[LeetCode] Largest Divisible Subset

Analysis:

we must sort the list first. According to the rule: if b%a == 0 c%b == 0, then c%a == 0
we can refer to the Longest Increasing Sequence and using a parents array to represent the subset

Time Complexity:

  • O(n^2)

Space Complexity：

  • O(n)

Code

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class Solution(object):
    def largestDivisibleSubset(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        n = len(nums)
        if n == 0:
            return []

        dp = [1 for i in xrange(n)]
        curnums = sorted(nums)
        parents = [-1 for i in xrange(n)]
        for i in xrange(1, n):
            for j in xrange(i):
                if curnums[i] % curnums[j] == 0:
                    if dp[i] < dp[j]+1:
                        dp[i] = dp[j]+1
                        parents[i] = j

        retlist = []
        maxidx, maxval = 0, dp[0]
        for i in xrange(n):
            if dp[i] > maxval:
                maxval = dp[i]
                maxidx = i

        while maxidx >= 0:
            retlist.append(curnums[maxidx])
            if parents[maxidx] == -1:
                break
            else:
                maxidx = parents[maxidx]
        return retlist

展开全文 >>

[LeetCode] Merge k Sorted Lists

Analysis:

define a heap to store the first node of lists, then pop the top node and push back the next node into heap, until the heap is empty

Time Complexity:

  • O(nlogn)

Space Complexity：

  • O(n)

Code

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class Solution(object):
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        if len(lists) == 0:
            return []
        heaplist = []
        for node in lists:
            if node:
                heaplist.append((node.val, node))

        heapq.heapify(heaplist)
        head = ListNode(-1)
        dummy = head
        while len(heaplist) > 0:
            topval, topnode = heapq.heappop(heaplist)
            dummy.next = topnode
            dummy = dummy.next

            if topnode.next:
                heapq.heappush(heaplist, (topnode.next.val, topnode.next))
        return head.next

展开全文 >>

[LeetCode] Remove Duplicate Letters

Analysis:

using a stack to keep the always increasing order of the results.
using a set to track the elements containing current answer result.
The key point is to ensure the order by comparing the current number with the tail number in the stack only if the counter of it is large than 1

Time Complexity:

  • O(n)

Space Complexity：

  • O(n)

Code

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class Solution(object):
    def removeDuplicateLetters(self, s):
        """
        :type s: str
        :rtype: str
        """
        store = []
        resultSet = set()

        count = collections.Counter(s)

        for n in s:
            count[n] -= 1
            if n in resultSet:
                continue
            while len(store) > 0 and store[-1] > n and count[store[-1]] > 0:
                resultSet.remove(store.pop())
            resultSet.add(n)
            store.append(n)

        return ''.join(store)

展开全文 >>

[LeetCode] Best Time to Buy and Sell Stock with Cooldown

Analysis:

we use sells to depict whether the maximum profit when selling stock at price[i]
and buys to depict the maximum profit when buying at prices[i]
so we can get
sells[i] = max(sells[i-1], buys[i-1]+prices[i])
buys[i] = max(buys[i-1], sells[i-2]-prices[i])

Time Complexity:

  • O(n)

Space Complexity：

  • O(n)

Code

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class Solution(object):
	def maxProfit(self, prices):
		"""
		:type prices: List[int]
		:rtype: int
		"""
		if len(prices) < 2:
			return 0

		sells = [0] * len(prices)
		buys = [0] * len(prices)

		sells[0], sells[1] = 0, max(0, prices[1]-prices[0])
		buys[0], buys[1] = -prices[0], max(-prices[0], -prices[1])

		for i in xrange(2, len(prices)):
			sells[i] = max(sells[i-1], buys[i-1]+prices[i])
			buys[i] = max(buys[i-1], sells[i-2]-prices[i])

		return sells[-1]


展开全文 >>

[LeetCode] Design Twitter

Analysis:

we can use a map with a set as value for followership. And a map with list as value for tweets storage. The important step is to use counter to determine the sequence of the tweets.

