Question

1671. Minimum Number of Removals to Make Mountain Array

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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 $nums$.

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)

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;
  }
};

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
  }
}