feat: add solutions to lc problem: No.2190

solution/2100-2199/2190.Most Frequent Number Following Key In an Array/README.md

@@ -69,13 +69,13 @@ target = 2 是紧跟着 key 之后出现次数最多的数字，所以我们返
 
 ### 方法一：遍历计数
 
-我们用一个哈希表或数组 $cnt$ 记录每个 $target$ 出现的次数，用一个变量 $mx$ 维护 $target$ 出现的最大次数，初始时 $mx = 0$。
+我们用一个哈希表或数组 $\textit{cnt}$ 记录每个 $\textit{target}$ 出现的次数，用一个变量 $\textit{mx}$ 维护 $\textit{target}$ 出现的最大次数，初始时 $\textit{mx} = 0$。
 
-遍历数组 $nums$，如果 $nums[i] = key$，则 $nums[i + 1]$ 出现的次数 $cnt[nums[i + 1]]$ 加一，如果此时 $mx \lt cnt[nums[i + 1]]$，则更新 $mx = cnt[nums[i + 1]]$，并更新答案 $ans = nums[i + 1]$。
+遍历数组 $\textit{nums}$，如果 $\textit{nums}[i] = \textit{key}$，则 $\textit{nums}[i + 1]$ 出现的次数 $\textit{cnt}[\textit{nums}[i + 1]]$ 加一，如果此时 $\textit{mx} \lt \textit{cnt}[\textit{nums}[i + 1]]$，则更新 $\textit{mx} = \textit{cnt}[\textit{nums}[i + 1]]$，并更新答案 $\textit{ans} = \textit{nums}[i + 1]$。
 
-遍历结束后，返回答案 $ans$。
+遍历结束后，返回答案 $\textit{ans}$。
 
-时间复杂度 $O(n)$，空间复杂度 $O(M)$。其中 $n$ 和 $M$ 分别为数组 $nums$ 的长度和数组 $nums$ 中元素的最大值。
+时间复杂度 $O(n)$，空间复杂度 $O(M)$。其中 $n$ 和 $M$ 分别为数组 $\textit{nums}$ 的长度和数组 $\textit{nums}$ 中元素的最大值。
 
 <!-- tabs:start -->
 
@@ -159,9 +159,8 @@ func mostFrequent(nums []int, key int) (ans int) {
 
 ```ts
 function mostFrequent(nums: number[], key: number): number {
-    const cnt: number[] = new Array(1001).fill(0);
-    let ans = 0;
-    let mx = 0;
+    const cnt: number[] = Array(Math.max(...nums) + 1).fill(0);
+    let [ans, mx] = [0, 0];
     for (let i = 0; i < nums.length - 1; ++i) {
         if (nums[i] === key) {
             if (mx < ++cnt[nums[i + 1]]) {
@@ -174,28 +173,47 @@ function mostFrequent(nums: number[], key: number): number {
 }
 ```
 
+#### JavaScript
+
+```js
+/**
+ * @param {number[]} nums
+ * @param {number} key
+ * @return {number}
+ */
+var mostFrequent = function (nums, key) {
+    const cnt = Array(Math.max(...nums) + 1).fill(0);
+    let [ans, mx] = [0, 0];
+    for (let i = 0; i < nums.length - 1; ++i) {
+        if (nums[i] === key) {
+            if (mx < ++cnt[nums[i + 1]]) {
+                mx = cnt[nums[i + 1]];
+                ans = nums[i + 1];
+            }
+        }
+    }
+    return ans;
+};
+```
+
 #### PHP
 
 ```php
 class Solution {
-    /**
-     * @param Integer[] $nums
-     * @param Integer $key
-     * @return Integer
-     */
     function mostFrequent($nums, $key) {
-        $max = $maxNum = 0;
-        for ($i = 0; $i < count($nums) - 1; $i++) {
-            if ($nums[$i] == $key) {
-                $hashtable[$nums[$i + 1]] += 1;
-                $tmp = $hashtable[$nums[$i + 1]];
-                if ($tmp > $max) {
-                    $max = $tmp;
-                    $maxNum = $nums[$i + 1];
+        $cnt = array_fill(0, max($nums) + 1, 0);
+        $ans = 0;
+        $mx = 0;
+        for ($i = 0; $i < count($nums) - 1; ++$i) {
+            if ($nums[$i] === $key) {
+                $cnt[$nums[$i + 1]]++;
+                if ($mx < $cnt[$nums[$i + 1]]) {
+                    $mx = $cnt[$nums[$i + 1]];
+                    $ans = $nums[$i + 1];
                 }
             }
         }
-        return $maxNum;
+        return $ans;
     }
 }
 ```

solution/2100-2199/2190.Most Frequent Number Following Key In an Array/README_EN.md

@@ -67,7 +67,15 @@ target = 2 has the maximum number of occurrences following an occurrence of key,
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Traversal and Counting
+
+We use a hash table or an array $\textit{cnt}$ to record the number of occurrences of each $\textit{target}$, and use a variable $\textit{mx}$ to maintain the maximum number of occurrences of $\textit{target}$. Initially, $\textit{mx} = 0$.
+
+Traverse the array $\textit{nums}$. If $\textit{nums}[i] = \textit{key}$, increment the count of $\textit{nums}[i + 1]$ in $\textit{cnt}[\textit{nums}[i + 1]]$. If $\textit{mx} \lt \textit{cnt}[\textit{nums}[i + 1]]$, update $\textit{mx} = \textit{cnt}[\textit{nums}[i + 1]]$ and update the answer $\textit{ans} = \textit{nums}[i + 1]$.
+
+After the traversal, return the answer $\textit{ans}$.
+
+The time complexity is $O(n)$, and the space complexity is $O(M)$. Here, $n$ and $M$ are the length of the array $\textit{nums}$ and the maximum value of the elements in the array $\textit{nums}$, respectively.
 
