Added word break problem

algorithms/CPlusPlus/Dynamic-Programming/word-break.cpp

@@ -0,0 +1,115 @@
+/*
+
+Given a string A and a dictionary of n words B. find out if A can be segmented into a 
+space-seprated sequence of dicrionary words.
+
+
+Note: From the dictionary B each word can be taken any number of times
+and in any order 
+
+Example 1:
+
+N = 12
+B = {"i", "like", "sam", "sung", "samsung", "mobile", "ice", "icecream", "man",
+"man", "go", "mango"}
+A = "ilike"
+
+Ouput:  1 (True)
+Explanation:
+String can be segmented as "i like"
+
+
+TOPICS: Dynamic Programing; Matrix Chain Multiplication Algo. 
+
+*/
+
+
+#include <iostream>
+#include <bits/stdc++.h>
+
+using namespace std;
+
+//Utility function to find if a string exist in dictionary (find in vector of string)
+inline bool findVec(vector<string>& vec, string str)
+{
+    return (find(vec.begin(), vec.end(), str) != vec.end());
+}
+
+// rec_fun with dictionary, indices of start and end of substring A and array to store previous rec. calls
+int rec_fun(string A, vector<string>& B, int start, int end, int** dp)
+{
+    
+    string partial_string = A.substr(start, end - start + 1);
+    if(dp[start][end] != -1)
+        return dp[start][end];
+    if(findVec(B, partial_string))  // if substring found in the dictionary
+    {
+        dp[start][end] = 1;
+        return dp[start][end];
+    }
+    if(start == end)    // Base case with substring of size 1 and the substring not present in dictionary
+    {
+        dp[start][end] = 0;
+        return dp[start][end];
+    }
+    
+    // Using Matrix Chain Multiplication fundamentals to partition string
+    // from k= start +1 to k = end
+    for(int k= start +1; k<=end; k++)
+    {
+        int l_subcall = rec_fun(A, B, start, k-1, dp);
+        if(!l_subcall)  // if left partition substring's recursive call not found in dictionary try different value of k
+            continue;
+        int r_subcall = rec_fun(A, B, k, end, dp);
+
+        if(l_subcall & r_subcall) // if both left and right subcalls are valid partitions then given string partition found in dictionary
+        {
+            dp[start][end] = 1;
+            return dp[start][end];
+        }
+
+    }
+
+    dp[start][end] = 0;
+    return dp[start][end];
+
+}
+
+
+bool wordBreak(string A, vector<string>& B )
+{
+    //dp array used for memoizing the recursive calls 
+    int end = A.size() -1;
+    int** dp = new int*[end+1];
+    for(int i=0; i<=end; i++)
+    {
+        dp[i] = new int[end+1];
+        for(int j=0; j<=end; j++)
+            dp[i][j] = -1;  // initializing array with -1;
+    }
+    int start = 0;
+    return rec_fun(A, B, start, end, dp);
+}
+
+int main()
+{
+    int t ;
+    cin>>t;
+    while(t--)
+    {
+        int n;
+        cin>>n;
+
+        vector<string> dict;
+        for(int i=0; i<n; i++)
+        {
+            string s;
+            cin>>s;
+            dict.push_back(s);
+        }
+        string line;
+        cin>>line;
+        cout<<wordBreak(line, dict)<<endl;
+    }
+
+}