0225_implement-stack-using-queues/README.md

@@ -1,40 +0,0 @@
-Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (`push`, `top`, `pop`, and `empty`).
-
-Implement the `MyStack` class:
-
-*   `void push(int x)` Pushes element x to the top of the stack.
-*   `int pop()` Removes the element on the top of the stack and returns it.
-*   `int top()` Returns the element on the top of the stack.
-*   `boolean empty()` Returns `true` if the stack is empty, `false` otherwise.
-
-**Notes:**
-
-*   You must use **only** standard operations of a queue, which means that only `push to back`, `peek/pop from front`, `size` and `is empty` operations are valid.
-*   Depending on your language, the queue may not be supported natively. You may simulate a queue using a list or deque (double-ended queue) as long as you use only a queue's standard operations.
-
-**Example 1:**
-
-    Input
-    ["MyStack", "push", "push", "top", "pop", "empty"]
-    [[], [1], [2], [], [], []]
-    Output
-    [null, null, null, 2, 2, false]
-    
-    Explanation
-    MyStack myStack = new MyStack();
-    myStack.push(1);
-    myStack.push(2);
-    myStack.top(); // return 2
-    myStack.pop(); // return 2
-    myStack.empty(); // return False
-    
-
-**Constraints:**
-
-*   `1 <= x <= 9`
-*   At most `100` calls will be made to `push`, `pop`, `top`, and `empty`.
-*   All the calls to `pop` and `top` are valid.
-
-**Follow-up:** Can you implement the stack using only one queue?
-
-https://leetcode.com/problems/implement-stack-using-queues/

0225_implement-stack-using-queues/python3/two-queues-pop-linear.py

@@ -1,45 +0,0 @@
-from collections import deque
-
-class MyStack:
-
-    def __init__(self):
-        self.q = deque()
-        self.tmp = deque()
-
-    def push(self, x: int) -> None:
-        self.q.append(x)
-        self.head = x
-
-    def pop(self) -> int:
-        tmp = deque()
-        popped = None
-        prev = None
-        
-        while True:
-            popped = self.q.popleft()
-            
-            if not self.q:
-                break
-            
-            prev = popped
-            tmp.append(prev)
-            
-        self.q = tmp
-        self.head = prev
-        
-        return popped
-        
-
-    def top(self) -> int:
-        return self.head
-
-    def empty(self) -> bool:
-        return len(self.q) == 0
-
-
-# Your MyStack object will be instantiated and called as such:
-# obj = MyStack()
-# obj.push(x)
-# param_2 = obj.pop()
-# param_3 = obj.top()
-# param_4 = obj.empty()

1798_max-number-of-k-sum-pairs/README.md

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-You are given an integer array `nums` and an integer `k`.
-
-In one operation, you can pick two numbers from the array whose sum equals `k` and remove them from the array.
-
-Return _the maximum number of operations you can perform on the array_.
-
-**Example 1:**
-
-    Input: nums = [1,2,3,4], k = 5
-    Output: 2
-    Explanation: Starting with nums = [1,2,3,4]:
-    - Remove numbers 1 and 4, then nums = [2,3]
-    - Remove numbers 2 and 3, then nums = []
-    There are no more pairs that sum up to 5, hence a total of 2 operations.
-
-**Example 2:**
-
-    Input: nums = [3,1,3,4,3], k = 6
-    Output: 1
-    Explanation: Starting with nums = [3,1,3,4,3]:
-    - Remove the first two 3's, then nums = [1,4,3]
-    There are no more pairs that sum up to 6, hence a total of 1 operation.
-
-**Constraints:**
-
-*   `1 <= nums.length <= 105`
-*   `1 <= nums[i] <= 109`
-*   `1 <= k <= 109`
-
-https://leetcode.com/problems/max-number-of-k-sum-pairs

1798_max-number-of-k-sum-pairs/python3/solution.py

@@ -1,21 +0,0 @@
-class Solution:
-    def maxOperations(self, nums: List[int], k: int) -> int:
-        nums.sort()
-        
-        left, right = 0, len(nums) - 1
-        
-        count = 0
-        while left < right:
-            total = nums[left] + nums[right]
-            
-            if total == k:
-                count += 1
-                left += 1
-                right -= 1
-            elif total < k:
-                left += 1
-            else:
-                right -= 1
-        
-        return count
-