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1671. Minimum Number of Removals to Make Mountain Array
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) with0 < i < arr.length - 1such 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 <= 10001 <= 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 $nums$.
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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; } } -
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; } }; -
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) -
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 } -
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; } -
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 } }