LeetCode in Kotlin

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

class Solution {
    fun minimumMountainRemovals(nums: IntArray): Int {
        val n = nums.size
        // lbs -> longest bitomic subsequence
        var lbs = 0
        val dp = IntArray(n)
        // dp[i] -> lis end at index i, dp2[i] -> lds end at index i
        val dp2 = IntArray(n)
        var lis: MutableList<Int> = ArrayList()
        // calculate longest increasing subsequence
        for (i in 0 until n - 1) {
            if (lis.isEmpty() || lis[lis.size - 1] < nums[i]) {
                lis.add(nums[i])
            } else {
                val idx = lis.binarySearch(nums[i])
                if (idx < 0) {
                    lis[-idx - 1] = nums[i]
                }
            }
            dp[i] = lis.size
        }
        lis = ArrayList()
        // calculate longest decreasing subsequence
        for (i in n - 1 downTo 1) {
            if (lis.isEmpty() || lis[lis.size - 1] < nums[i]) {
                lis.add(nums[i])
            } else {
                val idx = lis.binarySearch(nums[i])
                if (idx < 0) {
                    lis[-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
    }
}

This site is open source. Improve this page.