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Scramble_String.cpp
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80 lines (64 loc) · 2.03 KB
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/*
Author: Timon Cui, [email protected]
Title: Scramble String
Description:
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.
Difficulty rating:
Notes:
Divide and conquer, O(n^2).
32 ms for 281 test cases on JudgeLarge.
*/
class Solution {
public:
bool isScramble(string s1, string s2) {
if (!isAnagram(s1, s2)) return false;
if (s1.length() <= 3) return true;
for (int left = 1; left < s1.size(); ++left) {
if (isScramble(s1.substr(0, left), s2.substr(0, left)) &&
isScramble(s1.substr(left, s1.length()), s2.substr(left, s2.length())))
return true;
if (isScramble(s1.substr(0, left), s2.substr(s2.length() - left, left)) &&
isScramble(s1.substr(left, s1.length()), s2.substr(0, s2.length() - left)))
return true;
}
return false;
}
private:
bool isAnagram(const string& s1, const string& s2) {
if (s1.size() != s2.size()) return false;
vector<int> N(26, 0);
for (int i = 0; i < s1.size(); ++i) N[s1[i] - 'a'] ++;
for (int i = 0; i < s2.size(); ++i) {
if (N[s2[i] - 'a'] -- == 0) return false;
}
return true;
}
};