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.
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]) lis[i] = Math.Max(lis[i], lis[j] + 1);
11 }
12
13 for (int i = n - 1; i >= 0; i--) {
14 lds[i] = 1;
15 for (int j = n - 1; j > i; j--)
16 if (nums[i] > nums[j]) lds[i] = Math.Max(lds[i], lds[j] + 1);
17 }
18
19 int maxMountain = 0;
20 for (int i = 0; i < n; i++) {
21 if (lis[i] > 1 && lds[i] > 1)
22 maxMountain = Math.Max(maxMountain, lis[i] + lds[i] - 1);
23 }
24 return n - maxMountain;
25 }
26}
27This is a C# solution using arrays to compute LIS and LDS at each index. After identifying peaks, it calculates the maximum possible mountain's length and the number of elements that need to be removed thereafter.
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.
1import java.util.Arrays;
2class Solution {
3 public int minMountainRemovals(int[] nums) {
4 int n = nums.length;
5 int[] lis = new int[n];
6 int[] lds = new int[n];
7 Arrays.fill(lis, 1);
8 Arrays.fill(lds, 1);
9
10 for (int i = 0; i < n; i++) {
11 for (int j = 0; j < i; j++) {
12 if (nums[i] > nums[j]) {
13 lis[i] = Math.max(lis[i], lis[j] + 1);
14 }
15 }
16 }
17
18 for (int i = n - 1; i >= 0; i--) {
19 for (int j = n - 1; j > i; j--) {
20 if (nums[i] > nums[j]) {
21 lds[i] = Math.max(lds[i], lds[j] + 1);
22 }
23 }
24 }
25
26 int maxMountain = 0;
27 for (int i = 0; i < n; i++) {
28 if (lis[i] > 1 && lds[i] > 1) {
29 maxMountain = Math.max(maxMountain, lis[i] + lds[i] - 1);
30 }
31 }
32
33 return n - maxMountain;
34 }
35}
36By continuing to consolidate the increasing and decreasing sequence identification, this Java solution implements a dual-pass matrix, achieving intricate sequence stitching. Aliases within the Java language such as Arrays allow efficient array manipulation across sequences.