add isScramble.cpp

Author: kamyu104

C++/isScramble.cpp

@@ -0,0 +1,41 @@
+// Time Complexity: O(n^4),  due to enumeration of n, i, j, k in (f[n][i][j] and k)
+// Space Complexity: O(n^3), due to hash of f[n][i][j]
+
+class Solution {
+    public:
+        bool isScramble(string s1, string s2) {
+            if(s1.size() != s2.size())
+                return false;
+            return isScramble(s1.size(), s1.begin(), s2.begin());
+        }
+    private:
+        typedef string::const_iterator Iterator;
+        map<tuple<int, Iterator, Iterator>, bool> hash;
+
+        bool isScramble(int n, Iterator s1, Iterator s2) {
+            // hash
+            if( hash.find(make_tuple(n, s1, s2)) != hash.end()) return hash[make_tuple(n, s1, s2)];
+
+            if(n == 1) return *s1 == *s2;
+
+            // pruning
+            int value1 = 0, value2=0;
+            for (auto it1 = s1, it2 = s2; it1 != s1 + n; ++it1, ++it2) {
+                value1 += (*it1-'a');
+                value2 += (*it2-'a');
+            }
+            if (value1 != value2)
+                return hash[make_tuple(n, s1, s2)] = false; 
+
+            // recursive
+            for(int k = 1; k < n; ++k) {
+                if((isScramble(k, s1, s2) && isScramble(n - k, s1 + k, s2 + k))
+                        || (isScramble(k, s1, s2 + n - k) && isScramble(n - k, s1 + k, s2)) ) {
+                    return hash[make_tuple(n, s1, s2)] = true;
+                }
+            }
+
+            return hash[make_tuple(n, s1, s2)] = false;
+        }
+
+};