feat: update solutions to lc problems: No.2293~2295 (#3285)

Author: yanglbme

solution/2200-2299/2293.Min Max Game/README.md

@@ -77,7 +77,7 @@ tags:
 
 根据题意，我们可以模拟整个过程，最后剩下的数字即为答案。在实现上，我们不需要额外创建数组，直接在原数组上进行操作即可。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 `nums` 的长度。
+时间复杂度 $O(n)$，其中 $n$ 是数组 $\textit{nums}$ 的长度。空间复杂度 $O(1)$。
 
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solution/2200-2299/2293.Min Max Game/README_EN.md

@@ -69,7 +69,11 @@ Third: nums = [1]
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Simulation
+
+According to the problem statement, we can simulate the entire process, and the remaining number will be the answer. In implementation, we do not need to create an additional array; we can directly operate on the original array.
+
+The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.
 
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solution/2200-2299/2294.Partition Array Such That Maximum Difference Is K/README.md

@@ -83,9 +83,9 @@ tags:
 
 ### 方法一：贪心 + 排序
 
-题目是要求划分子序列，而不是子数组，因此子序列中的元素可以不连续。我们可以将数组 `nums` 排序，假设当前子序列的第一个元素为 $a$，则子序列中的最大值和最小值的差值不会超过 $k$。因此我们可以遍历数组 `nums`，如果当前元素 $b$ 与 $a$ 的差值大于 $k$，则更新 $a$ 为 $b$，并将子序列数目加 1。遍历结束后，即可得到最少子序列数目，注意初始时子序列数目为 $1$。
+题目要求划分子序列，而不是子数组，因此子序列中的元素可以不连续。我们可以将数组 $\textit{nums}$ 排序，假设当前子序列的第一个元素为 $a$，则子序列中的最大值和最小值的差值不会超过 $k$。因此我们可以遍历数组 $\textit{nums}$，如果当前元素 $b$ 与 $a$ 的差值大于 $k$，则更新 $a$ 为 $b$，并将子序列数目加 1。遍历结束后，即可得到最少子序列数目，注意初始时子序列数目为 $1$。
 
-时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组 `nums` 的长度。
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
 
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solution/2200-2299/2294.Partition Array Such That Maximum Difference Is K/README_EN.md

@@ -79,7 +79,11 @@ Since three subsequences were created, we return 3. It can be shown that 3 is th
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Greedy + Sorting
+
+The problem requires dividing into subsequences, not subarrays, so the elements in a subsequence can be non-continuous. We can sort the array $\textit{nums}$. Assuming the first element of the current subsequence is $a$, the difference between the maximum and minimum values in the subsequence will not exceed $k$. Therefore, we can iterate through the array $\textit{nums}$. If the difference between the current element $b$ and $a$ is greater than $k$, then update $a$ to $b$ and increase the number of subsequences by 1. After the iteration, we can obtain the minimum number of subsequences, noting that the initial number of subsequences is $1$.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $\textit{nums}$.
 
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solution/2200-2299/2295.Replace Elements in an Array/README.md

@@ -78,11 +78,11 @@ tags:
 
 ### 方法一：哈希表
 
-我们先用哈希表 $d$ 记录数组 `nums` 中每个数字的下标，然后遍历操作数组 `operations`，对于每个操作 $[a, b]$，我们将 $a$ 在 `nums` 中的下标 $d[a]$ 对应的数字替换为 $b$，并更新 $d$ 中 $b$ 的下标为 $d[a]$。
+我们先用哈希表 $d$ 记录数组 $\textit{nums}$ 中每个数字的下标，然后遍历操作数组 $\textit{operations}$，对于每个操作 $[x, y]$，我们将 $x$ 在 $\textit{nums}$ 中的下标 $d[x]$ 对应的数字替换为 $y$，并更新 $d$ 中 $y$ 的下标为 $d[x]$。
 
-最后返回 `nums` 即可。
+最后返回 $\textit{nums}$ 即可。
 
-时间复杂度 $O(n + m)$，空间复杂度 $O(n)$。其中 $n$ 和 $m$ 分别是数组 `nums` 和 `operations` 的长度。
+时间复杂度 $O(n + m)$，空间复杂度 $O(n)$。其中 $n$ 和 $m$ 分别是数组 $\textit{nums}$ 的长度和操作数组 $\textit{operations}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -91,10 +91,10 @@ tags:
 ```python
 class Solution:
     def arrayChange(self, nums: List[int], operations: List[List[int]]) -> List[int]:
-        d = {v: i for i, v in enumerate(nums)}
-        for a, b in operations:
-            nums[d[a]] = b
-            d[b] = d[a]
+        d = {x: i for i, x in enumerate(nums)}
+        for x, y in operations:
+            nums[d[x]] = y
+            d[y] = d[x]
         return nums
 ```
 
