1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
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
}
|