Word Break Problem using Backtracking

Difficulty Level : Medium
Last Updated : 19 Jan, 2021

Given a valid sentence without any spaces between the words and a dictionary of valid English words, find all possible ways to break the sentence in individual dictionary words.

Example

Consider the following dictionary 
{ i, like, sam, sung, samsung, mobile, ice, 
  and, cream, icecream, man, go, mango}

Input: "ilikesamsungmobile"
Output: i like sam sung mobile
        i like samsung mobile

Input: "ilikeicecreamandmango"
Output: i like ice cream and man go
        i like ice cream and mango
        i like icecream and man go
        i like icecream and mango

We have discussed a Dynamic Programming solution in the below post.
Dynamic Programming | Set 32 (Word Break Problem)

The Dynamic Programming solution only finds whether it is possible to break a word or not. Here we need to print all possible word breaks.
We start scanning the sentence from left. As we find a valid word, we need to check whether the rest of the sentence can make valid words or not. Because in some situations the first found word from the left side can leave a remaining portion which is not further separable. So, in that case, we should come back and leave the currently found word and keep on searching for the next word. And this process is recursive because to find out whether the right portion is separable or not, we need the same logic. So we will use recursion and backtracking to solve this problem. To keep track of the found words we will use a stack. Whenever the right portion of the string does not make valid words, we pop the top string from the stack and continue finding.

Below is the implementation of the above idea:

CPP

// A recursive program to print all possible
// partitions of a given string into dictionary
// words
#include <iostream>
using namespace std;
 
/* A utility function to check whether a word
  is present in dictionary or not.  An array of
  strings is used for dictionary.  Using array
  of strings for dictionary is definitely not
  a good idea. We have used for simplicity of
  the program*/
int dictionaryContains(string &word)
{
    string dictionary[] = {"mobile","samsung","sam","sung",
                            "man","mango", "icecream","and",
                            "go","i","love","ice","cream"};
    int n = sizeof(dictionary)/sizeof(dictionary[0]);
    for (int i = 0; i < n; i++)
        if (dictionary[i].compare(word) == 0)
            return true;
    return false;
}
 
// Prototype of wordBreakUtil
void wordBreakUtil(string str, int size, string result);
 
// Prints all possible word breaks of given string
void wordBreak(string str)
{
    // Last argument is prefix
    wordBreakUtil(str, str.size(), "");
}
 
// Result store the current prefix with spaces
// between words
void wordBreakUtil(string str, int n, string result)
{
    //Process all prefixes one by one
    for (int i=1; i<=n; i++)
    {
        // Extract substring from 0 to i in prefix
        string prefix = str.substr(0, i);
 
        // If dictionary conatins this prefix, then
        // we check for remaining string. Otherwise
        // we ignore this prefix (there is no else for
        // this if) and try next
        if (dictionaryContains(prefix))
        {
            // If no more elements are there, print it
            if (i == n)
            {
                // Add this element to previous prefix
                result += prefix;
                cout << result << endl;
                return;
            }
            wordBreakUtil(str.substr(i, n-i), n-i,
                                result + prefix + " ");
        }
    }     
}
 
//Driver Code
int main()
{
   
    // Function call
    cout << "First Test:\n";
    wordBreak("iloveicecreamandmango");
 
    cout << "\nSecond Test:\n";
    wordBreak("ilovesamsungmobile");
    return 0;
}

Output

First Test:
i love ice cream and man go
i love ice cream and mango
i love icecream and man go
i love icecream and mango

Second Test:
i love sam sung mobile
i love samsung mobile

Complexities:

Time Complexity: O(nⁿ). Because there are nⁿ combinations in The Worst Case.
Auxiliary Space: O(n²). Because of the Recursive Stack of wordBreakUtil(…) function in The Worst Case.

Where n is the length of the input string.

This article is contributed by Raghav Jajodia. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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