# Word Break Problem using Backtracking

• Difficulty Level : Hard
• Last Updated : 08 Nov, 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 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:

## C++

 `// A recursive program to print all possible``// partitions of a given string into dictionary``// words``#include ``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);``    ``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;``}`

## Java

 `// 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 dict, String s)``  ``{``    ``String ans=``""``;``    ``wordBreakUtil(n, s, dict, ans);``  ``}` `  ``static` `void` `wordBreakUtil(``int` `n, String s, List 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 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.`

## Python3

 `# 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`

## C#

 `// 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.`

## Javascript

 ``
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(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.

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