feat(work): add LeetCode solutions for problems 380, 82, 92

work/1/Medium/Array/380.insert-delete-getrandom-o1.py

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+"""
+380. Insert Delete GetRandom O(1)
+Difficulty: Medium
+https://leetcode.com/problems/insert-delete-getrandom-o1/
+
+──────────────────────────────────────────────────
+
+Implement the RandomizedSet class:
+
+        • RandomizedSet() Initializes the RandomizedSet object.
+
+• bool insert(int val) Inserts an item val into the set if not
+present. Returns true if the item was not present, false otherwise.
+
+• bool remove(int val) Removes an item val from the set if present.
+Returns true if the item was present, false otherwise.
+
+• int getRandom() Returns a random element from the current set of
+elements (it's guaranteed that at least one element exists when this
+method is called). Each element must have the same probability of
+being returned.
+
+You must implement the functions of the class such that each function
+works in average O(1) time complexity.
+
+
+
+Example 1:
+
+Input
+["RandomizedSet", "insert", "remove", "insert", "getRandom",
+"remove", "insert", "getRandom"]
+[[], [1], [2], [2], [], [1], [2], []]
+Output
+[null, true, false, true, 2, true, false, 2]
+
+Explanation
+RandomizedSet randomizedSet = new RandomizedSet();
+randomizedSet.insert(1); // Inserts 1 to the set. Returns true as 1
+was inserted successfully.
+randomizedSet.remove(2); // Returns false as 2 does not exist in the
+set.
+randomizedSet.insert(2); // Inserts 2 to the set, returns true. Set
+now contains [1,2].
+randomizedSet.getRandom(); // getRandom() should return either 1 or 2
+randomly.
+randomizedSet.remove(1); // Removes 1 from the set, returns true. Set
+now contains [2].
+randomizedSet.insert(2); // 2 was already in the set, so return false.
+randomizedSet.getRandom(); // Since 2 is the only number in the set,
+getRandom() will always return 2.
+
+
+
+Constraints:
+
+        • -2^31 <= val <= 2^31 - 1
+
+• At most 2 * 10^5 calls will be made to insert, remove, and
+getRandom.
+
+• There will be at least one element in the data structure when
+getRandom is called.
+"""
+
+from random import choice
+
+
+class RandomizedSet:
+    def __init__(self):
+        self.list = []
+        self.dict = {}
+
+    def insert(self, val: int) -> bool:
+        if val in self.dict:
+            return False
+        self.dict[val] = len(self.list)
+        self.list.append(val)
+        return True
+
+    def remove(self, val: int) -> bool:
+        if val in self.dict:
+            last, idx = self.list[-1], self.dict[val]
+            self.list[idx], self.dict[last] = last, idx
+            self.list.pop()
+            del self.dict[val]
+            return True
+        return False
+
+    def getRandom(self) -> int:
+        return choice(self.list)
+
+
+# Your RandomizedSet object will be instantiated and called as such:
+# obj = RandomizedSet()
+# param_1 = obj.insert(val)
+# param_2 = obj.remove(val)
+# param_3 = obj.getRandom()

work/1/Medium/Linked List/82.remove-duplicates-from-sorted-list-ii.py

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+"""
+82. Remove Duplicates from Sorted List II
+Difficulty: Medium
+https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/
+
+──────────────────────────────────────────────────
+
+You are given the head of a sorted linked list.
+
+Delete all nodes that have duplicate numbers, leaving only distinct
+numbers from the original list.
+
+Return the linked list sorted as well.
+
+
+
+Example 1:
+
+Input: head = [1,2,3,3,4,4,5]
+Output: [1,2,5]
+
+Example 2:
+
+Input: head = [1,1,1,2,3]
+Output: [2,3]
+
+
+
+Constraints:
+
+        • The number of nodes in the list is in the range [0, 300].
+
+        • -100 <= Node.val <= 100
+
+        • The list is guaranteed to be sorted in ascending order.
+"""
+
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def deleteDuplicates(self, head: ListNode | None) -> ListNode | None:
+        sentinel = ListNode(0, head)
+        pred = sentinel
+
+        while head:
+            if head.next and head.val == head.next.val:
+                while head.next and head.val == head.next.val:
+                    head = head.next
+
+                pred.next = head.next
+            else:
+                pred = pred.next
+
+            head = head.next
+
+        return sentinel.next

work/1/Medium/Linked List/92.reverse-linked-list-ii.py

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+"""
+92. Reverse Linked List II
+Difficulty: Medium
+https://leetcode.com/problems/reverse-linked-list-ii/
+
+──────────────────────────────────────────────────
+
+Given the head of a singly linked list and two integers left and
+right where left <= right, reverse the nodes of the list from position
+left to position right, and return the reversed list.
+
+
+
+Example 1:
+
+Input: head = [1,2,3,4,5], left = 2, right = 4
+Output: [1,4,3,2,5]
+
+Example 2:
+
+Input: head = [5], left = 1, right = 1
+Output: [5]
+
+
+
+Constraints:
+
+        • The number of nodes in the list is n.
+
+        • 1 <= n <= 500
+
+        • -500 <= Node.val <= 500
+
+        • 1 <= left <= right <= n
+
+
+
+Follow up: Could you do it in one pass?
+"""
+
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def reverseBetween(
+        self, head: ListNode | None, left: int, right: int
+    ) -> ListNode | None:
+        cur, prev = head, None
+        while left > 1:
+            prev = cur
+            cur = cur.next
+            left, right = left - 1, right - 1
+
+        tail, con = cur, prev
+        while right:
+            third = cur.next
+            cur.next = prev
+            prev = cur
+            cur = third
+            right -= 1
+
+        if con:
+            con.next = prev
+        else:
+            head = prev
+
+        tail.next = cur
+        return head