[Leetcode] 388. Longest Absolute File Path

Suppose we abstract our file system by a string in the following manner:

The string "dir\n\tsubdir1\n\tsubdir2\n\t\tfile.ext" represents:

dir
    subdir1
    subdir2
        file.ext

The directory dir contains an empty sub-directory subdir1 and a sub-directory subdir2 containing a file file.ext.

The string "dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext" represents:

dir
    subdir1
        file1.ext
        subsubdir1
    subdir2
        subsubdir2
            file2.ext

The directory dir contains two sub-directories subdir1 and subdir2. subdir1 contains a file file1.ext and an empty second-level sub-directory subsubdir1. subdir2 contains a second-level sub-directory subsubdir2 containing a file file2.ext.

We are interested in finding the longest (number of characters) absolute path to a file within our file system. For example, in the second example above, the longest absolute path is "dir/subdir2/subsubdir2/file2.ext", and its length is 32 (not including the double quotes).

Given a string representing the file system in the above format, return the length of the longest absolute path to file in the abstracted file system. If there is no file in the system, return 0.

Note:

• The name of a file contains at least a . and an extension.
• The name of a directory or sub-directory will not contain a ..

Time complexity required: O(n) where n is the size of the input string.

Notice that a/aa/aaa/file1.txt is not the longest file path, if there is another path aaaaaaaaaaaaaaaaaaaaa/sth.png.

---

I solved this problem about 7 months ago - when I first started to tackle these Leetcode problems. Back then, I wasn't proficient in any algorithms. But somehow, I managed to solve this problem with an extremely messy code. Ever since I got that job offer, I thought I should re-do some of the problems or focus those problems that require some modeling (i.e., Google problems).

This problem is on the easier side of this kind. It is very obvious files and folders are organized in tree structure. Only if we have a well-organized tree structure, then we can basically use backtracking to find out the answers. However, the input is a string. But, by looking at the string, we can see that the number of '\t' in the element is the same as the level of that elements. For example, '\tsubdir1', '\t\tfile.txt' are in the first, second level of the file organization, respectively.

We can't easily do a tree traverse on this type of string input, but hey, the elements order after input.split('\n') is similar to the  in-order traversal of the directory tree (here might be a forest).

Therefore, the abstraction of the problem will be:

Given the in-order traversal of a tree and each element's level, reconstruct the maximum length path from root to leaf where we only considers those leaves that contain '.' (file). 

Well, then we just use a stack, when the level is higher than the current level, meaning that the incoming element will be the children of current element; when lower, meaning that the incoming one is the parent; the same, siblings.

class Solution(object):
    def lengthLongestPath(self, input):
        """
        :type input: str
        :rtype: int
        """
        stack = []
        current_level = 0
        res = 0
        for name in input.split('\n'):
            #print stack, current_level
            tabs = name.split('\t')
            if len(tabs) - 1 == current_level:
                if stack:
                    stack.pop()
                stack.append(tabs[-1])
            elif len(tabs) -1 > current_level:
                stack.append(tabs[-1])
            else:
                for _ in range(current_level - len(tabs) + 2):
                    if stack: stack.pop()
                stack.append(tabs[-1])
            if '.' in tabs[-1]:
                res = max(res, len('/'.join(stack)))
            current_level = len(tabs) - 1
        return res

[Update] The end of a period

I got a job offer in a small company in town - it matches my expectation in almost every aspect: meaningful domain, the opportunity to learn and improve, and nice pay.

The interview did not use any of the stuff I have prepared for the last few month - days and nights going through, sometimes painfully, all these Leetcode problems. But I am glad I did it. By doing these algorithmic problems, I have gained a much deeper understanding of common algorithms, been more appreciative of good and efficient code, and cultivated a habit to endure some hard and lonely moments in life.

My progress in Leetcode - I actually never have imagined that I could have done so many problems.

I like doing these problems, so I will likely keep updating some interesting problems I have solved, but to a less intensity. This might be even better, as now I am in a much more relaxing mood - I might be able to think deeper than just finishing as much problems as possible.

I look forward to this new chapter of my life

[Leetcode] 334. Increasing Triplet Subsequence

This problem reminds me of the fancy version of longest increasing subsequence problem ($O(n\log n)$ solution).

Let's look at an example array: 1, 2, -1, 1, -2, 3. When we go through the first two element, we are all good. Then, we encounter -1, here we face a decision, should we update the current minimum to -1?

Let's see why we should and why we shouldn't. -1 is a safer move than 1, what if later our solution is -1, 0, 1? Then we definitely want to update it. However, you might say, if we update it to -1, then 2 shouldn't be in our solution anymore, according to the "subsequence" rule.

Well, the answer is YES, but we only update the minimum, keeping the second minimum (which is 2 here).The argument is as following: when we keep the second minimum (call it min2), this means that we only count increasing when the next value is greater than min2. You see, by changing min1 to a smaller value, it doesn't really affect the decision we are going to make about adding an increasing element. However, if we see the next value is smaller than min2 but greater than min1, we then want to update min2.

In this way, we always keep track the safest possible value yet maintain the order we have for now.

In actual implementation, I used a stack instead of min1, min2. It turned out to be much easier to handle. Well, who knows such a seeming simple question took me good two hours to think through - this kind of question is very tricky.

---

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.

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

Examples:
Given [1, 2, 3, 4, 5],
return true.

Given [5, 4, 3, 2, 1],
return false.

