Given an input string and a dictionary of words, find out if the input string can be segmented into a space-separated sequence of dictionary words. See following examples for more details.
This is a famous Google interview question, also being asked by many other companies now a days.
Consider the following dictionary
{ i, like, sam, sung, samsung, mobile, ice,
cream, icecream, man, go, mango}
Input: ilike
Output: Yes
The string can be segmented as "i like".
Input: ilikesamsung
Output: Yes
The string can be segmented as "i like samsung"
or "i like sam sung".
Recursive implementation:
The idea is simple, we consider each prefix and search for it in dictionary. If the prefix is present in dictionary, we recur for rest of the string (or suffix).
C++
#include <algorithm>#include <iostream>#include <string>#include <vector>using namespace std;// Function to check if the given word can be broken// down into words from the wordListbool WordBreak(const vector<string>& wordList, const string& word){ // If the word is empty, it can be broken down into // an empty list of words if (word.empty()) return true; int wordLen = word.length(); // Check if the word can be broken down into // words from the wordList for (int i = 1; i <= wordLen; ++i) { string prefix = word.substr(0, i); if (find(wordList.begin(), wordList.end(), prefix) != wordList.end() && WordBreak(wordList, word.substr(i))) { return true; } } return false;}int main(){ vector<string> wordList = { "mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i", "like", "ice", "cream" }; bool result = WordBreak(wordList, "ilikesamsung"); cout << boolalpha << result << endl; // Output: true return 0;} |
Java
import java.util.*;public class Main { // Function to check if the given word can be broken // down into words from the wordList static boolean wordBreak(List<String> wordList, String word) { // If the word is empty, it can be broken down into // an empty list of words if (word.isEmpty()) { return true; } int wordLen = word.length(); // Check if the word can be broken down into // words from the wordList for (int i = 1; i <= wordLen; ++i) { String prefix = word.substring(0, i); if (wordList.contains(prefix) && wordBreak(wordList, word.substring(i))) { return true; } } return false; } public static void main(String[] args) { List<String> wordList = Arrays.asList( "mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i", "like", "ice", "cream"); boolean result = wordBreak(wordList, "ilikesamsung"); System.out.println(result); // Output: true }} |
Python3
def wordBreak(wordList, word): if word == '': return True else: wordLen = len(word) return any([(word[:i] in wordList) and wordBreak(wordList, word[i:]) for i in range(1, wordLen+1)]) |
C#
using System;using System.Collections.Generic;using System.Linq;class GFG { // Function to check if the given word can be broken // down into words from the wordList public static bool WordBreak(List<string> wordList, string word) { // If the word is empty, it can be broken down into // an empty list of words if (word == "") return true; int wordLen = word.Length; // Check if the word can be broken down into // words from the wordList return Enumerable.Range(1, wordLen) .Any(i => wordList.Contains(word.Substring(0, i)) && WordBreak(wordList, word.Substring(i))); } public static void Main(string[] args) { Console.WriteLine(WordBreak( new List<string>(new string[] { "mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i", "like", "ice", "cream" }), "ilikesamsung")); }} |
Output
true
If the recursive call for suffix returns true, we return true, otherwise we try next prefix. If we have tried all prefixes and none of them resulted in a solution, we return false.
We strongly recommend to see substr function which is used extensively in following implementations.
C++
// A recursive program to test whether a given // string can be segmented into space separated // words in dictionary#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","like","ice","cream"}; int size = sizeof(dictionary)/sizeof(dictionary[0]); for (int i = 0; i < size; i++) if (dictionary[i].compare(word) == 0) return true; return false;}// returns true if string can be segmented into space // separated words, otherwise returns falsebool wordBreak(string str){ int size = str.size(); // Base case if (size == 0) return true; // Try all prefixes of lengths from 1 to size for (int i=1; i<=size; i++) { // The parameter for dictionaryContains is // str.substr(0, i) which is prefix (of input // string) of length 'i'. We first check whether // current prefix is in dictionary. Then we // recursively check for remaining string // str.substr(i, size-i) which is suffix of // length size-i if (dictionaryContains( str.substr(0, i) ) && wordBreak( str.substr(i, size-i) )) return true; } // If we have tried all prefixes and // none of them worked return false;}// Driver program to test above functionsint main(){ wordBreak("ilikesamsung")? cout <<"Yes\n": cout << "No\n"; wordBreak("iiiiiiii")? cout <<"Yes\n": cout << "No\n"; wordBreak("")? cout <<"Yes\n": cout << "No\n"; wordBreak("ilikelikeimangoiii")? cout <<"Yes\n": cout << "No\n"; wordBreak("samsungandmango")? cout <<"Yes\n": cout << "No\n"; wordBreak("samsungandmangok")? cout <<"Yes\n": cout << "No\n"; return 0;} |
Java
