Add Solution for Problem 087

C++/087_Scramble_String.cpp

@@ -0,0 +1,193 @@
+// 087. Scramble String
+/**
+ * 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.
+ * 
+ * Tags: Dynamic Programming, String
+ * 
+ * Author: Kuang Qin
+ */
+
+#include "stdafx.h"
+#include <string>
+
+using namespace std;
+
+// Recursive
+// To test if s2 is a scrambled string of s1, we can divide them in to two substrings
+//                                        s2[0...len-1]
+//                                       /             \
+//        s1[0...len-1]            s2[0...k-1]   s2[k...len-1]
+//       /             \
+// s1[0...k-1]   s1[k...len-1]            s2[0...len-1]
+//                                       /             \
+//                             s2[0...len-k-1]   s2[len-k...len-1]
+//
+// Then s1 and s2 scramble strings if:
+// a) isScramble(s1[0...k-1], s2[0...k-1]) && isScramble(s1[k...len-1], s2[k...len-1])
+// b) isScramble(s1[0...k-1], s2[len-k...len-1]) && isScramble(s1[k...len-1], s2[0...len-k-1])
+//
+// To speed up recursion, we can terminate our recursion early (pruning) by checking if s1 and s2 have the same character set
+class Solution {
+public:
+    bool isScramble(string s1, string s2) {
+        int len1 = s1.size(), len2 = s2.size();
+        if (len1 != len2)
+        {
+            return false;
+        }
+
+        if ((len1 == 0) || (s1 == s2))
+        {
+            return true;
+        }
+
+        // use a count array to store how many times a char appears in the string
+        int count[256] = {0};
+        for (int i = 0; i < len1; i++)
+        {
+            count[s1[i]]++;
+            count[s2[i]]--;
+        }
+
+        // if a char appears different times in s1 and s2, return false
+        for (int i = 0; i < len1; i++)
+        {
+            if (count[s1[i]] != 0)
+            {
+                return false;
+            }
+        }
+
+        // divide into substring and recursively check if they are scramble strings
+        for (int i = 1; i < len1; i++)
+        {
+            if (isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i)))
+            {
+                return true;
+            }
+
+            if (isScramble(s1.substr(0, i), s2.substr(len1 - i)) && isScramble(s1.substr(i), s2.substr(0, len1 - i)))
+            {
+                return true;
+            }
+
+        }
+
+        return false;
+    }
+};
+
+// Dynamic Programming
+// isScr[n-1][i][j] is a 3D DP array which indicates if s1[i...i+n-1] is the scrambled string of s2[j...j+n-1]
+// isScr[n-1][i][j] = (isScr[k-1][i][j] && isScr[n-k-1][i+k][j+k]) || (isScr[k-1][i][j+n-k] && isScr[n-k-1][i+k][j])
+class Solution_DP {
+public:
+    bool isScramble(string s1, string s2) {
+        int len1 = s1.size(), len2 = s2.size();
+        if (len1 != len2)
+        {
+            return false;
+        }
+
+        if ((len1 == 0) || (s1 == s2))
+        {
+            return true;
+        }
+
+        // allocate DP array
+        bool*** isScr = new bool**[len1];
+        for (int i = 0; i < len1; i++)
+        {
+            isScr[i] = new bool*[len1];
+            for (int j = 0; j < len1; j++)
+            {
+                isScr[i][j] = new bool[len2]();
+            }
+        }
+
+        // initialize DP array
+        for (int i = 0; i < len1; i++)
+        {
+            for (int j = 0; j < len2; j++)
+            {
+                isScr[0][i][j] = (s1[i] == s2[j]);
+            }
+        }
+
+        for (int n = 2; n <= len1; n++)
+        {
+            for (int i = 0; i <= len1 - n; i++)
+            {
+                for (int j = 0; j <= len2 - n; j++)
+                {
+                    bool bScr = false;
+                    // for any k that makes it scramble strings, isScr[n][i][j] = true
+                    for (int k = 1; (k < n) && (bScr == false); k++)
+                    {
+                        bScr = (isScr[k-1][i][j] && isScr[n-k-1][i+k][j+k]) || (isScr[k-1][i][j+n-k] && isScr[n-k-1][i+k][j]);
+                    }
+                    isScr[n-1][i][j] = bScr;
+                }
+            }
+        }
+
+        // save result, deallocate memory to prevent memory leak
+        bool res = isScr[len1-1][0][0];
+        for (int i = 0; i < len1; i++)
+        {
+            for (int j = 0; j < len2; j++)
+            {
+                delete[] isScr[i][j];
+            }
+            delete[] isScr[i];
+        }
+        delete[] isScr;
+
+        return res;
+    }
+};
+
+int _tmain(int argc, _TCHAR* argv[])
+{
+    string s1("abb");
+    string s2("bab");
+    Solution_DP mySolution;
+    bool res = mySolution.isScramble(s1, s2);
+    return 0;
+}
+