<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>. In one step, <strong>remove</strong> all elements <code>nums[i]</code> where <code>nums[i - 1] &gt; nums[i]</code> for all <code>0 &lt; i &lt; nums.length</code>.</p>

<p>Return <em>the number of steps performed until </em><code>nums</code><em> becomes a <strong>non-decreasing</strong> array</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [5,3,4,4,7,3,6,11,8,5,11]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The following are the steps performed:
- Step 1: [5,<strong><u>3</u></strong>,4,4,7,<u><strong>3</strong></u>,6,11,<u><strong>8</strong></u>,<u><strong>5</strong></u>,11] becomes [5,4,4,7,6,11,11]
- Step 2: [5,<u><strong>4</strong></u>,4,7,<u><strong>6</strong></u>,11,11] becomes [5,4,7,11,11]
- Step 3: [5,<u><strong>4</strong></u>,7,11,11] becomes [5,7,11,11]
[5,7,11,11] is a non-decreasing array. Therefore, we return 3.
</pre>

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

<pre>
<strong>Input:</strong> nums = [4,5,7,7,13]
<strong>Output:</strong> 0
<strong>Explanation:</strong> nums is already a non-decreasing array. Therefore, we return 0.
</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>

from typing import List

class Solution:
    def totalSteps(self, nums: List[int]) -> int:
        stack = []  # Stack to store (value, steps required to remove)
        max_steps = 0
        
        for num in nums:
            steps = 1  # Initially, we assume the current element might need 1 step to remove all previous decreasing elements
            while stack and stack[-1][0] <= num:
                # If the current element is greater than or equal to the top of the stack
                # update the steps required to remove the current element
                steps = max(steps, stack.pop()[1] + 1)
            
            if stack:
                # If there are elements in the stack, it means the current element is smaller than the top of the stack
                # so it will be removed in the future steps
                max_steps = max(max_steps, steps)
            else:
                # If the stack is empty, it means the current element is greater than all previous elements
                # so it won't be removed, reset steps to 0
                steps = 0
            
            stack.append((num, steps))
        
        return max_steps

# Example usage:
solution = Solution()
nums = [5,3,4,4,7,3,6,11,8,5,11]
print(solution.totalSteps(nums))  # Output: 3

nums = [4,5,7,7,13]
print(solution.totalSteps(nums))  # Output: 0
707