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feat: add solutions to lc problem: No.2863
No.2863.Maximum Length of Semi-Decreasing Subarrays
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# [2863. Maximum Length of Semi-Decreasing Subarrays](https://leetcode.cn/problems/maximum-length-of-semi-decreasing-subarrays)
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[English Version](/solution/2800-2899/2863.Maximum%20Length%20of%20Semi-Decreasing%20Subarrays/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>You are given an integer array <code>nums</code>.</p>
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<p>Return <em>the length of the <strong>longest semi-decreasing</strong> subarray of </em><code>nums</code><em>, and </em><code>0</code><em> if there are no such subarrays.</em></p>
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<ul>
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<li>A <b>subarray</b> is a contiguous non-empty sequence of elements within an array.</li>
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<li>A non-empty array is <strong>semi-decreasing</strong> if its first element is <strong>strictly greater</strong> than its last element.</li>
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</ul>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [7,6,5,4,3,2,1,6,10,11]
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<strong>Output:</strong> 8
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<strong>Explanation:</strong> Take the subarray [7,6,5,4,3,2,1,6].
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The first element is 7 and the last one is 6 so the condition is met.
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Hence, the answer would be the length of the subarray or 8.
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It can be shown that there aren&#39;t any subarrays with the given condition with a length greater than 8.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [57,55,50,60,61,58,63,59,64,60,63]
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<strong>Output:</strong> 6
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<strong>Explanation:</strong> Take the subarray [61,58,63,59,64,60].
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The first element is 61 and the last one is 60 so the condition is met.
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Hence, the answer would be the length of the subarray or 6.
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It can be shown that there aren&#39;t any subarrays with the given condition with a length greater than 6.
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,3,4]
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> Since there are no semi-decreasing subarrays in the given array, the answer is 0.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
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<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:哈希表 + 排序**
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题目实际上是求逆序对的最大长度,我们不妨用哈希表 $d$ 记录数组中每个数字 $x$ 对应的下标 $i$。
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接下来,我们按照数字从大到小的顺序遍历哈希表的键,用一个数字 $k$ 维护此前出现过的最小的下标,那么对于当前的数字 $x,ドル我们可以得到一个最大的逆序对长度 $d[x][|d[x]|-1]-k + 1,ドル其中 $|d[x]|$ 表示数组 $d[x]$ 的长度,即数字 $x$ 在原数组中出现的次数,我们更新答案即可。然后,我们将 $k$ 更新为 $d[x][0],ドル即数字 $x$ 在原数组中第一次出现的下标。继续遍历哈希表的键,直到遍历完所有的键。
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时间复杂度 $O(n \times \log n),ドル空间复杂度 $O(n)$。其中 $n$ 是数组 $nums$ 的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def maxSubarrayLength(self, nums: List[int]) -> int:
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d = defaultdict(list)
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for i, x in enumerate(nums):
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d[x].append(i)
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ans, k = 0, inf
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for x in sorted(d, reverse=True):
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ans = max(ans, d[x][-1] - k + 1)
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k = min(k, d[x][0])
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int maxSubarrayLength(int[] nums) {
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TreeMap<Integer, List<Integer>> d = new TreeMap<>(Comparator.reverseOrder());
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for (int i = 0; i < nums.length; ++i) {
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d.computeIfAbsent(nums[i], k -> new ArrayList<>()).add(i);
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}
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int ans = 0, k = 1 << 30;
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for (List<Integer> idx : d.values()) {
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ans = Math.max(ans, idx.get(idx.size() - 1) - k + 1);
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k = Math.min(k, idx.get(0));
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxSubarrayLength(vector<int>& nums) {
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map<int, vector<int>, greater<int>> d;
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for (int i = 0; i < nums.size(); ++i) {
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d[nums[i]].push_back(i);
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}
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int ans = 0, k = 1 << 30;
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for (auto& [_, idx] : d) {
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ans = max(ans, idx.back() - k + 1);
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k = min(k, idx[0]);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maxSubarrayLength(nums []int) (ans int) {
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d := map[int][]int{}
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for i, x := range nums {
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d[x] = append(d[x], i)
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}
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keys := []int{}
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for x := range d {
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keys = append(keys, x)
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}
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sort.Slice(keys, func(i, j int) bool { return keys[i] > keys[j] })
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k := 1 << 30
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for _, x := range keys {
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idx := d[x]
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ans = max(ans, idx[len(idx)-1]-k+1)
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k = min(k, idx[0])
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}
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return ans
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **TypeScript**
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```ts
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function maxSubarrayLength(nums: number[]): number {
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const d: Map<number, number[]> = new Map();
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for (let i = 0; i < nums.length; ++i) {
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if (!d.has(nums[i])) {
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d.set(nums[i], []);
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}
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d.get(nums[i])!.push(i);
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}
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const keys = Array.from(d.keys()).sort((a, b) => b - a);
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let ans = 0;
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let k = Infinity;
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for (const x of keys) {
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const idx = d.get(x)!;
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ans = Math.max(ans, idx.at(-1) - k + 1);
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k = Math.min(k, idx[0]);
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}
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return ans;
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [2863. Maximum Length of Semi-Decreasing Subarrays](https://leetcode.com/problems/maximum-length-of-semi-decreasing-subarrays)
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[中文文档](/solution/2800-2899/2863.Maximum%20Length%20of%20Semi-Decreasing%20Subarrays/README.md)
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## Description
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<p>You are given an integer array <code>nums</code>.</p>
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<p>Return <em>the length of the <strong>longest semi-decreasing</strong> subarray of </em><code>nums</code><em>, and </em><code>0</code><em> if there are no such subarrays.</em></p>
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<ul>
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<li>A <b>subarray</b> is a contiguous non-empty sequence of elements within an array.</li>
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<li>A non-empty array is <strong>semi-decreasing</strong> if its first element is <strong>strictly greater</strong> than its last element.</li>
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</ul>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [7,6,5,4,3,2,1,6,10,11]
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<strong>Output:</strong> 8
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<strong>Explanation:</strong> Take the subarray [7,6,5,4,3,2,1,6].
