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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

Solutions

Solution 1

Thinking

Each round deletes every element that is strictly smaller than its left neighbor; we want the number of rounds. \(n \le 10^5\) forbids simulating rounds. The deletion time of an index is determined by how long a decreasing suffix to its right takes to be absorbed by a larger left value.

A stack from the right holds indices not yet eaten. Popping updates \(dp[i] = \max(dp[i]+1, dp[\textit{top}])\). The answer is \(\max(dp)\).

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class Solution:
    def totalSteps(self, nums: List[int]) -> int:
        stk = []
        ans, n = 0, len(nums)
        dp = [0] * 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)

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class Solution {
    public int totalSteps(int[] nums) {
        Deque<Integer> stk = new ArrayDeque<>();
        int ans = 0;
        int n = nums.length;
        int[] dp = new int[n];
        for (int i = n - 1; i >= 0; --i) {
            while (!stk.isEmpty() && nums[i] > nums[stk.peek()]) {
                dp[i] = Math.max(dp[i] + 1, dp[stk.pop()]);
                ans = Math.max(ans, dp[i]);
            }
            stk.push(i);
        }
        return ans;
    }
}

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class Solution {
public:
    int totalSteps(vector<int>& nums) {
        stack<int> stk;
        int ans = 0, n = nums.size();
        vector<int> dp(n);
        for (int i = n - 1; i >= 0; --i) {
            while (!stk.empty() && nums[i] > nums[stk.top()]) {
                dp[i] = max(dp[i] + 1, dp[stk.top()]);
                ans = max(ans, dp[i]);
                stk.pop();
            }
            stk.push(i);
        }
        return ans;
    }
};

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func totalSteps(nums []int) int {
    stk := []int{}
    ans, n := 0, len(nums)
    dp := make([]int, n)
    for i := n - 1; i >= 0; i-- {
        for len(stk) > 0 && nums[i] > nums[stk[len(stk)-1]] {
            dp[i] = max(dp[i]+1, dp[stk[len(stk)-1]])
            stk = stk[:len(stk)-1]
            ans = max(ans, dp[i])
        }
        stk = append(stk, i)
    }
    return ans
}

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function totalSteps(nums: number[]): number {
    let ans = 0;
    let stack = [];
    for (let num of nums) {
        let max = 0;
        while (stack.length && stack[0][0] <= num) {
            max = Math.max(stack[0][1], max);
            stack.shift();
        }
        if (stack.length) max++;
        ans = Math.max(max, ans);
        stack.unshift([num, max]);
    }
    return ans;
}

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