Merge pull request #762 from ask2sm/patch-31

wordbreakingproblem.cpp

Author: probot-auto-merge[bot]

Dynamic Programming/wordbreakingproblem.cpp

@@ -0,0 +1,89 @@
+// A Dynamic Programming based program to test  whether a given string can be segmented into  space separated words in dictionary 
+
+
+#include <bits/stdc++.h> 
+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) 
+{ 
+
+//intialization of dictionary
+	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 s) 
+{ 
+	int n = s.size(); 
+	if (n == 0) 
+		return true; 
+
+	// Create the DP table to store results of subroblems. 
+	// The value dp[i] will be true if str[0..i] can be 
+	// segmented into dictionary words, otherwise false. 
+	vector<bool> dp(n + 1, 0); // Initialize all values 
+	// as false. 
+
+	// matched_index array represents the indexes for which 
+	// dp[i] is true. Initially only -1 element is present 
+	// in this array. 
+	vector<int> matched_index; 
+	matched_index.push_back(-1); 
+
+	for (int i = 0; i < n; i++) { 
+		int msize = matched_index.size(); 
+
+		// Flag value which tells that a substring matches 
+		// with given words or not. 
+		int f = 0; 
+
+		// Check all the substring from the indexes matched 
+		// earlier. If any of that substring matches than 
+		// make flag value = 1; 
+		for (int j = msize - 1; j >= 0; j--) { 
+
+			// sb is substring starting from matched_index[j] 
+			// + 1 and of length i - matched_index[j] 
+			string sb = s.substr(matched_index[j] + 1, i - matched_index[j]); 
+
+			if (dictionaryContains(sb)) { 
+				f = 1; 
+				break; 
+			} 
+		} 
+
+		// If substring matches than do dp[i] = 1 and 
+		// push that index in matched_index array. 
+		if (f == 1) { 
+			dp[i] = 1; 
+			matched_index.push_back(i); 
+		} 
+	} 
+	return dp[n - 1]; 
+} 
+
+// Driver code 
+int main() 
+{ 
+//test cases
+	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; 
+}