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].
Example 3:
Input: nums = [4,3,2,1,1,2,3,1]
Output: 4
Example 4:
Input: nums = [1,2,3,4,4,3,2,1]
Output: 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 {
public int minimumMountainRemovals(int[] nums) {
int n = nums.length;
int[] inc = new int[n];
int[] dec = new int[n];
inc[0] = 0;
dec[n - 1] = 0;
for (int i = 1; i < n; i ++) {
for (int j = 0; j < i; j ++) {
if (nums[i] > nums[j]) {
inc[i] = Math.max(inc[i], inc[j] + 1);
}
}
}
for (int i = n - 2; i >= 0; i --) {
for (int j = n - 1; j > i; j --) {
if (nums[i] > nums[j]) {
dec[i] = Math.max(dec[i], dec[j] + 1);
}
}
}
int min = n;
for (int i = 0; i < n; i ++) {
if (inc[i] > 0 && dec[i] > 0) {
min = Math.min(min, n - (inc[i] + dec[i] + 1));
}
}
// System.out.println(Arrays.toString(inc));
// System.out.println(Arrays.toString(dec));
return min;
}
}