This approach involves finding the longest increasing subsequence (LIS) ending at each index and the longest decreasing subsequence (LDS) starting from each index. The idea is to find a peak such that the sum of the longest increasing and decreasing subsequences is maximized, and then determine how many elements need to be removed such that only the peak and the subsequences are left.
Time Complexity: O(n^2), where n is the length of the array, due to the nested loops to calculate LIS and LDS.
Space Complexity: O(n) for the LIS and LDS arrays.
1def minMountainRemovals(nums):
2 n = len(nums)
3 lis = [1] * n
4 lds = [1] * n
5
6 for i in range(n):
7 for j in range(i):
8 if nums[i] > nums[j]:
9 lis[i] = max(lis[i], lis[j] + 1)
10
11 for i in range(n - 1, -1, -1):
12 for j in range(n - 1, i, -1):
13 if nums[i] > nums[j]:
14 lds[i] = max(lds[i], lds[j] + 1)
15
16 max_mountain = 0
17 for i in range(n):
18 if lis[i] > 1 and lds[i] > 1:
19 max_mountain = max(max_mountain, lis[i] + lds[i] - 1)
20
21 return n - max_mountain
22This Python solution calculates LIS and LDS for the array using nested loops. Then, it checks for valid peaks where both LIS and LDS are more than one. It returns minimum removals by subtracting the maximum mountain length from the total number of elements.
This method involves a greedy approach using two-pointer strategy to find potential mountain peaks. We employ two pointers to detect increasing and decreasing sequences, merging the results to form the largest mountain, iteratively removing non-peak elements.
Time Complexity: O(n^2), requiring traversal through sequences twice with nested loops.
Space Complexity: O(n) due to additional arrays retaining incremental results.
1public class Solution {
2 public int MinMountainRemovals(int[] nums) {
3 int n = nums.Length;
4 int[] lis = new int[n];
5 int[] lds = new int[n];
6
7 for (int i = 0; i < n; i++) {
8 lis[i] = 1;
9 for (int j = 0; j < i; j++)
10 if (nums[i] > nums[j])
11 lis[i] = Math.Max(lis[i], lis[j] + 1);
12 }
13
14 for (int i = n - 1; i >= 0; i--) {
15 lds[i] = 1;
16 for (int j = n - 1; j > i; j--)
17 if (nums[i] > nums[j])
18 lds[i] = Math.Max(lds[i], lds[j] + 1);
19 }
20
21 int maxMountain = 0;
22 for (int i = 0; i < n; i++) {
23 if (lis[i] > 1 && lds[i] > 1)
24 maxMountain = Math.Max(maxMountain, lis[i] + lds[i] - 1);
25 }
26
27 return n - maxMountain;
28 }
29}
30This C# solution revolves largely around the two-pointer strategy, closely coupled with sequence tracking. The code simplifies interactions and swiftly detects sequence gaps, allowing the resolution of mountain structure compliance with minimal removals.