Category Archives: Aha moments

Coding is not a game for me, which is painful but along with implicit happiness.

Word Break

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Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

Note:

  • The same word in the dictionary may be reused multiple times in the segmentation.
  • You may assume the dictionary does not contain duplicate words.

Example 1:

Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
             Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false

Recently, I am going to pick up my leetcode skills.

This problem that I still remember It token me more than two days to consider, but this time it was aced in 5 minutes, as well as just in nearly 1 line core code.

It seems like practising is really useful!

from functools import lru_cache

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        @lru_cache(None)
        def dfs(s):
            return True if not s else any(dfs(s[len(word):]) for word in wordDict if s.startswith(word))
            
        return dfs(s)

My Calendar I

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Implement a MyCalendar class to store your events. A new event can be added if adding the event will not cause a double booking.

Your class will have the method, book(int start, int end). Formally, this represents a booking on the half open interval [start, end), the range of real numbers x such that start <= x < end.

A double booking happens when two events have some non-empty intersection (ie., there is some time that is common to both events.)

For each call to the method MyCalendar.book, return true if the event can be added to the calendar successfully without causing a double booking. Otherwise, return false and do not add the event to the calendar.Your class will be called like this: MyCalendar cal = new MyCalendar();MyCalendar.book(start, end)

Example 1:
MyCalendar();
MyCalendar.book(10, 20); // returns true
MyCalendar.book(15, 25); // returns false
MyCalendar.book(20, 30); // returns true
Explanation: 
The first event can be booked.  The second can't because time 15 is already booked by another event.
The third event can be booked, as the first event takes every time less than 20, but not including 20.

Note:

  • The number of calls to MyCalendar.book per test case will be at most 1000.
  • In calls to MyCalendar.book(start, end), start and end are integers in the range [0, 10^9].
import bisect


class MyCalendar:

    def __init__(self):
        self.ints = []

    def book(self, start: int, end: int) -> bool:
        idx = bisect.bisect_left(self.ints, (start, end))

        is_left_valid = idx == 0 or self.ints[idx - 1][1] <= start
        is_right_valid = idx == len(self.ints) or end <= self.ints[idx][0]

        if is_left_valid and is_right_valid:
            self.ints.insert(idx, (start, end))
            return True
        return False

Find the Shortest Superstring

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Given an array A of strings, find any smallest string that contains each string in A as a substring.

We may assume that no string in A is substring of another string in A.

Example 1:
Input: ["alex","loves","leetcode"]
Output: "alexlovesleetcode"
Explanation: All permutations of "alex","loves","leetcode" would also be accepted.

Example 2:
Input: ["catg","ctaagt","gcta","ttca","atgcatc"]
Output: "gctaagttcatgcatc"

Note:
1 <= A.length <= 12
1 <= A[i].length <= 20

Solution 1 (Naive and TLE):
At the very first, I did this solution that naively using DFS to search each possible solution, and finally we can pick up the shortest one. It works on small test set but got TLE on large sets.

from functools import lru_cache

class Solution:
    @lru_cache(None)
    def combinateStrings(self, str1, str2):
        for i in range(len(str2), -1, -1):
            if str1.endswith(str2[:i]):
                return str1 + str2[i:]

    def shortestSuperstring(self, A):

        def func(current_list, result, results):
            if not current_list:
                results.append(result)

            for s in current_list:
                current_list.remove(s)
                func(current_list, self.combinateStrings(result, s), results)
                func(current_list, self.combinateStrings(s, result), results)
                current_list.append(s)

        results = []

        func(A, "", results)

        return min(results, key= lambda x: len(x))