README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/1600-1699/1671.Minimum%20Number%20of%20Removals%20to%20Make%20Mountain%20Array/README_EN.md	1912	Biweekly Contest 40 Q4	Greedy Array Binary Search Dynamic Programming

1671. Minimum Number of Removals to Make Mountain Array

中文文档

Description

You may recall that an array arr is a mountain array if and only if:

• arr.length >= 3
• There exists some index i (0-indexed) with 0 < i < arr.length - 1 such that:
  • arr[0] < arr[1] < ... < arr[i - 1] < arr[i]
  • arr[i] > arr[i + 1] > ... > arr[arr.length - 1]

Given an integer array nums​​​, return the minimum number of elements to remove to make nums​​​ a mountain array.

 

Example 1:

Input: nums = [1,3,1]
Output: 0
Explanation: The array itself is a mountain array so we do not need to remove any elements.

Example 2:

Input: nums = [2,1,1,5,6,2,3,1]
Output: 3
Explanation: One solution is to remove the elements at indices 0, 1, and 5, making the array nums = [1,5,6,3,1].

 

Constraints:

• 3 <= nums.length <= 1000
• 1 <= nums[i] <= 109
• It is guaranteed that you can make a mountain array out of nums.

Solutions

Solution 1: Dynamic Programming

This problem can be transformed into finding the longest increasing subsequence and the longest decreasing subsequence.

We define $left[i]$ as the length of the longest increasing subsequence ending with $nums[i]$, and define $right[i]$ as the length of the longest decreasing subsequence starting with $nums[i]$.

Then the final answer is $n - \max(left[i] + right[i] - 1)$, where $1 \leq i \leq n$, and $left[i] \gt 1$ and $right[i] \gt 1$.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

Python3

class Solution:
    def minimumMountainRemovals(self, nums: List[int]) -> int:
        n = len(nums)
        left = [1] * n
        right = [1] * n
        for i in range(1, n):
            for j in range(i):
                if nums[i] > nums[j]:
                    left[i] = max(left[i], left[j] + 1)
        for i in range(n - 2, -1, -1):
            for j in range(i + 1, n):
                if nums[i] > nums[j]:
                    right[i] = max(right[i], right[j] + 1)
        return n - max(a + b - 1 for a, b in zip(left, right) if a > 1 and b > 1)

Java

class Solution {
    public int minimumMountainRemovals(int[] nums) {
        int n = nums.length;
        int[] left = new int[n];
        int[] right = new int[n];
        Arrays.fill(left, 1);
        Arrays.fill(right, 1);
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[i] > nums[j]) {
                    left[i] = Math.max(left[i], left[j] + 1);
                }
            }
        }
        for (int i = n - 2; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                if (nums[i] > nums[j]) {
                    right[i] = Math.max(right[i], right[j] + 1);
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (left[i] > 1 && right[i] > 1) {
                ans = Math.max(ans, left[i] + right[i] - 1);
            }
        }
        return n - ans;
    }
}

C++

class Solution {
public:
    int minimumMountainRemovals(vector<int>& nums) {
        int n = nums.size();
        vector<int> left(n, 1), right(n, 1);
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[i] > nums[j]) {
                    left[i] = max(left[i], left[j] + 1);
                }
            }
        }
        for (int i = n - 2; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                if (nums[i] > nums[j]) {
                    right[i] = max(right[i], right[j] + 1);
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (left[i] > 1 && right[i] > 1) {
                ans = max(ans, left[i] + right[i] - 1);
            }
        }
        return n - ans;
    }
};

Go

func minimumMountainRemovals(nums []int) int {
	n := len(nums)
	left, right := make([]int, n), make([]int, n)
	for i := range left {
		left[i], right[i] = 1, 1
	}
	for i := 1; i < n; i++ {
		for j := 0; j < i; j++ {
			if nums[i] > nums[j] {
				left[i] = max(left[i], left[j]+1)
			}
		}
	}
	for i := n - 2; i >= 0; i-- {
		for j := i + 1; j < n; j++ {
			if nums[i] > nums[j] {
				right[i] = max(right[i], right[j]+1)
			}
		}
	}
	ans := 0
	for i := range left {
		if left[i] > 1 && right[i] > 1 {
			ans = max(ans, left[i]+right[i]-1)
		}
	}
	return n - ans
}

TypeScript

function minimumMountainRemovals(nums: number[]): number {
    const n = nums.length;
    const left = Array(n).fill(1);
    const right = Array(n).fill(1);
    for (let i = 1; i < n; ++i) {
        for (let j = 0; j < i; ++j) {
            if (nums[i] > nums[j]) {
                left[i] = Math.max(left[i], left[j] + 1);
            }
        }
    }
    for (let i = n - 2; i >= 0; --i) {
        for (let j = i + 1; j < n; ++j) {
            if (nums[i] > nums[j]) {
                right[i] = Math.max(right[i], right[j] + 1);
            }
        }
    }
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        if (left[i] > 1 && right[i] > 1) {
            ans = Math.max(ans, left[i] + right[i] - 1);
        }
    }
    return n - ans;
}

Rust

impl Solution {
    pub fn minimum_mountain_removals(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut left = vec![1; n];
        let mut right = vec![1; n];
        for i in 1..n {
            for j in 0..i {
                if nums[i] > nums[j] {
                    left[i] = left[i].max(left[j] + 1);
                }
            }
        }
        for i in (0..n - 1).rev() {
            for j in i + 1..n {
                if nums[i] > nums[j] {
                    right[i] = right[i].max(right[j] + 1);
                }
            }
        }

        let mut ans = 0;
        for i in 0..n {
            if left[i] > 1 && right[i] > 1 {
                ans = ans.max(left[i] + right[i] - 1);
            }
        }

        (n as i32) - ans
    }
}