The interleaving is s1 + t1 + s2 + t2 + s3 + t3 + ... or t1 + s1 + t2 + s2 + t3 + s3 + ...

Note: a + b is the concatenation of strings a and b.

Company Tags: Amazon, Google

Core Concept: Dynamic Programming, Recursion, Memoization

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68:T452, - This is a 2D dynamic programming problem. - First, a quick check: if `len(s1) + len(s2) != len(s3)`, it's impossible, so return `false`. - Let `dp[i][j]` be a boolean indicating whether the first `i+j` characters of `s3` can be formed by an interleaving of the first `i` characters of `s1` and the first `j` characters of `s2`. - **Recurrence Relation**: To determine `dp[i][j]`, we look at the character `s3[i+j-1]`. This character must match either the last character of the current `s1` prefix (`s1[i-1]`) or the last character of the `s2` prefix (`s2[j-1]`). - If `s3[i+j-1]` matches `s1[i-1]`, then `dp[i][j]` can be true if `dp[i-1][j]` was true. - If `s3[i+j-1]` matches `s2[j-1]`, then `dp[i][j]` can be true if `dp[i][j-1]` was true. - **Base Case**: `dp[0][0]` is true (two empty strings form an empty string). - **DP Table Fill**: You can fill a 2D DP table of size `(len(s1)+1) x (len(s2)+1)`. - The final answer is the value at `dp[len(s1)][len(s2)]`. - This can also be solved with a recursive DFS with memoization, where the state is `(i, j)`, the current pointers into `s1` and `s2`.69:T4f2,class TrieNode: def __init__(self): self.children = {} self.isWord = False self.word = None def addWord(self, word): cur = self for c in word: if c not in cur.children: cur.children[c] = TrieNode() cur = cur.children[c] cur.isWord = True

Meeting Rooms II (#253)

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