2289. Steps to Make Array Non-decreasing

Problem

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 <= 10^5

• 1 <= nums[i] <= 10^9

Solution (Java)

class Solution {
    public int totalSteps(int[] nums) {
        int max = 0;
        int[] pos = new int[nums.length + 1];
        int[] steps = new int[nums.length + 1];
        int top = -1;
        for (int i = 0; i <= nums.length; i++) {
            int val = i == nums.length ? Integer.MAX_VALUE : nums[i];
            while (top >= 0 && nums[pos[top]] <= val) {
                if (top == 0) {
                    max = Math.max(max, steps[pos[top--]]);
                } else {
                    steps[pos[--top]] = Math.max(steps[pos[top]] + 1, steps[pos[top + 1]]);
                }
            }
            pos[++top] = i;
        }
        return max;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).