty() && num < stack.last().unwrap() {
-                    stack.pop();
-                }
-                stack.push(max);
+        let mut stk = Vec::new();
+        for &v in arr.iter() {
+            if stk.is_empty() || v >= *stk.last().unwrap() {
+                stk.push(v);
             } else {
-                stack.push(*num);
+                let mut mx = stk.pop().unwrap();
+                while let Some(&top) = stk.last() {
+                    if top > v {
+                        stk.pop();
+                    } else {
+                        break;
+                    }
+                }
+                stk.push(mx);
             }
         }
-        stack.len() as i32
+        stk.len() as i32
     }
 }

solution/0700-0799/0768.Max Chunks To Make Sorted II/Solution.ts

@@ -1,15 +1,15 @@
 function maxChunksToSorted(arr: number[]): number {
-    const stack = [];
-    for (const num of arr) {
-        if (stack.length !== 0 && num < stack[stack.length - 1]) {
-            const max = stack.pop();
-            while (stack.length !== 0 && num < stack[stack.length - 1]) {
-                stack.pop();
-            }
-            stack.push(max);
+    const stk: number[] = [];
+    for (let v of arr) {
+        if (stk.length === 0 || v >= stk[stk.length - 1]) {
+            stk.push(v);
         } else {
-            stack.push(num);
+            let mx = stk.pop()!;
+            while (stk.length > 0 && stk[stk.length - 1] > v) {
+                stk.pop();
+            }
+            stk.push(mx);
         }
     }
-    return stack.length;
+    return stk.length;
 }

solution/0700-0799/0769.Max Chunks To Make Sorted/README.md

@@ -68,9 +68,9 @@ tags:
 
 ### 方法一：贪心 + 一次遍历
 
-由于 $arr$ 是 $[0,..,n-1]$ 的一个排列，若已遍历过的数中的最大值 $mx$ 与当前遍历到的下标 $i$ 相等，说明可以进行一次分割，累加答案。
+由于 $\textit{arr}$ 是 $[0,..,n-1]$ 的一个排列，若已遍历过的数中的最大值 $\textit{mx}$ 与当前遍历到的下标 $i$ 相等，说明可以进行一次分割，累加答案。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 $arr$ 的长度。
+时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 $\textit{arr}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -141,10 +141,10 @@ func maxChunksToSorted(arr []int) int {
 function maxChunksToSorted(arr: number[]): number {
     const n = arr.length;
     let ans = 0;
-    let max = 0;
+    let mx = 0;
     for (let i = 0; i < n; i++) {
-        max = Math.max(arr[i], max);
-        if (max == i) {
+        mx = Math.max(arr[i], mx);
+        if (mx == i) {
             ans++;
         }
     }
@@ -157,15 +157,15 @@ function maxChunksToSorted(arr: number[]): number {
 ```rust
 impl Solution {
     pub fn max_chunks_to_sorted(arr: Vec<i32>) -> i32 {
-        let mut res = 0;
-        let mut max = 0;
+        let mut ans = 0;
+        let mut mx = 0;
         for i in 0..arr.len() {
-            max = max.max(arr[i]);
-            if max == (i as i32) {
-                res += 1;
+            mx = mx.max(arr[i]);
+            if mx == (i as i32) {
+                ans += 1;
             }
         }
-        res
+        ans
     }
 }
 ```
@@ -176,15 +176,15 @@ impl Solution {
 #define max(a, b) (((a) > (b)) ? (a) : (b))
 
 int maxChunksToSorted(int* arr, int arrSize) {
-    int res = 0;
+    int ans = 0;
     int mx = -1;
     for (int i = 0; i < arrSize; i++) {
         mx = max(mx, arr[i]);
         if (mx == i) {
-            res++;
+            ans++;
         }
     }
-    return res;
+    return ans;
 }
 ```
 
@@ -202,7 +202,7 @@ int maxChunksToSorted(int* arr, int arrSize) {
 
 以上这种解法，不仅可以解决本题，也可以解决 [768. 最多能完成排序的块 II](https://github.com/doocs/leetcode/blob/main/solution/0700-0799/0768.Max%20Chunks%20To%20Make%20Sorted%20II/README.md) 这道困难题。大家可以自行尝试。
 
-时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $arr$ 的长度。
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{arr}$ 的长度。
 
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