<p>Given two integer arrays <code>pushed</code> and <code>popped</code> each with distinct values, return <code>true</code><em> if this could have been the result of a sequence of push and pop operations on an initially empty stack, or </em><code>false</code><em> otherwise.</em></p>

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

<pre>
<strong>Input:</strong> pushed = [1,2,3,4,5], popped = [4,5,3,2,1]
<strong>Output:</strong> true
<strong>Explanation:</strong> We might do the following sequence:
push(1), push(2), push(3), push(4),
pop() -&gt; 4,
push(5),
pop() -&gt; 5, pop() -&gt; 3, pop() -&gt; 2, pop() -&gt; 1
</pre>

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

<pre>
<strong>Input:</strong> pushed = [1,2,3,4,5], popped = [4,3,5,1,2]
<strong>Output:</strong> false
<strong>Explanation:</strong> 1 cannot be popped before 2.
</pre>

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

<ul>
	<li><code>1 &lt;= pushed.length &lt;= 1000</code></li>
	<li><code>0 &lt;= pushed[i] &lt;= 1000</code></li>
	<li>All the elements of <code>pushed</code> are <strong>unique</strong>.</li>
	<li><code>popped.length == pushed.length</code></li>
	<li><code>popped</code> is a permutation of <code>pushed</code>.</li>
</ul>

class Solution:
    def validateStackSequences(self, pushed: List[int], popped: List[int]) -> bool:
        i, j = 0, 0
        stack = []
        while i < len(pushed) and j < len(popped):
            if not stack:
                stack.append(pushed[i])
                i += 1
                continue
            if stack[-1] == popped[j]:
                stack.pop()
                j += 1
                continue
            stack.append(pushed[i])
            i += 1
        while j < len(popped):
            if stack[-1] != popped[j]:
                return False
            stack.pop()
            j += 1
        

        return j == len(popped)
            
16700