Split a string to a string of valid words using Dynamic Programming

Question

I need to find a dynamic programming algorithm to solve this problem. I tried but couldn\'t figure it out. Here is the problem:

You are given a string of n character

Answer 1

Below is an O(n^2) solution for this problem.

void findstringvalid() {
string s = "itwasthebestoftimes";
set dict;
dict.insert("it");
dict.insert("was");
dict.insert("the");
dict.insert("best");
dict.insert("of");
dict.insert("times");

vector b(s.size() + 1, false);
vector spacepos(s.size(), -1);
//Initialization phase
b[0] = true; //String of size 0 is always a valid string
for (int i = 1; i <= s.size(); i++) {
    for (int j = 0; j ::iterator it = dict.find(temp);
              if (it != dict.end()) {
                  b[i] = true;
                  spacepos[i-1] = j;
              }
           }
        }
    }
}
if(b[s.size()])
    for (int i = 1; i < spacepos.size(); i++) {
        if (spacepos[i] != -1) {
            string temp = s.substr(spacepos[i], i - spacepos[i] + 1);
            cout << temp << " ";
    }
    }
}