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) with0 < 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: DP / LIS
LIS[i] := longest increasing subsequence ends with nums[i]
LDS[i] := longest decreasing subsequence starts with nums[i]
Let nums[i] be the peak, the length of the mountain array is LIS[i] + LDS[i] – 1
Note: LIS[i] and LDS[i] must be > 1 to form a valid mountain array.
ans = min(n – (LIS[i] + LDS[i] – 1)) 0 <= i < n
Time complexity: O(n^2)
Space complexity: O(n)
C++
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 |
// Author: Huahua class Solution { public: int minimumMountainRemovals(vector<int>& nums) { const int n = nums.size(); vector<int> LIS(n, 1); // LIS[i] := Longest increasing subseq ends with nums[i] vector<int> LDS(n, 1); // LDS[i] := Longest decreasing subseq starts with nums[i] for (int i = 0; i < n; ++i) for (int j = 0; j < i; ++j) if (nums[i] > nums[j]) LIS[i] = max(LIS[i], LIS[j] + 1); for (int i = n - 1; i >= 0; --i) for (int j = n - 1; j > i; --j) if (nums[i] > nums[j]) LDS[i] = max(LDS[i], LDS[j] + 1); int ans = INT_MAX; for (int i = 0; i < n; ++i) { if (LIS[i] < 2 || LDS[i] < 2) continue; ans = min(ans, n - (LIS[i] + LDS[i] - 1)); } return ans; } }; |
请尊重作者的劳动成果,转载请注明出处!花花保留对文章/视频的所有权利。
如果您喜欢这篇文章/视频,欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.
Be First to Comment