// 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 &nums) { int n = nums.size(); vector 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 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 }