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package ScrambleString; import java.util.HashMap; import java.util.Map; /** * User: Danyang * Date: 1/28/2015 * Time: 19:06 * * 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. */ public class Solution { /** * Algorithm: recursive * treat as set of chars * partition the string * * Notice: * 1. prune, otherwise TLE * @param s1 * @param s2 * @return */ public boolean isScramble(String s1, String s2) { if(s1.equals(s2)) return true; if(!relaxedEqual(s1, s2)) return false; for(int l=1; l bucket = new HashMap<>(); for(char c: s1.toCharArray()) { if(!bucket.containsKey(c)) bucket.put(c, 0); bucket.put(c, bucket.get(c)+1); } for(char c: s2.toCharArray()) { if(!bucket.containsKey(c) || bucket.get(c)<1) return false; bucket.put(c, bucket.get(c)-1); } for(Map.Entry e: bucket.entrySet()) { if(e.getValue()!=0) return false; } return true; } }