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 into 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 the 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 that 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:
// 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 contains 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;
} |
// A recursive program to print all possible // partitions of a given string into dictionary // words import java.io.*;
import java.util.*;
class GFG {
// Prints all possible word breaks of given string
static void wordBreak( int n, List<String> dict, String s)
{
String ans= "" ;
wordBreakUtil(n, s, dict, ans);
}
static void wordBreakUtil( int n, String s, List<String> dict, String ans)
{
for ( int i = 1 ; i <= n; i++)
{
// Extract substring from 0 to i in prefix
String prefix=s.substring( 0 , i);
// If dictionary contains this prefix, then
// we check for remaining string. Otherwise
// we ignore this prefix (there is no else for
// this if) and try next
if (dict.contains(prefix))
{
// If no more elements are there, print it
if (i == n)
{
// Add this element to previous prefix
ans += prefix;
System.out.println(ans);
return ;
}
wordBreakUtil(n - i, s.substring(i,n), dict, ans+prefix+ " " );
}
}
}
// main function
public static void main(String args[])
{
String str1 = "iloveicecreamandmango" ; // for first test case
String str2 = "ilovesamsungmobile" ; // for second test case
int n1 = str1.length(); // length of first string
int n2 = str2.length(); // length of second string
// List of strings in dictionary
List <String> dict= Arrays.asList( "mobile" , "samsung" , "sam" , "sung" ,
"man" , "mango" , "icecream" , "and" ,
"go" , "i" , "love" , "ice" , "cream" );
System.out.println( "First Test:" );
// call to the method
wordBreak(n1,dict,str1);
System.out.println( "\nSecond Test:" );
// call to the method
wordBreak(n2,dict,str2);
}
} // This code is contributed by mohitjha727. |
# A recursive program to print all possible # partitions of a given string into dictionary # words # 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 def dictionaryContains(word):
dictionary = { "mobile" , "samsung" , "sam" , "sung" , "man" ,
"mango" , "icecream" , "and" , "go" , "i" , "love" , "ice" , "cream" }
return word in dictionary
# Prints all possible word breaks of given string def wordBreak(string):
# Last argument is prefix
wordBreakUtil(string, len (string), "")
# Result store the current prefix with spaces # between words def wordBreakUtil(string, n, result):
# Process all prefixes one by one
for i in range ( 1 , n + 1 ):
# Extract substring from 0 to i in prefix
prefix = string[:i]
# If dictionary contains 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
print (result)
return
wordBreakUtil(string[i:], n - i, result + prefix + " " )
# Driver Code if __name__ = = "__main__" :
print ( "First Test:" )
wordBreak( "iloveicecreamandmango" )
print ( "\nSecond Test:" )
wordBreak( "ilovesamsungmobile" )
# This code is contributed by harshitkap00r |
// A recursive program to print all possible // partitions of a given string into dictionary // words using System;
using System.Collections.Generic;
class GFG {
// Prints all possible word breaks of given string
static void wordBreak( int n, List< string > dict, string s)
{
string ans= "" ;
wordBreakUtil(n, s, dict, ans);
}
static void wordBreakUtil( int n, string s, List< string > dict, string ans)
{
for ( int i = 1; i <= n; i++)
{
// Extract substring from 0 to i in prefix
string prefix=s.Substring(0, i);
// If dictionary contains this prefix, then
// we check for remaining string. Otherwise
// we ignore this prefix (there is no else for
// this if) and try next
if (dict.Contains(prefix))
{
// If no more elements are there, print it
if (i == n)
{
// Add this element to previous prefix
ans += prefix;
Console.WriteLine(ans);
return ;
}
wordBreakUtil(n - i, s.Substring(i,n-i), dict, ans+prefix+ " " );
}
}
}
static void Main() {
string str1 = "iloveicecreamandmango" ; // for first test case
string str2 = "ilovesamsungmobile" ; // for second test case
int n1 = str1.Length; // length of first string
int n2 = str2.Length; // length of second string
// List of strings in dictionary
List< string > dict= new List< string >( new string []{ "mobile" , "samsung" , "sam" , "sung" ,
"man" , "mango" , "icecream" , "and" ,
"go" , "i" , "love" , "ice" , "cream" });
Console.WriteLine( "First Test:" );
// call to the method
wordBreak(n1,dict,str1);
Console.WriteLine();
Console.WriteLine( "Second Test:" );
// call to the method
wordBreak(n2,dict,str2);
}
} // This code is contributed by divyeshrabadiya07. |
<script> // A recursive program to print all possible // partitions of a given string into dictionary // words // Prints all possible word breaks of given string function wordBreak(n,dict,s)
{ let ans= "" ;
wordBreakUtil(n, s, dict, ans);
} function wordBreakUtil(n,s,dict,ans)
{ for (let i = 1; i <= n; i++)
{
// Extract substring from 0 to i in prefix
let prefix=s.substring(0, i);
// If dictionary contains this prefix, then
// we check for remaining string. Otherwise
// we ignore this prefix (there is no else for
// this if) and try next
if (dict.includes(prefix))
{
// If no more elements are there, print it
if (i == n)
{
// Add this element to previous prefix
ans += prefix;
document.write(ans+ "<br>" );
return ;
}
wordBreakUtil(n - i, s.substring(i,n), dict, ans+prefix+ " " );
}
}
} // main function
let str1 = "iloveicecreamandmango" ; // for first test case
let str2 = "ilovesamsungmobile" ; // for second test case
let n1 = str1.length; // length of first string
let n2 = str2.length; // length of second string
// List of strings in dictionary let dict= [ "mobile" , "samsung" , "sam" , "sung" ,
"man" , "mango" , "icecream" , "and" ,
"go" , "i" , "love" , "ice" , "cream" ];
document.write( "First Test:<br>" );
// call to the method wordBreak(n1,dict,str1); document.write( "<br>Second Test:<br>" );
// call to the method wordBreak(n2,dict,str2); // This code is contributed by avanitrachhadiya2155 </script> |
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(2n). Because there are 2n 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.