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Formatted question description: https://leetcode.ca/all/2289.html

# 2289. Steps to Make Array Non-decreasing

• Difficulty: Medium.
• Related Topics: Array, Linked List, Stack, Monotonic Stack.
• Similar Questions: Remove One Element to Make the Array Strictly Increasing.

## 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 <= 105

• 1 <= nums[i] <= 109

## Solution (Java, C++, Python)

• 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;
}
}

• Todo

• class Solution:
def totalSteps(self, nums: List[int]) -> int:
stk = []
ans, n = 0, len(nums)
dp =  * n
for i in range(n - 1, -1, -1):
while stk and nums[i] > nums[stk[-1]]:
dp[i] = max(dp[i] + 1, dp[stk.pop()])
stk.append(i)
return max(dp)

############

# 2289. Steps to Make Array Non-decreasing
# https://leetcode.com/problems/steps-to-make-array-non-decreasing

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)])
p = set()

for i, (a, b) in enumerate(zip(nums, nums[1:])):
if a > b:

while p:
newP = set()
res += 1

for j, b in p:
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 p:

p = newP

return res



Explain:

nope.

Complexity:

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