@@ -103,14 +103,15 @@ class Solution:
 ```java
 class Solution {
     public int[] arrayChange(int[] nums, int[][] operations) {
-        Map<Integer, Integer> d = new HashMap<>();
-        for (int i = 0; i < nums.length; ++i) {
+        int n = nums.length;
+        Map<Integer, Integer> d = new HashMap<>(n);
+        for (int i = 0; i < n; ++i) {
             d.put(nums[i], i);
         }
         for (var op : operations) {
-            int a = op[0], b = op[1];
-            nums[d.get(a)] = b;
-            d.put(b, d.get(a));
+            int x = op[0], y = op[1];
+            nums[d.get(x)] = y;
+            d.put(y, d.get(x));
         }
         return nums;
     }
@@ -128,9 +129,9 @@ public:
             d[nums[i]] = i;
         }
         for (auto& op : operations) {
-            int a = op[0], b = op[1];
-            nums[d[a]] = b;
-            d[b] = d[a];
+            int x = op[0], y = op[1];
+            nums[d[x]] = y;
+            d[y] = d[x];
         }
         return nums;
     }
@@ -142,13 +143,13 @@ public:
 ```go
 func arrayChange(nums []int, operations [][]int) []int {
 	d := map[int]int{}
-	for i, v := range nums {
-		d[v] = i
+	for i, x := range nums {
+		d[x] = i
 	}
 	for _, op := range operations {
-		a, b := op[0], op[1]
-		nums[d[a]] = b
-		d[b] = d[a]
+		x, y := op[0], op[1]
+		nums[d[x]] = y
+		d[y] = d[x]
 	}
 	return nums
 }
@@ -158,10 +159,10 @@ func arrayChange(nums []int, operations [][]int) []int {
 
 ```ts
 function arrayChange(nums: number[], operations: number[][]): number[] {
-    const d = new Map(nums.map((v, i) => [v, i]));
-    for (const [a, b] of operations) {
-        nums[d.get(a)] = b;
-        d.set(b, d.get(a));
+    const d: Map<number, number> = new Map(nums.map((x, i) => [x, i]));
+    for (const [x, y] of operations) {
+        nums[d.get(x)!] = y;
+        d.set(y, d.get(x)!);
     }
     return nums;
 }

solution/2200-2299/2295.Replace Elements in an Array/README_EN.md

@@ -78,11 +78,11 @@ We return the array [2,1].
 
 ### Solution 1: Hash Table
 
-First, we use a hash table $d$ to record the index of each number in the array `nums`. Then, we iterate through the operation array `operations`. For each operation $[a, b]$, we replace the number at index $d[a]$ in `nums` with $b$, and update the index of $b$ in $d$ to $d[a]$.
+First, we use a hash table $d$ to record the indices of each number in the array $\textit{nums}$. Then, we iterate through the operation array $\textit{operations}$. For each operation $[x, y]$, we replace the number at index $d[x]$ in $\textit{nums}$ with $y$, and update the index of $y$ in $d$ to $d[x]$.
 
-Finally, we return `nums`.
+Finally, we return $\textit{nums}$.
 
-The time complexity is $O(n + m)$, and the space complexity is $O(n)$. Here, $n$ and $m$ are the lengths of the arrays `nums` and `operations`, respectively.
+The time complexity is $O(n + m)$, and the space complexity is $O(n)$. Here, $n$ and $m$ are the lengths of the array $\textit{nums}$ and the operation array $\textit{operations}$, respectively.
 
 <!-- tabs:start -->
 
@@ -91,10 +91,10 @@ The time complexity is $O(n + m)$, and the space complexity is $O(n)$. Here, $n$
 ```python
 class Solution:
     def arrayChange(self, nums: List[int], operations: List[List[int]]) -> List[int]:
-        d = {v: i for i, v in enumerate(nums)}
-        for a, b in operations:
-            nums[d[a]] = b
-            d[b] = d[a]
+        d = {x: i for i, x in enumerate(nums)}
+        for x, y in operations:
+            nums[d[x]] = y
+            d[y] = d[x]
         return nums
 ```
 
@@ -103,14 +103,15 @@ class Solution:
 ```java
 class Solution {
     public int[] arrayChange(int[] nums, int[][] operations) {
-        Map<Integer, Integer> d = new HashMap<>();
-        for (int i = 0; i < nums.length; ++i) {
+        int n = nums.length;
+        Map<Integer, Integer> d = new HashMap<>(n);
+        for (int i = 0; i < n; ++i) {
             d.put(nums[i], i);
         }
         for (var op : operations) {
-            int a = op[0], b = op[1];
-            nums[d.get(a)] = b;
-            d.put(b, d.get(a));
+            int x = op[0], y = op[1];
+            nums[d.get(x)] = y;
+            d.put(y, d.get(x));
         }
         return nums;
     }
@@ -128,9 +129,9 @@ public:
             d[nums[i]] = i;
         }
         for (auto& op : operations) {
-            int a = op[0], b = op[1];
-            nums[d[a]] = b;
-            d[b] = d[a];
+            int x = op[0], y = op[1];
+            nums[d[x]] = y;
+            d[y] = d[x];
         }
         return nums;
     }
@@ -142,13 +143,13 @@ public:
 ```go
 func arrayChange(nums []int, operations [][]int) []int {
 	d := map[int]int{}
-	for i, v := range nums {
-		d[v] = i
+	for i, x := range nums {
+		d[x] = i
 	}
 	for _, op := range operations {
-		a, b := op[0], op[1]
-		nums[d[a]] = b
-		d[b] = d[a]
+		x, y := op[0], op[1]
+		nums[d[x]] = y
+		d[y] = d[x]
 	}
 	return nums
 }
@@ -158,10 +159,10 @@ func arrayChange(nums []int, operations [][]int) []int {
 
