LeetCode-in-Java

1671. Minimum Number of Removals to Make Mountain Array

Hard

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.

Solution

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class Solution {
    public int minimumMountainRemovals(int[] nums) {
        int n = nums.length;
        // lbs -> longest bitomic subsequence
        int lbs = 0;
        int[] dp = new int[n];
        // dp[i] -> lis end at index i, dp2[i] -> lds end at index i
        int[] dp2 = new int[n];
        List<Integer> lis = new ArrayList<>();
        // calculate longest increasing subsequence
        for (int i = 0; i < n - 1; i++) {
            if (lis.isEmpty() || lis.get(lis.size() - 1) < nums[i]) {
                lis.add(nums[i]);
            } else {
                int idx = Collections.binarySearch(lis, nums[i]);
                if (idx < 0) {
                    lis.set(-idx - 1, nums[i]);
                }
            }
            dp[i] = lis.size();
        }
        lis = new ArrayList<>();
        // calculate longest decreasing subsequence
        for (int i = n - 1; i >= 1; i--) {
            if (lis.isEmpty() || lis.get(lis.size() - 1) < nums[i]) {
                lis.add(nums[i]);
            } else {
                int idx = Collections.binarySearch(lis, nums[i]);
                if (idx < 0) {
                    lis.set(-idx - 1, nums[i]);
                }
            }
            dp2[i] = lis.size();
            if (dp[i] > 1 && dp2[i] > 1) {
                lbs = Math.max(lbs, dp[i] + dp2[i] - 1);
            }
        }
        return n - lbs;
    }
}

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