2289. Steps to Make Array Non-decreasing

Description

You are given a 0-indexed integer array nums. In one step, remove all elements nums[i] where nums[i - 1] > nums[i] for all 0 < i < nums.length.

Return the number of steps performed until nums becomes a non-decreasing array.

Example 1:

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

Example 2:

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

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Solution

steps-to-make-array-non-decreasing.py

from sortedcontainers import SortedList

class Solution:
    def totalSteps(self, nums: List[int]) -> int:
        nums = [float('-inf')] + nums + [float('inf')]
        res = 0

        sl = SortedList([(i, x) for i, x in enumerate(nums)])
        to_process = set()

        for index, (a, b) in enumerate(zip(nums, nums[1:])):
            if a > b:
                to_process.add((index + 1, b))

        while to_process:
            new_process = set()
            res += 1

            for j, b in to_process:
                index = sl.bisect_left((j, b))
                i, a = sl[index - 1]
                k, c = sl[index + 1]

                del sl[index]

                if a > c and (k, c) not in to_process:
                    new_process.add((k, c))


            to_process = new_process

        return res