import java.util.*;// Recursive implementation of // word break problem in javapublic class WordBreakProblem{ // set to hold dictionary values private static Set<String> dictionary = new HashSet<>(); public static void main(String []args) { // array of strings to be added in dictionary set. String temp_dictionary[] = {"mobile","samsung","sam","sung", "man","mango","icecream","and", "go","i","like","ice","cream"}; // loop to add all strings in dictionary set for (String temp :temp_dictionary) { dictionary.add(temp); } // sample input cases System.out.println(wordBreak("ilikesamsung")); System.out.println(wordBreak("iiiiiiii")); System.out.println(wordBreak("")); System.out.println(wordBreak("ilikelikeimangoiii")); System.out.println(wordBreak("samsungandmango")); System.out.println(wordBreak("samsungandmangok")); } // returns true if the word can be segmented into parts such // that each part is contained in dictionary public static boolean wordBreak(String word) { int size = word.length(); // base case if (size == 0) return true; //else check for all words for (int i = 1; i <= size; i++) { // Now we will first divide the word into two parts , // the prefix will have a length of i and check if it is // present in dictionary ,if yes then we will check for // suffix of length size-i recursively. if both prefix and // suffix are present the word is found in dictionary. if (dictionary.contains(word.substring(0,i)) && wordBreak(word.substring(i,size))) return true; } // if all cases failed then return false return false; }}// This code is contributed by Sparsh Singhal |
Python3
# Recursive implementation of # word break problem in Python# returns True if the word can be segmented into parts such# that each part is contained in dictionarydef wordBreak(word): global dictionary size = len(word) # base case if (size == 0): return True # else check for all words for i in range(1,size + 1): # Now we will first divide the word into two parts , # the prefix will have a length of i and check if it is # present in dictionary ,if yes then we will check for # suffix of length size-i recursively. if both prefix and # suffix are present the word is found in dictionary. if (word[0:i] in dictionary and wordBreak(word[i: size])): return True # if all cases failed then return False return False# set to hold dictionary valuesdictionary = set() # array of strings to be added in dictionary set.temp_dictionary = [ "mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i","like", "ice", "cream" ]# loop to add all strings in dictionary setfor temp in temp_dictionary: dictionary.add(temp)# sample input casesprint("Yes" if wordBreak("ilikesamsung") else "No")print("Yes" if wordBreak("iiiiiiii") else "No")print("Yes" if wordBreak("") else "No")print("Yes" if wordBreak("ilikelikeimangoiii") else "No")print("Yes" if wordBreak("samsungandmango") else "No")print("Yes" if wordBreak("samsungandmangok") else "No")# This code is contributed by shinjanpatra |
C#
using System;using System.Collections.Generic;// Recursive implementation of word break problem in C#public class WordBreakProblem{ // HashSet to hold dictionary values private static HashSet<string> dictionary = new HashSet<string>(); // Returns true if the word can be segmented into parts such // that each part is contained in dictionary public static bool wordBreak(string word) { int size = word.Length; // Base case if (size == 0) return true; // Else check for all words for (int i = 1; i <= size; i++) { // Divide the word into two parts, the prefix will have a length of i // and check if it is present in dictionary, if yes then we will check // for suffix of length size-i recursively. If both prefix and // suffix are present the word is found in dictionary. if (dictionary.Contains(word.Substring(0, i)) && wordBreak(word.Substring(i, size - i))) return true; } // If all cases failed then return false return false; } static void Main(string[] args) { // Array of strings to be added in dictionary HashSet string[] temp_dictionary = { "mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i", "like", "ice", "cream" }; // Loop to add all strings in dictionary HashSet foreach (string temp in temp_dictionary) { dictionary.Add(temp); } // Sample input cases Console.WriteLine(wordBreak("ilikesamsung")); Console.WriteLine(wordBreak("iiiiiiii")); Console.WriteLine(wordBreak("")); Console.WriteLine(wordBreak("ilikelikeimangoiii")); Console.WriteLine(wordBreak("samsungandmango")); Console.WriteLine(wordBreak("samsungandmangok")); }} |
Javascript
<script>// Recursive implementation of // word break problem in java // set to hold dictionary values var dictionary = new Set(); // array of strings to be added in dictionary set. var temp_dictionary = [ "mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i", "like", "ice", "cream" ]; // loop to add all strings in dictionary set for (var temp of temp_dictionary) { dictionary.add(temp); } // sample input cases document.write(((wordBreak("ilikesamsung"))?"Yes":"No")+"<br/>"); document.write(((wordBreak("iiiiiiii"))?"Yes":"No")+"<br/>"); document.write(((wordBreak(""))?"Yes":"No")+"<br/>"); document.write(((wordBreak("ilikelikeimangoiii"))?"Yes":"No")+"<br/>"); document.write(((wordBreak("samsungandmango"))?"Yes":"No")+"<br/>"); document.write(((wordBreak("samsungandmangok"))?"Yes":"No")+"<br/>"); // returns true if the word can be segmented into parts such // that each part is contained in dictionary function wordBreak( word) { var size = word.length; // base case if (size == 0) return true; // else check for all words for (var i = 1; i <= size; i++) { // Now we will first divide the word into two parts , // the prefix will have a length of i and check if it is // present in dictionary ,if yes then we will check for // suffix of length size-i recursively. if both prefix and // suffix are present the word is found in dictionary. if (dictionary.has(word.substring(0, i)) && wordBreak(word.substring(i, size))) return true; } // if all cases failed then return false return false;}// This code is contributed by Rajput-Ji</script> |
Output
Yes Yes Yes Yes Yes No
Time Complexity: The time complexity of the above code will be O(2^n).
