Tag Archives: Leetcode

Wiggle Sort II

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Given an unsorted array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3]....

Example 1:

Input: nums = [1, 5, 1, 1, 6, 4] 
Output: One possible answer is [1, 4, 1, 5, 1, 6].

Example 2:

Input: nums = [1, 3, 2, 2, 3, 1] 
Output: One possible answer is [2, 3, 1, 3, 1, 2].

Note:
You may assume all input has valid answer.

Follow Up:
Can you do it in O(n) time and/or in-place with O(1) extra space?

class Solution:
    def wiggleSort(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        nums.sort()
        half = len(nums[::2])
        nums[::2], nums[1::2] = nums[:half][::-1], nums[half:][::-1]

Increasing Triplet Subsequence

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Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.

Formally the function should:

Return true if there exists i, j, k
such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.

Note: Your algorithm should run in O(n) time complexity and O(1) space complexity.

Example 1:

Input: [1,2,3,4,5] 
Output: true

Example 2:

Input: [5,4,3,2,1] 
Output: false

Since this problem is literally similar another one (longest-increasing-subsequence), I have tried the traditional dynamic programming at very first. However, this question is not as easy as it looks be, I got a Time Limit Exceeded.

After that, I noticed that we actually can iterate the entire array by one pass with maintaining two pointers.

class Solution:
    def increasingTriplet(self, nums: List[int]) -> bool:
#         # TLE
#         if not nums:
#             return False
        
#         dp = [1] * len(nums)
        
#         for i in range(1, len(nums)):
#             for j in range(i):
#                 if nums[i] > nums[j]:
#                     dp[i] = max(dp[i], dp[j] + 1)
#                 if dp[i] == 3:
#                     return True
        
#         return False
    
        first = second = float('inf')
        
        for n in nums:
            if n <= first:
                first = n
                
            elif n <= second:
                second = n
            else:
                return True
            
        return False

Flip Equivalent Binary Trees

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For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees.

A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.

Write a function that determines whether two binary trees are flip equivalent.  The trees are given by root nodes root1 and root2.

Flipped Trees Diagram

Example 1:

Input: 
root1 = [1,2,3,4,5,6,null,null,null,7,8], 
root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7] 
Output: true 
Explanation: We flipped at nodes with values 1, 3, and 5.

Note:

  1. Each tree will have at most 100 nodes.
  2. Each value in each tree will be a unique integer in the range [0, 99].
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def flipEquiv(self, root1: TreeNode, root2: TreeNode) -> bool:
        if root1 == root2 == None:
            return True
        
        elif root1 and root2 and root1.val == root2.val:
            return (self.flipEquiv(root1.left, root2.left) and self.flipEquiv(root1.right, root2.right)) or (self.flipEquiv(root1.right, root2.left) and self.flipEquiv(root1.left, root2.right))
        
        else:
            return False

Remove Zero Sum Consecutive Nodes from Linked List

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Firstly, I can’t help myself to say: “Lee, you are a fucking genius!”

Given the head of a linked list, we repeatedly delete consecutive sequences of nodes that sum to 0 until there are no such sequences.

After doing so, return the head of the final linked list.  You may return any such answer.

(Note that in the examples below, all sequences are serializations of ListNode objects.)

Example 1:

Input: head = [1,2,-3,3,1]
Output: [3,1]
Note: The answer [1,2,1] would also be accepted.

Example 2:

Input: head = [1,2,3,-3,4]
Output: [1,2,4]

Example 3:

Input: head = [1,2,3,-3,-2]
Output: [1]
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeZeroSumSublists(self, head: ListNode) -> ListNode:
        cur = dummy = ListNode(0)
        dummy.next = head
        
        prefix = 0
        seen = collections.OrderedDict()
        
        while cur:
            prefix += cur.val
            node = seen.get(prefix, cur)
            while prefix in seen:
                seen.popitem()
            seen[prefix] = node
            node.next = cur = cur.next
            
        return dummy.next

Printing a pyramid matrix

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How to print a pyramid matrix like that:

n = 2
[1, 1, 1]
[1, 2, 1]
[1, 1, 1]

n = 3
[1, 1, 1, 1]
[1, 2, 2, 1]
[1, 2, 2, 1]
[1, 1, 1, 1]

n = 4
[1, 1, 1, 1, 1]
[1, 2, 2, 2, 1]
[1, 1, 3, 2, 1]
[1, 2, 2, 2, 1]
[1, 1, 1, 1, 1]
def func(N):
    N += 1
    matrix = [[1 for _ in range(N)] for _ in range(N)]
    cnt = 0

    while cnt < N:
        # UP
        for i in range(cnt, N - cnt - 1):
            matrix[cnt][i] = cnt + 1

        # RIGHT
        for i in range(cnt, N - cnt - 1):
            matrix[i][N - cnt - 1] = cnt + 1

        # DOWN
        for i in range(N - cnt - 1, cnt, -1):
            matrix[N - cnt - 1][i] = cnt + 1

        # LEFT
        for i in range(N - cnt, cnt, -1):
            matrix[N - cnt - 1][cnt] = cnt + 1

        cnt += 1

    return matrix


if __name__ == "__main__":
    matrix = func(N=4)

    for line in matrix:
        print(line)

Path In Zigzag Labelled Binary Tree

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In an infinite binary tree where every node has two children, the nodes are labelled in row order.

