leetcode

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = “great”:

great

/ \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node “gr” and swap its two children, it produces a scrambled string “rgeat”.

rgeat

/ \ rg eat / \ / \ r g e at / \ a t We say that “rgeat” is a scrambled string of “great”.

Similarly, if we continue to swap the children of nodes “eat” and “at”, it produces a scrambled string “rgtae”.

rgtae

/ \ rg tae / \ / \ r g ta e / \ t a We say that “rgtae” is a scrambled string of “great”.

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

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O(n^4) dp

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class Solution { public: bool isScramble(string s1, string s2) { if (s1.size() != s2.size()) return false; if (s1 == s2) return true; int cnt[256] = {0}; for (auto &ch : s1) ++cnt[ch]; for (auto &ch : s2) --cnt[ch]; for (int i = 0; i < 256; ++i) if (cnt[i]) return false; for (int i = 1; i < s1.size(); ++i) { string s11 = s1.substr(0, i); string s12 = s1.substr(i); string s21, s22; // forward order s21 = s2.substr(0, i); s22 = s2.substr(i); if (isScramble(s11, s21) && isScramble(s12, s22)) return true; // reverse order s21 = s2.substr(s1.size() - i); s22 = s2.substr(0, s1.size() - i); if (isScramble(s11, s21) && isScramble(s12, s22)) return true; } return false; } };