implement-stack-using-queues py3

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

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+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

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+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

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+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
+