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The first element is 7 and the last one is 6 so the condition is met.
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Hence, the answer would be the length of the subarray or 8.
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It can be shown that there aren&#39;t any subarrays with the given condition with a length greater than 8.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [57,55,50,60,61,58,63,59,64,60,63]
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<strong>Output:</strong> 6
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<strong>Explanation:</strong> Take the subarray [61,58,63,59,64,60].
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The first element is 61 and the last one is 60 so the condition is met.
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Hence, the answer would be the length of the subarray or 6.
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It can be shown that there aren&#39;t any subarrays with the given condition with a length greater than 6.
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,3,4]
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> Since there are no semi-decreasing subarrays in the given array, the answer is 0.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
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<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def maxSubarrayLength(self, nums: List[int]) -> int:
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d = defaultdict(list)
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for i, x in enumerate(nums):
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d[x].append(i)
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ans, k = 0, inf
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for x in sorted(d, reverse=True):
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ans = max(ans, d[x][-1] - k + 1)
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k = min(k, d[x][0])
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int maxSubarrayLength(int[] nums) {
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TreeMap<Integer, List<Integer>> d = new TreeMap<>(Comparator.reverseOrder());
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for (int i = 0; i < nums.length; ++i) {
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d.computeIfAbsent(nums[i], k -> new ArrayList<>()).add(i);
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}
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int ans = 0, k = 1 << 30;
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for (List<Integer> idx : d.values()) {
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ans = Math.max(ans, idx.get(idx.size() - 1) - k + 1);
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k = Math.min(k, idx.get(0));
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxSubarrayLength(vector<int>& nums) {
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map<int, vector<int>, greater<int>> d;
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for (int i = 0; i < nums.size(); ++i) {
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d[nums[i]].push_back(i);
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}
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int ans = 0, k = 1 << 30;
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for (auto& [_, idx] : d) {
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ans = max(ans, idx.back() - k + 1);
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k = min(k, idx[0]);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maxSubarrayLength(nums []int) (ans int) {
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d := map[int][]int{}
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for i, x := range nums {
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d[x] = append(d[x], i)
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}
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keys := []int{}
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for x := range d {
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keys = append(keys, x)
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}
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sort.Slice(keys, func(i, j int) bool { return keys[i] > keys[j] })
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k := 1 << 30
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for _, x := range keys {
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idx := d[x]
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ans = max(ans, idx[len(idx)-1]-k+1)
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k = min(k, idx[0])
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}
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return ans
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **TypeScript**
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```ts
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function maxSubarrayLength(nums: number[]): number {
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const d: Map<number, number[]> = new Map();
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for (let i = 0; i < nums.length; ++i) {
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if (!d.has(nums[i])) {
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d.set(nums[i], []);
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}
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d.get(nums[i])!.push(i);
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}
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const keys = Array.from(d.keys()).sort((a, b) => b - a);
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let ans = 0;
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let k = Infinity;
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for (const x of keys) {
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const idx = d.get(x)!;
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ans = Math.max(ans, idx.at(-1) - k + 1);
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k = Math.min(k, idx[0]);
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}
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return ans;
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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class Solution {
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public:
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int maxSubarrayLength(vector<int>& nums) {
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map<int, vector<int>, greater<int>> d;
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for (int i = 0; i < nums.size(); ++i) {
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d[nums[i]].push_back(i);
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}
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int ans = 0, k = 1 << 30;
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for (auto& [_, idx] : d) {
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ans = max(ans, idx.back() - k + 1);
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k = min(k, idx[0]);
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}
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return ans;
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}
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};

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