Auxiliary Space: The space complexity will be O(n) as we are using recursion and the recursive call stack will take O(n) space.
Dynamic Programming
Why Dynamic Programming? The above problem exhibits overlapping sub-problems. For example, see the following partial recursion tree for string “abcde” in the worst case.
CPP
// A Dynamic Programming based program to test whether a given string can// be segmented into space separated words in dictionary#include <iostream>#include <string.h>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","like","ice","cream"}; int size = sizeof(dictionary)/sizeof(dictionary[0]); for (int i = 0; i < size; i++) if (dictionary[i].compare(word) == 0) return true; return false;}// Returns true if string can be segmented into space separated// words, otherwise returns falsebool wordBreak(string str){ int size = str.size(); if (size == 0) return true; // Create the DP table to store results of subproblems. The value wb[i] // will be true if str[0..i-1] can be segmented into dictionary words, // otherwise false. bool wb[size+1]; memset(wb, 0, sizeof(wb)); // Initialize all values as false. for (int i=1; i<=size; i++) { // if wb[i] is false, then check if current prefix can make it true. // Current prefix is "str.substr(0, i)" if (wb[i] == false && dictionaryContains( str.substr(0, i) )) wb[i] = true; // wb[i] is true, then check for all substrings starting from // (i+1)th character and store their results. if (wb[i] == true) { // If we reached the last prefix if (i == size) return true; for (int j = i+1; j <= size; j++) { // Update wb[j] if it is false and can be updated // Note the parameter passed to dictionaryContains() is // substring starting from index 'i' and length 'j-i' if (wb[j] == false && dictionaryContains( str.substr(i, j-i) )) wb[j] = true; // If we reached the last character if (j == size && wb[j] == true) return true; } } } /* Uncomment these lines to print DP table "wb[]" for (int i = 1; i <= size; i++) cout << " " << wb[i]; */ // If we have tried all prefixes and none of them worked return false;}// Driver program to test above functionsint main(){ wordBreak("ilikesamsung")? cout <<"Yes\n": cout << "No\n"; wordBreak("iiiiiiii")? cout <<"Yes\n": cout << "No\n"; wordBreak("")? cout <<"Yes\n": cout << "No\n"; wordBreak("ilikelikeimangoiii")? cout <<"Yes\n": cout << "No\n"; wordBreak("samsungandmango")? cout <<"Yes\n": cout << "No\n"; wordBreak("samsungandmangok")? cout <<"Yes\n": cout << "No\n"; return 0;} |
Java
// A Dynamic Programming based program to test whether a given String can// be segmented into space separated words in dictionaryimport java.util.*;class GFG{/* 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*/static boolean dictionaryContains(String word){ String dictionary[] = {"mobile","samsung","sam","sung","man","mango", "icecream","and","go","i","like","ice","cream"}; int size = dictionary.length; for (int i = 0; i < size; i++) if (dictionary[i].compareTo(word) == 0) return true; return false;}// Returns true if String can be segmented into space separated// words, otherwise returns falsestatic boolean wordBreak(String str){ int size = str.length(); if (size == 0) return true; // Create the DP table to store results of subproblems. The value wb[i] // will be true if str[0..i-1] can be segmented into dictionary words, // otherwise false. boolean []wb = new boolean[size+1]; for (int i=1; i<=size; i++) { // if wb[i] is false, then check if current prefix can make it true. // Current prefix is "str.substring(0, i)" if (wb[i] == false && dictionaryContains( str.substring(0, i) )) wb[i] = true; // wb[i] is true, then check for all subStrings starting from // (i+1)th character and store their results. if (wb[i] == true) { // If we reached the last prefix if (i == size) return true; for (int j = i+1; j <= size; j++) { // Update wb[j] if it is false and can be updated // Note the parameter passed to dictionaryContains() is // subString starting from index 'i' and length 'j-i' if (wb[j] == false && dictionaryContains( str.substring(i, j) )) wb[j] = true; // If we reached the last character if (j == size && wb[j] == true) return true; } } } /* Uncomment these lines to print DP table "wb[]" for (int i = 1; i <= size; i++) System.out.print(" " + wb[i]; */ // If we have tried all prefixes and none of them worked return false;}// Driver program to test above functionspublic static void main(String[] args){ if(wordBreak("ilikesamsung")) System.out.print("Yes\n"); else System.out.print("No\n"); if(wordBreak("iiiiiiii")) System.out.print("Yes\n"); else System.out.print("No\n"); if(wordBreak("")) System.out.print("Yes\n"); else System.out.print("No\n"); if(wordBreak("ilikelikeimangoiii")) System.out.print("Yes\n"); else System.out.print("No\n"); if(wordBreak("samsungandmango")) System.out.print("Yes\n"); else System.out.print("No\n"); if(wordBreak("samsungandmangok")) System.out.print("Yes\n"); else System.out.print("No\n"); }}// This code is contributed by Rajput-Ji |
