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+// 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
+}