检查一个字符串是否与其他两个字符串交织

Question

I came across multiple links to the solution of the problem - "How to check if a string is interleaving of two other strings"

Two solutions looked particularly interesting to me which work but I have doubts in both of them.

FIRST I did not get the hashing part in this where author is saying "A pure recursive solution will cause time limit exceed. We can optimize it by caching the false visited solutions in the visited set. That will short circuit many repeated search path"

SECOND I did not the the "else condition" on line 18 in recursive. Won't one of the conditions (line 14th and line 16th) will always be true as they are inside else of line 11th if condition which is if(s2.charAt(0) != s3.charAt(0) && s1.charAt(0) != s3.charAt(0)) {

Answer 1

First

This is actually space-time tradeoff (the computation time can be reduced at the cost of increased memory use). Why does the author say pure recursive solution slow (in fact it's exponential time complexity)? It comes from repeated recursion and because of that, it computes the same values again and again.

So what can you do? You can store the value you already computed. Next time you want this value again, just look up in a table. This is called caching, when the values are cached, you can treat every recursive call inside the function as it would run in O(1) time complexity. The core idea is don't calculate the same things twice.

Second

In the case s2.charAt(0) == s3.charAt(0) && s1.charAt(0) == s3.charAt(0) .