Created WordBreak.cpp

Author: SHIVANATHRIHARIKRISHNA

Diff for: WordBreak.cpp

@@ -0,0 +1,62 @@
+ #include<bits/stdc++.h>
+ using namespace std;
+ 
+ vector<int> dp;//for memorization of strings which can be formed using dictionary
+int func(string A,set<string> s)
+{
+    int n=A.length();
+    dp.resize(n,0);
+    for(int i=0;i<n;i++)
+    {
+        string temp="";
+        for(int j=i;j>=0;j--)//Checking for all possible strings before the present string which can form this string
+        {
+            temp+=A[j];
+            if(j==0)
+            {
+                if(s.find(temp)!=s.end())
+                {
+                 dp[i]=1;//If string can be formed
+                 break;
+                }
+            }
+            else
+            {
+                if((s.find(temp)!=s.end())&&(dp[j-1]==1))
+                {
+                    dp[i]=1;
+                    break;
+                }
+            }
+        }
+    }
+    int val=dp[n-1];
+     dp.clear();
+    return val;
+}
+
+int wordBreak(string A, vector<string> &B) {
+   set<string> s;
+  for(int i=0;i<B.size();i++)
+  {
+      string temp=B[i];
+      reverse(temp.begin(),temp.end());
+      s.insert(temp);//Storing reverse of the dictionary words to make search easier in next function
+  }
+  return func(A,s);
+}
+
+int main()
+{
+    int n;
+    cin>>n;//No of words in dictionary
+    string A;//Main word
+    vector<string> B(n);//Words in dictionary
+    cin>>A;
+    for(int i=0;i<n;i++)
+    {
+        cin>>B[i];
+    }
+    cout<<wordBreak(A,B)<<endl;//This will tell whether the main word can be formed using combination of dictionary words
+     return 0;
+}