 <!-- tabs:start -->
 
@@ -151,9 +159,8 @@ func mostFrequent(nums []int, key int) (ans int) {
 
 ```ts
 function mostFrequent(nums: number[], key: number): number {
-    const cnt: number[] = new Array(1001).fill(0);
-    let ans = 0;
-    let mx = 0;
+    const cnt: number[] = Array(Math.max(...nums) + 1).fill(0);
+    let [ans, mx] = [0, 0];
     for (let i = 0; i < nums.length - 1; ++i) {
         if (nums[i] === key) {
             if (mx < ++cnt[nums[i + 1]]) {
@@ -166,28 +173,47 @@ function mostFrequent(nums: number[], key: number): number {
 }
 ```
 
+#### JavaScript
+
+```js
+/**
+ * @param {number[]} nums
+ * @param {number} key
+ * @return {number}
+ */
+var mostFrequent = function (nums, key) {
+    const cnt = Array(Math.max(...nums) + 1).fill(0);
+    let [ans, mx] = [0, 0];
+    for (let i = 0; i < nums.length - 1; ++i) {
+        if (nums[i] === key) {
+            if (mx < ++cnt[nums[i + 1]]) {
+                mx = cnt[nums[i + 1]];
+                ans = nums[i + 1];
+            }
+        }
+    }
+    return ans;
+};
+```
+
 #### PHP
 
 ```php
 class Solution {
-    /**
-     * @param Integer[] $nums
-     * @param Integer $key
-     * @return Integer
-     */
     function mostFrequent($nums, $key) {
-        $max = $maxNum = 0;
-        for ($i = 0; $i < count($nums) - 1; $i++) {
-            if ($nums[$i] == $key) {
-                $hashtable[$nums[$i + 1]] += 1;
-                $tmp = $hashtable[$nums[$i + 1]];
-                if ($tmp > $max) {
-                    $max = $tmp;
-                    $maxNum = $nums[$i + 1];
+        $cnt = array_fill(0, max($nums) + 1, 0);
+        $ans = 0;
+        $mx = 0;
+        for ($i = 0; $i < count($nums) - 1; ++$i) {
+            if ($nums[$i] === $key) {
+                $cnt[$nums[$i + 1]]++;
+                if ($mx < $cnt[$nums[$i + 1]]) {
+                    $mx = $cnt[$nums[$i + 1]];
+                    $ans = $nums[$i + 1];
                 }
             }
         }
-        return $maxNum;
+        return $ans;
     }
 }
 ```

solution/2100-2199/2190.Most Frequent Number Following Key In an Array/Solution.js

@@ -0,0 +1,18 @@
+/**
+ * @param {number[]} nums
+ * @param {number} key
+ * @return {number}
+ */
+var mostFrequent = function (nums, key) {
+    const cnt = Array(Math.max(...nums) + 1).fill(0);
+    let [ans, mx] = [0, 0];
+    for (let i = 0; i < nums.length - 1; ++i) {
+        if (nums[i] === key) {
+            if (mx < ++cnt[nums[i + 1]]) {
+                mx = cnt[nums[i + 1]];
+                ans = nums[i + 1];
+            }
+        }
+    }
+    return ans;
+};

solution/2100-2199/2190.Most Frequent Number Following Key In an Array/Solution.php

@@ -1,21 +1,17 @@
 class Solution {
-    /**
-     * @param Integer[] $nums
-     * @param Integer $key
-     * @return Integer
-     */
     function mostFrequent($nums, $key) {
-        $max = $maxNum = 0;
-        for ($i = 0; $i < count($nums) - 1; $i++) {
-            if ($nums[$i] == $key) {
-                $hashtable[$nums[$i + 1]] += 1;
-                $tmp = $hashtable[$nums[$i + 1]];
-                if ($tmp > $max) {
-                    $max = $tmp;
-                    $maxNum = $nums[$i + 1];
+        $cnt = array_fill(0, max($nums) + 1, 0);
+        $ans = 0;
+        $mx = 0;
+        for ($i = 0; $i < count($nums) - 1; ++$i) {
+            if ($nums[$i] === $key) {
+                $cnt[$nums[$i + 1]]++;
+                if ($mx < $cnt[$nums[$i + 1]]) {
+                    $mx = $cnt[$nums[$i + 1]];
+                    $ans = $nums[$i + 1];
                 }
             }
         }
-        return $maxNum;
+        return $ans;
     }
 }

solution/2100-2199/2190.Most Frequent Number Following Key In an Array/Solution.ts

@@ -1,7 +1,6 @@
 function mostFrequent(nums: number[], key: number): number {
-    const cnt: number[] = new Array(1001).fill(0);
-    let ans = 0;
-    let mx = 0;
+    const cnt: number[] = Array(Math.max(...nums) + 1).fill(0);
+    let [ans, mx] = [0, 0];
     for (let i = 0; i < nums.length - 1; ++i) {
         if (nums[i] === key) {
             if (mx < ++cnt[nums[i + 1]]) {