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Word Break Problem | DP-32 | Set – 2
  • Difficulty Level : Hard
  • Last Updated : 21 Apr, 2021

Given a non-empty sequence S and a dictionary dict[] containing a list of non-empty words, print all possible ways to break the sentence in individual dictionary words.
Examples: 
 

Input: S = “catsanddog” 
dict[] = {“cat”, “cats”, “and”, “sand”, “dog”} 
Output: 
“cats and dog” 
“cat sand dog”
Input: S = “pineapplepenapple” 
dict[] = {“apple”, “pen”, “applepen”, “pine”, “pineapple”} 
Output: 
“pine apple pen apple” 
“pineapple pen apple” 
“pine applepen apple” 
 

 

A similar problem to this is discussed in this article, where the task is to check that is there any solution such that the sequence can be broken into the dictionary words.
Approach: The idea is to check for every substring starting from any position I , such that it ends at the length of the string which is present in the dictionary then simply recurse for the substring [0, I]. Meanwhile, store the overlapping subproblems for each substring to avoid the computation of the subproblem again. Overlapping subproblems can be shown as follows – 
 



Below is the implementation of the above approach: 
 

C++




// C++ implementation to break
// a sequence into the words of
// the dictionary
 
#include <bits/stdc++.h>
 
using namespace std;
 
// Unordered_map used for storing
// the sentences the key string
// can be broken into
unordered_map<string,
            vector<string> > mp;
 
// Unordered_set used
// to store the dictionary.
unordered_set<string> dict;
 
// Utility funtion used for
// printing the obtained result
void printResult(vector<string> A)
{
    for (int i = 0; i < A.size(); i++)
        cout << A[i] << '\n';
}
 
// Utility function for
// appending new words
// to already existing strings
vector<string> combine(
     vector<string> prev, string word){
     
    // Loop to find the append string
    // which can be broken into
    for (int i = 0; i < prev.size(); i++) {
        prev[i] += " " + word;
    }
    return prev;
}
 
// Utility funtion for word Break
vector<string> wordBreakUtil(string s)
{  
    // Condition to check if the
    // subproblem is already computed
    if (mp.find(s) != mp.end())
        return mp[s];
    vector<string> res;
     
    // If the whole word is a dictionary
    // word then directly append into
    // the result array for the string
    if (dict.find(s) != dict.end())
        res.push_back(s);
     
    // Loop to iterate over the substring
    for (int i = 1; i < s.length(); i++) {
        string word = s.substr(i);
         
        // If the substring is present into
        // the dictionary then recurse for
        // other substring of the string
        if (dict.find(word) != dict.end()) {
            string rem = s.substr(0, i);
            vector<string> prev =
             combine(wordBreakUtil(rem), word);
            res.insert(res.end(),
                 prev.begin(), prev.end());
        }
    }
     
    // Store the subproblem
    // into the map
    mp[s] = res;
    return res;
}
 
// Master wordBreak function converts
// the string vector to unordered_set
vector<string> wordBreak(string s,
             vector<string>& wordDict)
{
    // Clear the previous stored data
    mp.clear();
    dict.clear();
    dict.insert(wordDict.begin(), wordDict.end());
    return wordBreakUtil(s);
}
 
// Driver Code
int main()
{
    vector<string> wordDict1 = {
        "cat", "cats", "and", "sand", "dog" };
    printResult(wordBreak("catsanddog", wordDict1));
    return 0;
}

Python3




# Python3 implementation to break
# a sequence into the words of
# the dictionary
  
# Unordered_map used for storing
# the sentences the key string
# can be broken into
mp = dict()
  
# Unordered_set used
# to store the dictionary.
dict_t = set()
  
# Utility funtion used for
# printing the obtained result
def printResult(A):
 
    for i in range(len(A)):
        print(A[i])
  
# Utility function for
# appending new words
# to already existing strings
def combine( prev, word):
      
    # Loop to find the append string
    # which can be broken into
    for i in range(len(prev)):
     
        prev[i] += " " + word;
     
    return prev;
 
# Utility funtion for word Break
def wordBreakUtil(s):
 
    # Condition to check if the
    # subproblem is already computed
    if (s in mp):
        return mp[s];
     
    res = []
      
    # If the whole word is a dictionary
    # word then directly append into
    # the result array for the string
    if (s in dict_t):
        res.append(s);
      
    # Loop to iterate over the substring
    for i in range(1, len(s)):
         
        word = s[i:];
          
        # If the substring is present into
        # the dictionary then recurse for
        # other substring of the string
        if (word in dict_t):
             
            rem = s[:i]
            prev = combine(wordBreakUtil(rem), word);
            for i in prev:
                res.append(i)
      
    # Store the subproblem
    # into the map
    #res is an resference so we need to assign an referece to something if its keep on changing
    #res values changes after it start going through combine method
    #you can check if you had a doubt so here we just clone res
    x=[]
    for i in res:
      x.append(i)
    mp[s] = x;
    return res;
  
# Master wordBreak function converts
# the string vector to unordered_set
def wordBreak(s,  wordDict):
 
    # Clear the previous stored data
    mp.clear();
    dict_t.clear();
    for i in wordDict:
        dict_t.add(i)
    return wordBreakUtil(s);
 
# Driver Code
if __name__=='__main__':
 
    wordDict1 = ["cat", "cats", "and", "sand", "dog" ]
    printResult(wordBreak("catsanddog", wordDict1));
 
# This code is contributed by rutvik_56
Output: 
cat sand dog
cats and dog

 

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