<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>. You have to partition the array into one or more <strong>contiguous</strong> subarrays.</p>

<p>We call a partition of the array <strong>valid</strong> if each of the obtained subarrays satisfies <strong>one</strong> of the following conditions:</p>

<ol>
	<li>The subarray consists of <strong>exactly</strong> <code>2,</code> equal elements. For example, the subarray <code>[2,2]</code> is good.</li>
	<li>The subarray consists of <strong>exactly</strong> <code>3,</code> equal elements. For example, the subarray <code>[4,4,4]</code> is good.</li>
	<li>The subarray consists of <strong>exactly</strong> <code>3</code> consecutive increasing elements, that is, the difference between adjacent elements is <code>1</code>. For example, the subarray <code>[3,4,5]</code> is good, but the subarray <code>[1,3,5]</code> is not.</li>
</ol>

<p>Return <code>true</code><em> if the array has <strong>at least</strong> one valid partition</em>. Otherwise, return <code>false</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,4,4,5,6]
<strong>Output:</strong> true
<strong>Explanation:</strong> The array can be partitioned into the subarrays [4,4] and [4,5,6].
This partition is valid, so we return true.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,1,2]
<strong>Output:</strong> false
<strong>Explanation:</strong> There is no valid partition for this array.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution {
public:
    // int rec(int i, int &n, vector<int>& nums, vector<int> &dp)
    // {
    //     if(i==n)    return 1;

    //     if(dp[i] != -1) return dp[i];

    //     if(i < n-1)
    //     {
    //         if(nums[i] == nums[i+1])
    //         {
    //             if(rec(i+2, n, nums, dp)==1)   return dp[i] = 1;

    //             if(i<n-2 && nums[i+1]==nums[i+2])
    //             {
    //                 return dp[i] = rec(i+3, n, nums, dp);
    //             }
    //         }
    //     }
    //     if(i<n-2 && nums[i]+1 == nums[i+1] && nums[i+1]+1== nums[i+2])
    //     {
    //         return dp[i] = rec(i+3, n, nums, dp);
    //     }
    //     return dp[i] = 0;
    // }
    bool validPartition(vector<int>& nums) {
        int n = nums.size();
        vector<int> dp(n+1, 0);
        dp[n]=1;
        for(int i = n-2; i>=0; i--)
        {
            if(nums[i]==nums[i+1])
            {
                dp[i] = max(dp[i], dp[i+2]);
                if(dp[i])   continue;
                if(i<n-2 && nums[i+1]==nums[i+2])
                {
                    dp[i] = dp[i+3];
                    continue;
                }   
            }
            if(i<n-2 && nums[i]+1 == nums[i+1] && nums[i+1]+1== nums[i+2])
            {
                dp[i] = dp[i+3];
            }
        }
        return dp[0];
    }
};
90000