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  • Leetcode: Best time to buy and sell stock

    The task is to find out of a series of stock prices the best time to buy and sell the stock. Only one stock can be owned at the time and it can be sold and bought at the same time.

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  • Leetcode: Repeated substring pattern

    The task is to find out if a given string is comprised of multiple identical substrings. The first observation is that the substring cannot be longer than half of the length of the original string. The second observation is that the length of the total string has to be a multiple of the string that is tested, if it is not it can be discarded right away. Lastly, a string with the length of one can be discarded immediately.

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  • Leetcode: String in String

    A classic find the needle in the haystack problem. The task is to find if a string is contained within another string and return the index of the first position. If the string is not contained return -1 and if the needle is empty, return 0.

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  • Leetcode: Merge two lists

    class Solution:
      def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        root = ListNode()
        current = root
        while list1 is not None and list2 is not None:
          if list1.val < list2.val:
   = ListNode(list1.val)
            current =
            list1 =
   = ListNode(list2.val)
            current =
            list2 =
        if list1 is not None:
 = list1
        if list2 is not None:
 = list2


    Runtime: 34 ms, faster than 97.81% of Python3 online submissions for Merge Two Sorted Lists. Memory Usage: 13.9 MB, less than 79.56% of Python3 online submissions for Merge Two Sorted Lists.

  • Leetcode: All possible subsets

    A subset of a list with the elements {a,b,c} is {a,b}. The task is to find all possible subsets. Sets in programming languages are usually not ordered. Therefore, the set {a,b,c} is equivalent to {b,c,a}. The idea for the solution is a nested for-loop that iterates over the results again and again, starting with one empty ([]) result.

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