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Diffstat (limited to 'min_num_remove_mountain_arr.cpp')
-rw-r--r-- | min_num_remove_mountain_arr.cpp | 51 |
1 files changed, 51 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 +} |