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author | Jack Sangdahl <[email protected]> | 2024-10-29 19:53:49 -0600 |
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committer | Jack Sangdahl <[email protected]> | 2024-10-29 19:53:49 -0600 |
commit | 80c009131135eef4b5c611a9d8091abdf94a8f30 (patch) | |
tree | 3a33269c7fafe47d1d8fe9b20709aa7cd7dac12c | |
parent | bf0ef281bee5004aee7ac459d42d7bab526c6d8a (diff) | |
download | leetcode-80c009131135eef4b5c611a9d8091abdf94a8f30.tar.gz leetcode-80c009131135eef4b5c611a9d8091abdf94a8f30.zip |
1671. minimum number of removals to make mountain array
-rw-r--r-- | min_num_remove_mountain_arr.cpp | 51 | ||||
-rw-r--r-- | min_num_remove_mountain_arr.py | 39 |
2 files changed, 90 insertions, 0 deletions
diff --git a/min_num_remove_mountain_arr.cpp b/min_num_remove_mountain_arr.cpp new file mode 100644 index 0000000..007c8c6 --- /dev/null +++ b/min_num_remove_mountain_arr.cpp @@ -0,0 +1,51 @@ +// 1671. Minimum Number of Removals to Make Mountain Array +#include "leetcode.h" + +/* + * you may recall that an array 'arr' is a mountain array if and only if + * 'arr.length >= 3', there exists soome index 'i' with '0 < i < arr.length - 1' + * such that 'arr[0] < arr[1] <... < arr[i - 1]' and 'arr[i] > arr[i + 1] > ... + * > arr[length - 1]'. given an integer array 'nums', return the minimum number + * of elements to remove to make 'nums' a mountain array + */ + +class Solution { +public: + int minimumMountainRemovals(vector<int> &nums) { + int n = nums.size(); + vector<int> left(n), right(n), stk; + for (int i = 0; i < n; i++) { + if (stk.empty() || stk.back() < nums[i]) { + stk.push_back(nums[i]); + left[i] = stk.size(); + } else { + auto iter = lower_bound(stk.begin(), stk.end(), nums[i]); + *iter = nums[i]; + left[i] = iter - stk.begin() + 1; + } + } + stk.clear(); + for (int i = n - 1; i >= 0; i--) { + if (stk.empty() || stk.back() < nums[i]) { + stk.push_back(nums[i]); + right[i] = stk.size(); + } else { + auto iter = lower_bound(stk.begin(), stk.end(), nums[i]); + *iter = nums[i]; + right[i] = iter - stk.begin() + 1; + } + } + int max_item = 0; + for (int i = 0; i < n; i++) + if (left[i] >= 2 && right[i] >= 2) + max_item = max(max_item, left[i] + right[i] - 1); + return n - max_item; + } +}; + +int main() { + Solution obj; + vector<int> n1 = {1, 3, 1}, n2 = {2, 1, 1, 5, 6, 2, 3, 1}; + printf("%d\n", obj.minimumMountainRemovals(n1)); // expect: 0 + printf("%d\n", obj.minimumMountainRemovals(n2)); // expect: 3 +} diff --git a/min_num_remove_mountain_arr.py b/min_num_remove_mountain_arr.py new file mode 100644 index 0000000..e2e0752 --- /dev/null +++ b/min_num_remove_mountain_arr.py @@ -0,0 +1,39 @@ +# 1671. Minimum Number of Removals to Make Mountain Array +from bisect import bisect_left + +""" +you may recall that an array 'arr' is a mountain array if and only if +'arr.length >= 3', there exists soome index 'i' with '0 < i < arr.length - 1' +such that 'arr[0] < arr[1] <... < arr[i - 1]' and 'arr[i] > arr[i + 1] > ... +> arr[length - 1]'. given an integer array 'nums', return the minimum number +of elements to remove to make 'nums' a mountain array +""" + + +class Solution(object): + def minimumMountainRemovals(self, nums): + """ + :type nums: List[int] + :rtype: int + """ + + def lis(nums): + dp = [10**10] * (len(nums) + 1) + lens = [0] * len(nums) + for i, j in enumerate(nums): + lens[i] = bisect_left(dp, j) + 1 + dp[lens[i] - 1] = j + return lens + + l1, l2 = lis(nums), lis(nums[::-1])[::-1] + ans, n = 0, len(nums) + for i in range(n): + if l1[i] >= 2 and l2[i] >= 2: + ans = max(ans, l1[i] + l2[i] - 1) + return n - ans + + +if __name__ == "__main__": + obj = Solution() + print(obj.minimumMountainRemovals(nums=[1, 3, 1])) + print(obj.minimumMountainRemovals(nums=[2, 1, 1, 5, 6, 2, 3, 1])) |