In the odd numbered rows (ie., the first, third, fifth,…), the labelling is left to right, while in the even numbered rows (second, fourth, sixth,…), the labelling is right to left.

Given the label of a node in this tree, return the labels in the path from the root of the tree to the node with that label.

Example 1:

Input: label = 14
Output: [1,3,4,14]

Example 2:

Input: label = 26
Output: [1,2,6,10,26]
from math import log

class Solution:
    def pathInZigZagTree(self, label: int) -> List[int]:
        ans = [label]
        
        def findParent(label):
            level =  math.floor(math.log(label,2))            
            diff = math.ceil((label - 2**level + 1) / 2)       
            parent = 2**(level) - diff
            return parent
            
        while label != 1:
            label = findParent(label)
            ans.append(label)
            
        return ans[::-1]

Remove Duplicates from Sorted Array

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Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.











Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.





Example 1:





Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn't matter what you leave beyond the returned length.





Example 2:





Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.





class Solution:
    def removeDuplicates(self, nums: List[int]):
        if not nums:
            return 0

        newTail = 0

        for i in range(1, len(nums)):
            if nums[i] != nums[newTail]:
                newTail += 1
                nums[newTail] = nums[i]

        return newTail + 1

					

		
		
	


				
					
	

		

						
Permutations II

			
						

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Given a collection of numbers that might contain duplicates, return all possible unique permutations.





Example:





Input: [1,1,2]
Output:
[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]





class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        self.result = []

        def perm(data, begin, end):
            if begin == end:
                self.result += [[item for item in data]]
            else:
                i = begin
                for j in range(begin, end):
                    data[i], data[j] = data[j], data[i]
                    perm(data, begin + 1, end)
                    data[i], data[j] = data[j], data[i]
                    
        perm(nums, 0, len(nums))

        return self.result

					

		
		
	


				
					
	

		

						
Permutation Sequence

			
						

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The set [1,2,3,...,n] contains a total of n! unique permutations.





By listing and labeling all of the permutations in order, we get the following sequence for n = 3:





  1. "123"
  2. "132"
  3. "213"
  4. "231"
  5. "312"
  6. "321"





Given n and k, return the kth permutation sequence.





Note:





  • Given n will be between 1 and 9 inclusive.
  • Given k will be between 1 and n! inclusive.





Example 1:





Input: n = 3, k = 3
Output: "213"






Example 2:





Input: n = 4, k = 9
Output: "2314"





class Solution:
    def getPermutation(self, n: int, k: int) -> str:
        res, nums = "", list(range(1, n + 1))

        k -= 1

        for n in range(n, 0, -1):
            index, k = divmod(k, math.factorial(n-1))
            res += str(nums.pop(index))

        return res

					

		
		
	


				
					
	

		

						
Lowest Common Ancestor with N nodes

			
						

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Similar with finding lowest common ancestor of two nodes, we can extend this function to N nodes.





The method buildTree in the solution is in order to building a tree instantly by given strings.





# Definition for a binary tree node.
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None


class Solution:
    def buildTree(self, inorder, postorder):
        if not inorder or not postorder:
            return None

        while postorder:
            if postorder[-1] not in inorder:
                postorder.pop()
            else:
                break

        if len(postorder) == 1:
            return TreeNode(postorder[0])

        else:
            node = TreeNode(postorder[-1])
            in_index = inorder.index(postorder[-1])
            node.left = self.buildTree(inorder[:in_index], postorder[:-1])
            node.right = self.buildTree(inorder[in_index + 1:], postorder[:-1])
            return node

    def find(self, root, nodes):
        if not root:
            return None
        else:
            if root.val in nodes:
                return root

            left = self.find(root.left, nodes)
            right = self.find(root.right, nodes)

            if left and right:
                return root

            return right if not left else left


if __name__ == "__main__":
    obj = Solution()
    root = obj.buildTree([6, 5, 7, 2, 4, 3, 0, 1, 8], [6, 7, 4, 2, 5, 0, 8, 1, 3])
    nodes = [0, 1, 8]
    result = obj.find(root, nodes)
    print(result.val)