 ```ts
 function arrayChange(nums: number[], operations: number[][]): number[] {
-    const d = new Map(nums.map((v, i) => [v, i]));
-    for (const [a, b] of operations) {
-        nums[d.get(a)] = b;
-        d.set(b, d.get(a));
+    const d: Map<number, number> = new Map(nums.map((x, i) => [x, i]));
+    for (const [x, y] of operations) {
+        nums[d.get(x)!] = y;
+        d.set(y, d.get(x)!);
     }
     return nums;
 }

solution/2200-2299/2295.Replace Elements in an Array/Solution.cpp

@@ -6,9 +6,9 @@ class Solution {
             d[nums[i]] = i;
         }
         for (auto& op : operations) {
-            int a = op[0], b = op[1];
-            nums[d[a]] = b;
-            d[b] = d[a];
+            int x = op[0], y = op[1];
+            nums[d[x]] = y;
+            d[y] = d[x];
         }
         return nums;
     }

solution/2200-2299/2295.Replace Elements in an Array/Solution.go

@@ -1,12 +1,12 @@
 func arrayChange(nums []int, operations [][]int) []int {
 	d := map[int]int{}
-	for i, v := range nums {
-		d[v] = i
+	for i, x := range nums {
+		d[x] = i
 	}
 	for _, op := range operations {
-		a, b := op[0], op[1]
-		nums[d[a]] = b
-		d[b] = d[a]
+		x, y := op[0], op[1]
+		nums[d[x]] = y
+		d[y] = d[x]
 	}
 	return nums
 }

solution/2200-2299/2295.Replace Elements in an Array/Solution.java

@@ -1,13 +1,14 @@
 class Solution {
     public int[] arrayChange(int[] nums, int[][] operations) {
-        Map<Integer, Integer> d = new HashMap<>();
-        for (int i = 0; i < nums.length; ++i) {
+        int n = nums.length;
+        Map<Integer, Integer> d = new HashMap<>(n);
+        for (int i = 0; i < n; ++i) {
             d.put(nums[i], i);
         }
         for (var op : operations) {
-            int a = op[0], b = op[1];
-            nums[d.get(a)] = b;
-            d.put(b, d.get(a));
+            int x = op[0], y = op[1];
+            nums[d.get(x)] = y;
+            d.put(y, d.get(x));
         }
         return nums;
     }

solution/2200-2299/2295.Replace Elements in an Array/Solution.py

@@ -1,7 +1,7 @@
 class Solution:
     def arrayChange(self, nums: List[int], operations: List[List[int]]) -> List[int]:
-        d = {v: i for i, v in enumerate(nums)}
-        for a, b in operations:
-            nums[d[a]] = b
-            d[b] = d[a]
+        d = {x: i for i, x in enumerate(nums)}
+        for x, y in operations:
+            nums[d[x]] = y
+            d[y] = d[x]
         return nums

solution/2200-2299/2295.Replace Elements in an Array/Solution.ts

@@ -1,8 +1,8 @@
 function arrayChange(nums: number[], operations: number[][]): number[] {
-    const d = new Map(nums.map((v, i) => [v, i]));
-    for (const [a, b] of operations) {
-        nums[d.get(a)] = b;
-        d.set(b, d.get(a));
+    const d: Map<number, number> = new Map(nums.map((x, i) => [x, i]));
+    for (const [x, y] of operations) {
+        nums[d.get(x)!] = y;
+        d.set(y, d.get(x)!);
     }
     return nums;
 }