Word Break Problem

What is Word Break Problem?

The Word Break Problem is a logical puzzle in computer science where we are asked to check whether a given string can be segmented into a sequence of words from a dictionary. For example, if the given string is "ramhaspenandapple" and the dictionary is ["ram", "and", "has", "apple", "pen"], the answer is true because the string can be segmented as "ram has pen and apple".

Solving Word Break Problem using Backtracking Approach

In the input of this problem, one sentence is given without spaces, another dictionary is also provided with some valid English words. We have to find the possible ways to break the sentence into individual dictionary words. The given string and dictionary are as follows −

Dictionary: {ram, samuel, winter, man, mango, icecream, and, i, love, ice, cream}
Given String: "ilovewinterandicecream"

We will try to search from the left of the string to find a valid word when a valid word is found, we will search for words in the next part of that string. All possible ways to break the string into given words are −

i love winter and ice cream
i love winter and icecream

One way to solve this problem is to use a backtracking approach. It is a technique that tries different combinations and backtracks if a partial solution is not valid. The basic idea is to start from the beginning of the string and check whether the prefix is a word in the specified dictionary or not. If it is, then we recursively check if the remaining suffix can be segmented into words. If both the prefix and the suffix are valid, then we return true and mark it as a part of the solution. Otherwise, we backtrack and try a different prefix.

Pseudocode

Following is the pseudocode for solving a word break problem using the backtracking approach −

Begin
   for i := 0 to n, do
      subStr := substring of given string from (0..i)
      if subStr is in dictionary, then
         if i = n, then
            result := result + subStr
            display the result
            return
         wordBreak(substring from (i..n-i), n-i, result, subStr, ‘space’)
   done
End

Example

In the following example, we will practically demonstrate how to solve the word break problem.

C C++ Java Python 
#include <stdio.h>
#include <string.h>
#define N 13
char *dictionary[N] = {"mobile","samsung","sam","sung","man","mango", "icecream","and", "go","i","love","ice","cream"};
//check whether the word is in dictionary or not
int isInDict(char *word){
   for (int i = 0; i < N; i++)
      if (strcmp(dictionary[i], word) == 0)
         return 1;
   return 0;
}
void wordBreak(char *str, int n, char *result) {
   for (int i=1; i<=n; i++) {
      //get string from 0 to ith location of the string 
      char subStr[100];
      strncpy(subStr, str, i);
      subStr[i] = '\0';

      //if subStr is found in the dictionary
      if (isInDict(subStr)) {       
         if (i == n) {
            //add substring in the result
            strcat(result, subStr); 
            printf("%s\n", result);
            return;
         }
         //otherwise break rest part
         char newResult[100];
         strcpy(newResult, result);
         strcat(newResult, subStr);
         strcat(newResult, " ");
         wordBreak(str + i, n-i, newResult);    
      }
   }
}
int main() {
   char str[] = "iloveicecreamandmango";
   char result[100] = "";
   wordBreak(str, strlen(str), result);
   return 0;
}

#include <iostream>
#define N 13
using namespace std;
string dictionary[N] = {"mobile","samsung","sam","sung","man","mango", "icecream","and", "go","i","love","ice","cream"};
//check whether the word is in dictionary or not
int isInDict(string word){    
   for (int i = 0; i < N; i++)
      if (dictionary[i].compare(word) == 0)
         return true;
   return false;
}
void wordBreak(string str, int n, string result) {
   for (int i=1; i<=n; i++) {
      //get string from 0 to ith location of the string 
      string subStr = str.substr(0, i);       
      //if subStr is found in the dictionary
      if (isInDict(subStr)) {       
         if (i == n) {
            //add substring in the result
            result += subStr; 
            cout << result << endl;
            return;
         }
         //otherwise break rest part
         wordBreak(str.substr(i, n-i), n-i, result + subStr + " ");    
      }
   }
}
int main() {
   string str="iloveicecreamandmango";
   wordBreak(str, str.size(),"");
}

import java.util.Arrays;
import java.util.List;
public class Main {
   static final List<String> DICTIONARY = Arrays.asList("mobile","samsung","sam","sung","man","mango", "icecream","and", "go","i","love","ice","cream");
   //check whether the word is in dictionary or not
   public static boolean isInDict(String word){
      return DICTIONARY.contains(word);
   }
   public static void wordBreak(String str, int n, String result) {
      for (int i=1; i<=n; i++) {
         //get string from 0 to ith location of the string 
         String subStr = str.substring(0, i);
         //if subStr is found in the dictionary
         if (isInDict(subStr)) {
            if (i == n) {
               //add substring in the result
               result += subStr;
               System.out.println(result);
               return;
            }
            //otherwise break rest part
            wordBreak(str.substring(i, n), n-i, result + subStr + " ");
         }
      }
   }
   public static void main(String[] args) {
      String str = "iloveicecreamandmango";
      wordBreak(str, str.length(), "");
   }
}

# Define the dictionary
dictionary = ["mobile","samsung","sam","sung","man","mango", "icecream","and", "go","i","love","ice","cream"]

# Check whether the word is in dictionary or not
def isInDict(word):    
   return word in dictionary

# Function to break the word
def wordBreak(str, n, result):
   for i in range(1, n+1):
      # Get string from 0 to ith location of the string 
      subStr = str[:i]

      # If subStr is found in the dictionary
      if isInDict(subStr):       
         if i == n:
            # Add substring in the result
            result += subStr
            print(result)
            return

         # Otherwise break rest part
         wordBreak(str[i:], n-i, result + subStr + " ")

# Main function
def main():
   str = "iloveicecreamandmango"
   wordBreak(str, len(str), "")

# Call the main function
if __name__ == "__main__":
    main()

Output

i love ice cream and man go
i love ice cream and mango
i love icecream and man go
i love icecream and mango
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