README.md

2289. Steps to Make Array Non-decreasing

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

Solutions (Rust)

1. Solution

impl Solution {
    pub fn total_steps(nums: Vec<i32>) -> i32 {
        let mut stack = vec![];
        let mut ret = 0;

        for &num in &nums {
            let mut x = 0;

            while stack.last().unwrap_or(&(i32::MAX, 0)).0 <= num {
                x = x.max(stack.pop().unwrap().1);
            }

            if stack.is_empty() {
                stack.push((num, 0));
            } else {
                stack.push((num, x + 1));
                ret = ret.max(x + 1);
            }
        }

        ret
    }
}