Code

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class Twitter(object):
    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.followship = collections.defaultdict(set)
        self.tweets = collections.defaultdict(list)
        self.cnt = 0

    def postTweet(self, userId, tweetId):
        """
        Compose a new tweet.
        :type userId: int
        :type tweetId: int
        :rtype: void
        """
        self.tweets[userId].append([tweetId, self.cnt])
        self.cnt += 1

    def getNewsFeed(self, userId):
        """
        Retrieve the 10 most recent tweet ids in the user's news feed. Each item in the news feed must be posted by users who the user followed or by the user herself. Tweets must be ordered from most recent to least recent.
        :type userId: int
        :rtype: List[int]
        """
        userlist = list(self.followship[userId]) + [userId]
        user_tweet_index = [[userId, len(self.tweets[userId])-1] for userId in userlist if userId in self.tweets]
        retlist = []

        for _ in xrange(10):
            max_index = max_tweet_id = max_user_id = -1
            for i, (user_id, tweet_index) in enumerate(user_tweet_index):
                if tweet_index >= 0:
                    tweet_info = self.tweets[user_id][tweet_index]
                    if tweet_info[1] > max_tweet_id:
                        max_index, max_user_id, max_tweet_id = i, user_id, tweet_info[1]
            if max_index == -1:
                break

            retlist.append(self.tweets[max_user_id][user_tweet_index[max_index][1]][0])
            user_tweet_index[max_index][1] -= 1
        return retlist

    def follow(self, followerId, followeeId):
        """
        Follower follows a followee. If the operation is invalid, it should be a no-op.
        :type followerId: int
        :type followeeId: int
        :rtype: void
        """
        if followeeId == followerId: return
        self.followship[followerId].add(followeeId)

    def unfollow(self, followerId, followeeId):
        """
        Follower unfollows a followee. If the operation is invalid, it should be a no-op.
        :type followerId: int
        :type followeeId: int
        :rtype: void
        """
        if followerId in self.followship and followeeId in self.followship[followerId]:
            self.followship[followerId].remove(followeeId)

展开全文 >>

[LeetCode] Russian Doll Envelopes

Analysis:

Another form of Longest Increasing Sequence.

Time Complexity:

  • O(nlogn)

Space Complexity：

  • O(n)

Code

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class Solution {
public:
    int maxEnvelopes(vector<pair<int, int>>& envelopes) {
    	auto cmp = [](pair<int, int> a, pair<int, int> b) {return a.first < b.first;};

    	std::sort(envelopes.begin(), envelopes.end(), cmp);

    	int maxnum = 0;
    	vector<int> dp (envelopes.size(), 1);
    	for (int i = 1; i < envelopes.size();i++) {
    		for (int j = 0; j < i; j++) {
				pair<int, int> prev = envelopes[j];
				pair<int, int> cur = envelopes[i];

				if (prev.first < cur.first && prev.second < cur.second) {
					dp[i] = max(dp[i], 1+dp[j]);
				}
    		}
    	}

    	for (int p : dp) {
    		maxnum = max(maxnum, p);
    	}
    	return maxnum;

    }
};

展开全文 >>

[LeetCode] Count of Smaller Numbers After Self

Analysis:

Use custom TreeNode with leftcnt and cnt attributes. leftcnt depict the numbers smaller than the current node. cnt represent the nodes has the same value before current node

Time Complexity:

  • O(n)

Space Complexity：

  • O(n)

