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author | Jack Sangdahl <[email protected]> | 2025-06-27 12:23:02 -0600 |
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committer | Jack Sangdahl <[email protected]> | 2025-06-27 12:23:02 -0600 |
commit | a96b65cbfad7f62f3db6b2eab105e4e25c0414c5 (patch) | |
tree | e2a2b3f49f9fb741be81415266ac93e74cff42ec /other/min_num_remove_mountain_arr.cpp | |
parent | 694ad5ed71edcb1deccab54b72d706b779918276 (diff) | |
download | leetcode-a96b65cbfad7f62f3db6b2eab105e4e25c0414c5.tar.gz leetcode-a96b65cbfad7f62f3db6b2eab105e4e25c0414c5.zip |
chore: remove old cpp and rust solutions
Diffstat (limited to 'other/min_num_remove_mountain_arr.cpp')
-rw-r--r-- | other/min_num_remove_mountain_arr.cpp | 51 |
1 files changed, 0 insertions, 51 deletions
diff --git a/other/min_num_remove_mountain_arr.cpp b/other/min_num_remove_mountain_arr.cpp deleted file mode 100644 index 007c8c6..0000000 --- a/other/min_num_remove_mountain_arr.cpp +++ /dev/null @@ -1,51 +0,0 @@ -// 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 -} |