From a1edcc4705fa1c4620048069f6832a867cfa0660 Mon Sep 17 00:00:00 2001
From: Kuang <david.qinkuang@gmail.com>
Date: Wed, 20 Apr 2016 22:54:45 -0500
Subject: [PATCH] Add Solution for Problem 097

---
 C++/097_Interleaving_String.cpp | 83 +++++++++++++++++++++++++++++++++
 1 file changed, 83 insertions(+)
 create mode 100644 C++/097_Interleaving_String.cpp

diff --git a/C++/097_Interleaving_String.cpp b/C++/097_Interleaving_String.cpp
new file mode 100644
index 00000000..54c0443d
--- /dev/null
+++ b/C++/097_Interleaving_String.cpp
@@ -0,0 +1,83 @@
+// 097. Interleaving String
+/**
+ * Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
+ * 
+ * For example,
+ * Given:
+ * s1 = "aabcc",
+ * s2 = "dbbca",
+ * 
+ * When s3 = "aadbbcbcac", return true.
+ * When s3 = "aadbbbaccc", return false.
+ * 
+ * Tags: Dynamic Programming, String
+ * 
+ * Author: Kuang Qin
+ */
+
+#include "stdafx.h"
+#include <string>
+
+using namespace std;
+
+// DP[i][j] indicates if s3[0...i + j - 1] is the interleaving of s1[0...i - 1] and s2[0..j - 1]
+// DP[i][j] = (DP[i - 1][j] && s1[i - 1] == s3[i + j - 1]) || (DP[i][j - 1] && s2[j - 1] == s3[i + j - 1])
+class Solution {
+public:
+    bool isInterleave(string s1, string s2, string s3) {
+        int len1 = s1.size(), len2 = s2.size(), len3 = s3.size();
+        if (len3 != len1 + len2)    // length not match
+        {
+            return false;
+        }
+
+        if ((len1 == 0 || len2 == 0) && (s3 != s1 + s2))    // at least one of s1 and s2 is empty
+        {
+            return false;
+        }
+
+        // 2D DP array
+        bool **DP = new bool*[len1 + 1];
+        for (int i = 0; i <= len1; i++)
+        {
+            DP[i] = new bool[len2 + 1];
+        }
+        DP[0][0] = true;    // s3 is the interleaving of empty string s1 and s2
+
+        for (int i = 1; i <= len1; i++)
+        {
+            // corner cases if s2 is empty
+            DP[i][0] = DP[i - 1][0] && s1[i - 1] == s3[i - 1];
+        }
+
+        for (int j = 1; j <= len2; j++)
+        {
+            // corner cases if s1 is empty
+            DP[0][j] = DP[0][j - 1] && s2[j - 1] == s3[j - 1];
+            for (int i = 1; i <= len1; i++)
+            {
+                // update DP array
+                DP[i][j] = (DP[i - 1][j] && s1[i - 1] == s3[i + j - 1]) 
+                    || (DP[i][j - 1] && s2[j - 1] == s3[i + j - 1]);
+            }
+        }
+
+        bool res = DP[len1][len2];
+        for (int i = 0; i <= len1; i++)
+        {
+            delete[] DP[i];
+        }
+        delete[] DP;
+
+        return res;
+    }
+};
+
+int _tmain(int argc, _TCHAR* argv[])
+{
+    string s1(""), s2(""), s3("");
+    Solution mySolution;
+    bool res = mySolution.isInterleave(s1, s2, s3);
+	return 0;
+}
+