Code

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class TreeNode(object):
	def __init__(self, val):
		self.left = None
		self.right = None
		self.leftcnt = 0
		self.cnt = 1
		self.val = val

class BinaryTree(object):
	def __init__(self):
		self.root = None

	def insert(self, val):
		if self.root is None:
			self.root = TreeNode(val)
			return 0
		root = self.root
		cnt = 0

		while root:
			if val < root.val:
				root.leftcnt += 1
				if root.left is None:
					root.left = TreeNode(val)
					break
				root = root.left
			elif val > root.val:
				cnt += root.leftcnt + root.cnt
				if root.right is None:
					root.right = TreeNode(val)
					break
				root = root.right
			else:
				cnt += root.leftcnt
				root.cnt += 1
				break
		return cnt


class Solution(object):
	def countSmaller(self, nums):
		"""
		:type nums: List[int]
		:rtype: List[int]
		"""
		retlist = [0] * len(nums)
		tree = BinaryTree()
		for i in xrange(len(nums)-1, -1, -1):
			retlist[i] = tree.insert(nums[i])
		return retlist

展开全文 >>

[Repost] Radix Sort

The lower bound for Comparison based sorting algorithm (Merge Sort, Heap Sort, Quick-Sort .. etc) is Ω(nLogn), i.e., they cannot do better than nLogn.

Counting sort is a linear time sorting algorithm that sort in O(n+k) time when elements are in range from 1 to k.

What if the elements are in range from 1 to n2?

We can’t use counting sort because counting sort will take O(n2) which is worse than comparison based sorting algorithms. Can we sort such an array in linear time?
Radix Sort is the answer. The idea of Radix Sort is to do digit by digit sort starting from least significant digit to most significant digit. Radix sort uses counting sort as a subroutine to sort.

The Radix Sort Algorithm

1) Do following for each digit i where i varies from least significant digit to the most significant digit.
………….a) Sort input array using counting sort (or any stable sort) according to the i’th digit.

Example:
Original, unsorted list:

170, 45, 75, 90, 802, 24, 2, 66

Sorting by least significant digit (1s place) gives: [*Notice that we keep 802 before 2, because 802 occurred before 2 in the original list, and similarly for pairs 170 & 90 and 45 & 75.]

170, 90, 802, 2, 24, 45, 75, 66

Sorting by next digit (10s place) gives: [*Notice that 802 again comes before 2 as 802 comes before 2 in the previous list.]

802, 2, 24, 45, 66, 170, 75, 90

Sorting by most significant digit (100s place) gives:

2, 24, 45, 66, 75, 90, 170, 802

What is the running time of Radix Sort?

Let there be d digits in input integers. Radix Sort takes O(d(n+b)) time where b is the base for representing numbers, for example, for decimal system, b is 10. What is the value of d? If k is the maximum possible value, then d would be O(logb(k)). So overall time complexity is O((n+b) logb(k)). Which looks more than the time complexity of comparison based sorting algorithms for a large k. Let us first limit k. Let k <= nc where c is a constant. In that case, the complexity becomes O(nLogb(n)). But it still doesn’t beat comparison based sorting algorithms.
What if we make value of b larger?. What should be the value of b to make the time complexity linear? If we set b as n, we get the time complexity as O(n). In other words, we can sort an array of integers with range from 1 to nc if the numbers are represented in base n (or every digit takes log2(n) bits).

Is Radix Sort preferable to Comparison based sorting algorithms like Quick-Sort?

If we have log2n bits for every digit, the running time of Radix appears to be better than Quick Sort for a wide range of input numbers. The constant factors hidden in asymptotic notation are higher for Radix Sort and Quick-Sort uses hardware caches more effectively. Also, Radix sort uses counting sort as a subroutine and counting sort takes extra space to sort numbers.

