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authorJack Sangdahl <[email protected]>2024-10-29 19:53:49 -0600
committerJack Sangdahl <[email protected]>2024-10-29 19:53:49 -0600
commit80c009131135eef4b5c611a9d8091abdf94a8f30 (patch)
tree3a33269c7fafe47d1d8fe9b20709aa7cd7dac12c
parentbf0ef281bee5004aee7ac459d42d7bab526c6d8a (diff)
downloadleetcode-80c009131135eef4b5c611a9d8091abdf94a8f30.tar.gz
leetcode-80c009131135eef4b5c611a9d8091abdf94a8f30.zip
1671. minimum number of removals to make mountain array
-rw-r--r--min_num_remove_mountain_arr.cpp51
-rw-r--r--min_num_remove_mountain_arr.py39
2 files changed, 90 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
+}
diff --git a/min_num_remove_mountain_arr.py b/min_num_remove_mountain_arr.py
new file mode 100644
index 0000000..e2e0752
--- /dev/null
+++ b/min_num_remove_mountain_arr.py
@@ -0,0 +1,39 @@
+# 1671. Minimum Number of Removals to Make Mountain Array
+from bisect import bisect_left
+
+"""
+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(object):
+ def minimumMountainRemovals(self, nums):
+ """
+ :type nums: List[int]
+ :rtype: int
+ """
+
+ def lis(nums):
+ dp = [10**10] * (len(nums) + 1)
+ lens = [0] * len(nums)
+ for i, j in enumerate(nums):
+ lens[i] = bisect_left(dp, j) + 1
+ dp[lens[i] - 1] = j
+ return lens
+
+ l1, l2 = lis(nums), lis(nums[::-1])[::-1]
+ ans, n = 0, len(nums)
+ for i in range(n):
+ if l1[i] >= 2 and l2[i] >= 2:
+ ans = max(ans, l1[i] + l2[i] - 1)
+ return n - ans
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.minimumMountainRemovals(nums=[1, 3, 1]))
+ print(obj.minimumMountainRemovals(nums=[2, 1, 1, 5, 6, 2, 3, 1]))