feat: solve No.2369

Author: BaffinLee

2301-2400/2369. Check if There is a Valid Partition For The Array.md

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+# 2369. Check if There is a Valid Partition For The Array
+
+- Difficulty: Medium.
+- Related Topics: Array, Dynamic Programming.
+- Similar Questions: .
+
+## Problem
+
+You are given a **0-indexed** integer array `nums`. You have to partition the array into one or more **contiguous** subarrays.
+
+We call a partition of the array **valid** if each of the obtained subarrays satisfies **one** of the following conditions:
+
+
+	
+- The subarray consists of **exactly** `2` equal elements. For example, the subarray `[2,2]` is good.
+	
+- The subarray consists of **exactly** `3` equal elements. For example, the subarray `[4,4,4]` is good.
+	
+- The subarray consists of **exactly** `3` consecutive increasing elements, that is, the difference between adjacent elements is `1`. For example, the subarray `[3,4,5]` is good, but the subarray `[1,3,5]` is not.
+
+
+Return `true`** if the array has **at least** one valid partition**. Otherwise, return `false`.
+
+ 
+Example 1:
+
+```
+Input: nums = [4,4,4,5,6]
+Output: true
+Explanation: The array can be partitioned into the subarrays [4,4] and [4,5,6].
+This partition is valid, so we return true.
+```
+
+Example 2:
+
+```
+Input: nums = [1,1,1,2]
+Output: false
+Explanation: There is no valid partition for this array.
+```
+
+ 
+**Constraints:**
+
+
+	
+- `2 <= nums.length <= 105`
+	
+- `1 <= nums[i] <= 106`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} nums
+ * @return {boolean}
+ */
+var validPartition = function(nums) {
+    var dp = Array(nums.length + 1);
+    dp[nums.length] = true;
+    for (var i = nums.length - 1; i >= 0; i--) {
+        dp[i] = (nums[i] === nums[i + 1] && dp[i + 2])
+            || (nums[i] === nums[i + 1] && nums[i + 1] === nums[i + 2] && dp[i + 3])
+            || (nums[i] + 1 === nums[i + 1] && nums[i + 1] + 1 === nums[i + 2] && dp[i + 3]);
+    }
+    return dp[0];
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(n).