---

class Solution(object):
    def increasingTriplet(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        if len(nums) < 3: return False
        
        counter = 1
        stack = []
        for i in nums:
            if len(stack) == 0:
                stack.append(i)
            elif i < stack[0]:
                stack[0] = i
            elif len(stack) == 1 and i > stack[0]:
                stack.append(i)
            elif len(stack) == 2 and stack[0] < i < stack[1]:
                stack[1] = i
            elif len(stack) == 2 and i > stack[1]:
                return True
        return False

[Leetcode] 164. Maximum Gap

A really smart solution using pigeon hole principle and bucket sort...

---

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.

Try to solve it in linear time/space.

Return 0 if the array contains less than 2 elements.

You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.

---

class Solution(object):
    def maximumGap(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums: return 0
        min_v, max_v = min(nums), max(nums)
        if min_v == max_v: return 0
        
        buckets = [[] for _ in range(len(nums)+1)]
        for num in nums:
            bkt = (num - min_v) * len(nums) / (max_v - min_v)
            buckets[bkt].append(num)
        
        i = 0
        res = 0
        while i < len(buckets) - 1:
            k = i + 1
            while len(buckets[k]) == 0:
                k = k + 1
            res = max(res, min(buckets[k]) - max(buckets[i]))
            i = k
        return res

[MOOC] Implement TrieMatching

This is the third problem in assignment 1 for this class. It has been a long time since I worked on the first problem. I was confused why we have to model the trie as the graph where node is the number, edge the alphabet. Why can't we model it where node is the alphabet?

An counter-example will be: Consider a pattern 'ATAC', and we want to see if it matches to 'ATAT'. If we model the pattern as the first graph representation, it will be {'A': ['T', 'C'], 'T': ['A'], 'C': ['$']}. Then we will say 'ATAT' matches 'ATAC', which is obviously wrong.

---

Here is the code.

def solve (text, n, patterns):

    result = []
    # write your code here
    trie = build_trie(patterns)
    for i in range(len(text)):
        if match_pattern(text[i:], trie):
            result.append(i)
    return result

def build_trie(patterns):
    """Build trie for patterns"""
    trie = {0: {}}
    counter = 1
    for p in patterns:
        p = p + '$'
        current_node = 0
        for c in p:
            if c in trie[current_node]:
                current_node = trie[current_node][c]
            else:
                trie[current_node][c] = counter
                trie[counter] = {}
                current_node = counter
                counter += 1
    return trie

def match_pattern(text, trie):
    cur_node = 0
    for ch in text:
        if ch in trie[cur_node]:
            cur_node = trie[cur_node][ch]
            if '$' in trie[cur_node]:
                return True
        else:
            return False
    if '$' in trie[cur_node]:
        return True
    else:
        return False

[Leetcode] Sudoku Solver

Okay, this is a pretty classical backtracking problem. There are some details of writing recursion in python that I can't say I fully understand.

---

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

A sudoku puzzle...

---

I referred to this post to finally get my program running. One thing I did not do is to "return True/False", but handled it like permutation that kind of questions. I do think "return True/False" make sense, but I don't know why I did not think it necessary.

class Solution(object):
    def solveSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: void Do not return anything, modify board in-place instead.
        """
        
        self.solve(board, 0, 0)
    
    def solve(self, board, i, j):
        
        def candidates(i, j):
            not_available = []
            for x in range(9):
                if board[i][x] != '.':
                    not_available.append(board[i][x])
                if board[x][j] != '.':
                    not_available.append(board[x][j])
            grid_x, grid_y = i / 3, j / 3
            for x in (0, 1, 2):
                for y in (0, 1, 2):
                    if board[grid_x * 3 + x][grid_y * 3 + y] != '.':
                        not_available.append(board[grid_x * 3 + x][grid_y * 3 + y])
            return [i for i in '123456789' if i not in not_available]
        
        if i == 9: 
            return True
        elif j == 9:
            return self.solve(board, i+1, 0)
        else:            
            if board[i][j] == '.':
                cands = candidates(i, j)
                if not cands:
                    return False
                for cand in cands:
                    board[i][j] = cand
                    if self.solve(board, i, j + 1):
                        return True
                    board[i][j] = '.'
                return False
            else:
                return self.solve(board, i, j+1)

[Leetcode] Smallest Range

This is a smart use of heap.

You have k lists of sorted integers in ascending order. Find the smallest range that includes at least one number from each of the klists.

We define the range [a,b] is smaller than range [c,d] if b-a < d-c or a < c if b-a == d-c.

Example 1:

Input:[[4,10,15,24,26], [0,9,12,20], [5,18,22,30]]
Output: [20,24]
Explanation: 
List 1: [4, 10, 15, 24,26], 24 is in range [20,24].
List 2: [0, 9, 12, 20], 20 is in range [20,24].
List 3: [5, 18, 22, 30], 22 is in range [20,24].

Note:

1. The given list may contain duplicates, so ascending order means >= here.
2. 1 <= k <= 3500
3. -105 <= value of elements <= 105.

class Solution(object):
    def smallestRange(self, nums):
        """
        :type nums: List[List[int]]
        :rtype: List[int]
        """
        h = []
        max_v = -10**5 - 1
        min_v = 10**5 + 1
        for i, num in enumerate(nums):
            h.append([num[0], i, 0])
            if max_v < num[0]: max_v = num[0]
            if min_v > num[0]: min_v = num[0]
        res = [min_v, max_v]
        
        heapq.heapify(h)
        
        while True:
            v, list_index, element_index = heapq.heappop(h)
            if element_index == len(nums[list_index]) - 1:
                return res
            heapq.heappush(h, [nums[list_index][element_index+1], list_index, element_index+1]) 
            if nums[list_index][element_index+1] > max_v:
                max_v = nums[list_index][element_index+1]
            if max_v - h[0][0] < res[1] - res[0]:
                res = [h[0][0], max_v]