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Commit 03dbd1e

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Added Leetcode easy problems
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"""
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# CONTAINS DUPLICATE II
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Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.
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Example 1:
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Input: nums = [1,2,3,1], k = 3
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Output: true
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Example 2:
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Input: nums = [1,0,1,1], k = 1
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Output: true
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Example 3:
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Input: nums = [1,2,3,1,2,3], k = 2
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Output: false
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"""
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class Solution:
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def containsNearbyDuplicate(self, nums, k: int) -> bool:
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d={}
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for i,v in enumerate(nums):
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if v in d:
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if abs(d[v]-i)>k:
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d[v]=i
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else:
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return True
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else:
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d[v]=i
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return False
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"""
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# IMPLEMENT STACKS USING QUEUES
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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 queue (push, top, pop, and empty).
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Implement the MyStack class:
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void push(int x) Pushes element x to the top of the stack.
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int pop() Removes the element on the top of the stack and returns it.
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int top() Returns the element on the top of the stack.
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boolean empty() Returns true if the stack is empty, false otherwise.
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Notes:
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You must use only standard operations of a queue, which means only push to back, peek/pop from front, size, and is empty operations are valid.
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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.
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Example 1:
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Input
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["MyStack", "push", "push", "top", "pop", "empty"]
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[[], [1], [2], [], [], []]
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Output
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[null, null, null, 2, 2, false]
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Explanation
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MyStack myStack = new MyStack();
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myStack.push(1);
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myStack.push(2);
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myStack.top(); // return 2
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myStack.pop(); // return 2
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myStack.empty(); // return False
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Constraints:
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1 <= x <= 9
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At most 100 calls will be made to push, pop, top, and empty.
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All the calls to pop and top are valid.
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Follow-up: Can you implement the stack such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer. You can use more than two queues.
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"""
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class MyStack:
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def __init__(self):
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"""
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Initialize your data structure here.
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"""
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self.q = []
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self.q2 = []
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def push(self, x: int) -> None:
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"""
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Push element x onto stack.
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"""
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while self.q:
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self.q2.append(self.q.pop(0))
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self.q.append(x)
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while self.q2:
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self.q.append(self.q2.pop(0))
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def pop(self) -> int:
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"""
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Removes the element on top of the stack and returns that element.
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"""
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return self.q.pop(0)
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def top(self) -> int:
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"""
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Get the top element.
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"""
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return self.q[0]
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def empty(self) -> bool:
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"""
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Returns whether the stack is empty.
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"""
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return len(self.q) == 0
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# Your MyStack object will be instantiated and called as such:
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# obj = MyStack()
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# obj.push(x)
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# param_2 = obj.pop()
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# param_3 = obj.top()
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# param_4 = obj.empty()
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"""
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# INVERT A BINARY TREE
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Invert a binary tree.
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Example:
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Input:
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4
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- -
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2 7
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- - - -
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1 3 6 9
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Output:
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4
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- -
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7 2
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- - - -
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9 6 3 1
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Trivia:
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This problem was inspired by this original tweet by Max Howell:
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Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.
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"""
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# Definition for a binary tree node.
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class TreeNode:
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def __init__(self, val=0, left=None, right=None):
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self.val = val
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self.left = left
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self.right = right
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class Solution:
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def invertTree(self, root: TreeNode) -> TreeNode:
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if not root:
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return None
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if not root.left and not root.right:
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return root
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temp = self.invertTree(root.right)
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root.right = self.invertTree(root.left)
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root.left = temp
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return root

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