Python3
# A Dynamic Programming based program to test whether a given String can# be segmented into space separated words in dictionary# 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# programdef dictionaryContains(word): dictionary = [ "mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i", "like", "ice", "cream" ] size = len(dictionary) for i in range(size): if (dictionary[i]== word): return True return False# Returns True if String can be segmented into space separated# words, otherwise returns Falsedef wordBreak(Str): size = len(Str) if (size == 0): return True # Create the DP table to store results of subproblems. The value wb[i] # will be True if str[0..i-1] can be segmented into dictionary words, # otherwise False. wb = [False for i in range(size + 1)] for i in range(1,size + 1): # if wb[i] is False, then check if current prefix can make it True. # Current prefix is "str.substring(0, i)" if (wb[i] == False and dictionaryContains(Str[0: i])): wb[i] = True # wb[i] is True, then check for all subStrings starting from # (i+1)th character and store their results. if (wb[i] == True): # If we reached the last prefix if (i == size): return True for j in range(i + 1,size + 1): # Update wb[j] if it is False and can be updated # Note the parameter passed to dictionaryContains() is # subString starting from index 'i' and length 'j-i' if (wb[j] == False and dictionaryContains(Str[i: j])): wb[j] = True # If we reached the last character if (j == size and wb[j] == True): return True # If we have tried all prefixes and none of them worked return False# Driver program to test above functions if (wordBreak("ilikesamsung")): print("Yes")else: print("No")if (wordBreak("iiiiiiii")): print("Yes")else: print("No")if (wordBreak("")): print("Yes")else: print("No")if (wordBreak("ilikelikeimangoiii")): print("Yes")else: print("No")if (wordBreak("samsungandmango")): print("Yes")else: print("No")if (wordBreak("samsungandmangok")): print("Yes")else: print("No")# This code is contributed by shinjanpatra |
C#
// A Dynamic Programming based program to test whether a given String can// be segmented into space separated words in dictionary// Include namespace systemusing System;public class GFG{ // 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 public static bool dictionaryContains(String word) { String[] dictionary = {"mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i", "like", "ice", "cream"}; var size = dictionary.Length; for (int i = 0; i < size; i++) { if (string.CompareOrdinal(dictionary[i],word) == 0) { return true; } } return false; } // Returns true if String can be segmented into space separated // words, otherwise returns false public static bool wordBreak(String str) { var size = str.Length; if (size == 0) { return true; } // Create the DP table to store results of subproblems. The value wb[i] // will be true if str[0..i-1] can be segmented into dictionary words, // otherwise false. bool[] wb = new bool[size + 1]; for (int i = 1; i <= size; i++) { // if wb[i] is false, then check if current prefix can make it true. // Current prefix is "str.substring(0, i)" if (wb[i] == false && GFG.dictionaryContains(str.Substring(0,i-0))) { wb[i] = true; } // wb[i] is true, then check for all subStrings starting from // (i+1)th character and store their results. if (wb[i] == true) { // If we reached the last prefix if (i == size) { return true; } for (int j = i + 1; j <= size; j++) { // Update wb[j] if it is false and can be updated // Note the parameter passed to dictionaryContains() is // subString starting from index 'i' and length 'j-i' if (wb[j] == false && GFG.dictionaryContains(str.Substring(i,j-i))) { wb[j] = true; } // If we reached the last character if (j == size && wb[j] == true) { return true; } } } } // Uncomment these lines to print DP table "wb[]" // for (int i = 1; i <= size; i++) // System.out.print(" " + wb[i]; // If we have tried all prefixes and none of them worked return false; } // Driver program to test above functions public static void Main(String[] args) { if (GFG.wordBreak("ilikesamsung")) { Console.Write("Yes\n"); } else { