<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>. A subarray of <code>nums</code> is called <strong>continuous</strong> if:</p>

<ul>
	<li>Let <code>i</code>, <code>i + 1</code>, ..., <code>j</code><sub> </sub>be the indices in the subarray. Then, for each pair of indices <code>i &lt;= i<sub>1</sub>, i<sub>2</sub> &lt;= j</code>, <code><font face="monospace">0 &lt;=</font> |nums[i<sub>1</sub>] - nums[i<sub>2</sub>]| &lt;= 2</code>.</li>
</ul>

<p>Return <em>the total number of <strong>continuous</strong> subarrays.</em></p>

<p>A subarray is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>

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

<pre>
<strong>Input:</strong> nums = [5,4,2,4]
<strong>Output:</strong> 8
<strong>Explanation:</strong> 
Continuous subarray of size 1: [5], [4], [2], [4].
Continuous subarray of size 2: [5,4], [4,2], [2,4].
Continuous subarray of size 3: [4,2,4].
Thereare no subarrys of size 4.
Total continuous subarrays = 4 + 3 + 1 = 8.
It can be shown that there are no more continuous subarrays.
</pre>

<p>&nbsp;</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> 6
<strong>Explanation:</strong> 
Continuous subarray of size 1: [1], [2], [3].
Continuous subarray of size 2: [1,2], [2,3].
Continuous subarray of size 3: [1,2,3].
Total continuous subarrays = 3 + 2 + 1 = 6.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

import heapq
class Solution:
    def continuousSubarrays(self, L: List[int]) -> int:
        ans=0
        maxi=[]
        mini=[]
        idx=-1
        for j in range (0,len(L)):
            heapq.heappush(maxi,[-L[j],j])
            heapq.heappush(mini,[L[j],j])
            idx1=idx
            while len(mini)!=0 and (abs(L[j]-mini[0][0])>2 or mini[0][1]<idx1):
                a,b=heapq.heappop(mini)
                idx1=max(idx1,b)
            idx2=idx
            while len(maxi)!=0 and (abs(L[j]+maxi[0][0]) or maxi[0][1]<idx2)>2:
                a,b=heapq.heappop(maxi)
                idx2=max(idx2,b)
            idx=max(idx1,idx2)
            ans=ans+j-idx
        return ans


        
46300