Dynamic Programming - Word Break

Question

I am trying to solve this Problem.The question is as follows
Given an input string and a dictionary of words, find out if the input string can be segmented into a space-

Answer 1

Your code is actually wrong. To fail your code, try input like "likeman"

Note that there are two different return values possible from function searchSentence, 0 or 1. So if you initialize the r array with 0 there's no guarantee it's a new state when r[x][y] = 0. Initialize r array with some impossible value like -1 or 2 for this program and test again. Now you can easily confirm that if r[qbegin][qend] != -1 then this state has already been checked so you can return r[qbegin][qend] from here

Updated code :

#include <iostream>
#include <string.h>
using namespace std;

int r[100][100];  //To Store the calculated values

bool searchWord(char q[], char D[][20], int start, int end)
{
    cout << "In Search Word Loop with " << start << " " << end << endl;



    char temp[end - start + 1];
    int j = 0;

    for (int i = start; i <= end ; ++i)
    {
        //cout << "Looping i " << i << endl;
        temp[j] = q[i];
        j++;
    }
    temp[j] = '\0';

    //cout << "For Word " << temp << endl;

    for (int i = 0; i < 12; ++i)
    {
        // cout << "Comparing with " << D[i] << endl;
        if (!strcmp(temp, D[i]))
        {
            cout << "Found Word" << temp << " " << D[i] << endl;
            return 1;
        }
    }

    return 0;
}

bool searchSentence(char q[], char D[][20], int qstart, int qend)
{
    cout << "In Search Sentence Loop" << endl;

    if (r[qstart][qend] != -1)
    {
        cout << "DP Helped!!!" << endl;
        return r[qstart][qend];
    }


    if (qstart == qend)
    {
        if (searchWord(q, D, qstart, qstart))
            return 1;
        else return 0;
    }
    if (qstart > qend) return 1;

    int i;

    for (i = qstart; i <= qend; i++)
    {
        if (searchWord(q, D, qstart, i))
        {
            r[i + 1][qend] = searchSentence(q, D, i + 1, qend);
            if (r[i + 1][qend] == 1) return 1;
        }
    }
    return 0;
}

int main()
{
    char D[20][20] = { "i", "like", "sam", "sung", "samsung", "mobile", "ice", "cream", "icecream", "man", "go", "mango"};
    char q[100] = "ilike";

    int index = 0; char ch;
    ch = q[0];
    memset(r, -1, sizeof(r));
    while (ch != '\0')
    {
        index++;
        ch = q[index];
    }

    if (searchSentence(q, D, 0, index - 1))
    cout << "Yes" << endl;
    else cout << "No" << endl;
}

P.S : There are some redundant lines of codes but I didn't change them and I added a null character in the end of the character array temp in function searchWord

Answer 2

Is recursion mandatory? I see, iterative DP-solution is easiest and compact:

#include <stdio.h>
#include <string.h>

int main() {
  const char *D[] = { "i", "like", "sam", "sung", "samsung", "mobile", "ice", "cream", "icecream", "man", "go", "mango", NULL};
  const char q[] = "samsungmango";
  char dp[100];
  short d_len[20];
  memset(dp, 0, sizeof(dp));
  dp[0] = 1; // 0 element is always reacheable
  int i, j;
  // compute dict string lengths
  for(i = 0; D[i]; i++)
      d_len[i] = strlen(D[i]);

  // Compute splits using DP array
  for(i = 0; q[i] != 0; i++)
      if(dp[i])  // this index is reacheable
          for(j = 0; D[j]; j++) // try to make next reacheable indexes
              if(strncmp(&q[i], D[j], d_len[j]) == 0)
                  dp[i + d_len[j]] = 1; // That position is reacheable, too

  // if EOLN(q) is reached, then yes
  printf("Answer is %s\n", dp[i]? "YES" : "NO");

} // main