Vinit Khandelwal

  Approach #1: Insert Into Sorted Structure [Accepted] Intuition Let's add flowers in the order they bloom. When each flower blooms, we check it's neighbors to see if they can satisfy the condition with the current flower. Algorithm We'll maintain  active , a sorted data structure containing every flower that has currently bloomed. When we add a flower to  active , we should check it's lower and higher neighbors. If some neighbor satisfies the condition, we know the condition occurred first on this day. Complexity Analysis Time Complexity (Java):  O(N \log N) O ( N lo g N ) , where  N N  is the length of  flowers . Every insertion and search is  O(\log N) O ( lo g N ) . Time Complexity (Python):  O(N^2) O ( N 2 ) . As above, except  list.insert  is  O(N) O ( N ) . Space Complexity:  O(N) O ( N ) , the size of  active . Approach #2: Min Queue [Accepted] Intuition For each contiguous block ("window") of  k  po...

  Solution This article assumes you already have some confidence with  graph algorithms , such as  breadth-first search  and  depth-first searching . If you're familiar with those, but not with  topological sort  (the topic tag for this problem), don't panic, as you should still be able to make sense of it. It is one of the many more advanced algorithms that keen programmers tend to "invent" themselves before realizing it's already a widely known and used algorithm. There are a couple of approaches to topological sort;  Kahn's Algorithm  and DFS. A few things to keep in mind: The letters  within a word  don't tell us anything about the relative order. For example, the presence of the word  kitten  in the list does  not  tell us that the letter  k  is before the letter  i . The input can contain words followed by their prefix, for example,  abcd  and then  ab . These cases will never ...