One way to check that a vector is an eigenvector is to simply apply the matrix transformation and see if this is the same as multiplying by a scalar. A
class Solution: def countPrimes(self, n: int) -> int: if n < 3: return 0 primes = [0, 0] +  * (n - 2) for i in range(2, int(n ** 0.5) + 1): if primes[i]: primes[i * i: n: i] =  * len(primes[i * i: n: i]) return sum(primes)
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
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();
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.
- The number of calls to
MyCalendar.bookper test case will be at most
- In calls to
endare integers in the range
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] <= start is_right_valid = idx == len(self.ints) or end <= self.ints[idx] if is_left_valid and is_right_valid: self.ints.insert(idx, (start, end)) return True return False
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
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"
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))
Your are given an array of integers
prices, for which the
i-th element is the price of a given stock on day
i; and a non-negative integer
fee representing a transaction fee.
You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)
Return the maximum profit you can make.
Example 1: Input: prices = [1, 3, 2, 8, 4, 9], fee = 2 Output: 8 Explanation: The maximum profit can be achieved by: Buying at prices = 1Selling at prices = 8Buying at prices = 4Selling at prices = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
0 < prices.length <= 50000.
0 < prices[i] < 50000.
0 <= fee < 50000.
class Solution: def maxProfit(self, prices: List[int], fee: int) -> int: sell = 0 buy = -0x7777777 for price in prices: # Not buy or pay current price and fee buy = max(buy, sell - price - fee) # Not sell or get current price sell = max(sell, buy + price) return sell