Create WordBreakProblem.cpp

Algorithms/Dynamic Programming/C++/WordBreakProblem.cpp

@@ -0,0 +1,63 @@
+// A recursive program to test whether a given 
+// string can be segmented into space separated 
+// words in dictionary 
+#include <iostream> 
+using namespace std; 
+
+/* A utility function to check whether a word is 
+present in dictionary or not. An array of strings 
+is used for dictionary. Using array of strings for 
+dictionary is definitely not a good idea. We have 
+used for simplicity of the program*/
+int dictionaryContains(string word) 
+{ 
+	string dictionary[] = {"mobile","samsung","sam","sung", 
+							"man","mango","icecream","and", 
+							"go","i","like","ice","cream"}; 
+	int size = sizeof(dictionary)/sizeof(dictionary[0]); 
+	for (int i = 0; i < size; i++) 
+		if (dictionary[i].compare(word) == 0) 
+		return true; 
+	return false; 
+} 
+
+// returns true if string can be segmented into space 
+// separated words, otherwise returns false 
+bool wordBreak(string str) 
+{ 
+	int size = str.size(); 
+
+	// Base case 
+	if (size == 0) return true; 
+
+	// Try all prefixes of lengths from 1 to size 
+	for (int i=1; i<=size; i++) 
+	{ 
+		// The parameter for dictionaryContains is 
+		// str.substr(0, i) which is prefix (of input 
+		// string) of length 'i'. We first check whether 
+		// current prefix is in dictionary. Then we 
+		// recursively check for remaining string 
+		// str.substr(i, size-i) which is suffix of 
+		// length size-i 
+		if (dictionaryContains( str.substr(0, i) ) && 
+			wordBreak( str.substr(i, size-i) )) 
+			return true; 
+	} 
+
+	// If we have tried all prefixes and 
+	// none of them worked 
+	return false; 
+} 
+
+// Driver program to test above functions 
+int main() 
+{ 
+	wordBreak("ilikesamsung")? cout <<"Yes\n": cout << "No\n"; 
+	wordBreak("iiiiiiii")? cout <<"Yes\n": cout << "No\n"; 
+	wordBreak("")? cout <<"Yes\n": cout << "No\n"; 
+	wordBreak("ilikelikeimangoiii")? cout <<"Yes\n": cout << "No\n"; 
+	wordBreak("samsungandmango")? cout <<"Yes\n": cout << "No\n"; 
+	wordBreak("samsungandmangok")? cout <<"Yes\n": cout << "No\n"; 
+	return 0; 
+}