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[] left=new int[n];
int[] right=new int[n];
int result=0;
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>-1;i--)
{
for (int j=n-1;j>i;j--)
{
if (nums[i]>nums[j])
right[i]=Math.max(right[i],right[j]+1);
}
}
for (int i=1;i<n;i++)
{
// System.out.println("left["+i+"]"+left[i]);
// System.out.println("right["+i+"]"+right[i]);
if (left[i] != 0 && right[i] != 0)
result = Math.max(result, left[i] +right[i]);
}
return n - (result + 1);
}
}