Code

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// C++ implementation of Radix Sort
#include<iostream>
using namespace std;
 
// A utility function to get maximum value in arr[]
int getMax(int arr[], int n)
{
    int mx = arr[0];
    for (int i = 1; i < n; i++)
        if (arr[i] > mx)
            mx = arr[i];
    return mx;
}
 
// A function to do counting sort of arr[] according to
// the digit represented by exp.
void countSort(int arr[], int n, int exp)
{
    int output[n]; // output array
    int i, count[10] = {0};
 
    // Store count of occurrences in count[]
    for (i = 0; i < n; i++)
        count[ (arr[i]/exp)%10 ]++;
 
    // Change count[i] so that count[i] now contains actual
    //  position of this digit in output[]
    for (i = 1; i < 10; i++)
        count[i] += count[i - 1];
 
    // Build the output array
    for (i = n - 1; i >= 0; i--)
    {
        output[count[ (arr[i]/exp)%10 ] - 1] = arr[i];
        count[ (arr[i]/exp)%10 ]--;
    }
 
    // Copy the output array to arr[], so that arr[] now
    // contains sorted numbers according to current digit
    for (i = 0; i < n; i++)
        arr[i] = output[i];
}
 
// The main function to that sorts arr[] of size n using 
// Radix Sort
void radixsort(int arr[], int n)
{
    // Find the maximum number to know number of digits
    int m = getMax(arr, n);
 
    // Do counting sort for every digit. Note that instead
    // of passing digit number, exp is passed. exp is 10^i
    // where i is current digit number
    for (int exp = 1; m/exp > 0; exp *= 10)
        countSort(arr, n, exp);
}
 
// A utility function to print an array
void print(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
 
// Driver program to test above functions
int main()
{
    int arr[] = {170, 45, 75, 90, 802, 24, 2, 66};
    int n = sizeof(arr)/sizeof(arr[0]);
    radixsort(arr, n);
    print(arr, n);
    return 0;
}

展开全文 >>

[LeetCode] Construct Binary Tree from Inorder and Postorder Traversal

Analysis:

find the root node and divide them into two parts, and build recursively into subtrees
the difference is that root is always the last one

Time Complexity:

  • O(n)

Space Complexity：

  • O(n)

Code

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

class Solution(object):

    def build(self, instart, inend, poststart, postend):
        if poststart >= postend:
            return None

        root = TreeNode(self.postorder[postend-1])
        idx = self.inorder[instart:inend+1].index(root.val)
        root.left = self.build(instart, instart+idx, poststart, poststart+idx)
        root.right = self.build(instart+idx+1, inend, poststart+idx, postend-1)
        return root

    def buildTree(self, inorder, postorder):
        """
        :type inorder: List[int]
        :type postorder: List[int]
        :rtype: TreeNode
        """
        if len(inorder) == 0:
            return None
        self.inorder = inorder
        self.postorder = postorder
        return self.build(0, len(inorder), 0, len(postorder))

展开全文 >>

[LeetCode] Construct Binary Tree from Preorder and Inorder Traversal

Analysis:

find the root node and divide them into two parts, and build recursively into subtrees

Time Complexity:

  • O(n)

Space Complexity：

  • O(n)

Code

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

class Solution(object):
	def build(self, prestart, preend, instart,inend):
		if prestart >= preend:
			return None
		root = TreeNode(self.preorder[prestart])
		idx = self.inorder[instart:inend+1].index(root.val)
		root.left = self.build(prestart+1, prestart+idx+1, instart, instart+idx)
		root.right = self.build(prestart+idx+1, preend, instart+idx+1, inend)
		return root


	def buildTree(self, preorder, inorder):
		"""
		:type preorder: List[int]
		:type inorder: List[int]
		:rtype: TreeNode
		"""
		if (len(preorder) == 0):
			return None
		self.preorder = preorder
		self.inorder = inorder
		return self.build(0, len(preorder), 0, len(inorder))

展开全文 >>

tag:

    缺失模块。
    1、请确保node版本大于6.2
    2、在博客根目录（注意不是yilia根目录）执行以下命令：
    npm i hexo-generator-json-content --save

    3、在根目录_config.yml里添加配置： 

      jsonContent:
    meta: false
    pages: false
    posts:
      title: true
      date: true
      path: true
      text: false
      raw: false
      content: false
      slug: false
      updated: false
      comments: false
      link: false
      permalink: false
      excerpt: false
      categories: false
      tags: true

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