1671 - Minimum Number of Removals to Make Mountain Array

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.

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;
      }
  };

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;
      }
  }

PYTHON

• class Solution:
      def minimumMountainRemovals(self, nums):
          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)
  
          ans = 0
          for i in range(n):
              if left[i] > 1 and right[i] > 1:
                  ans = max(ans, left[i] + right[i] - 1)
  
          return n - ans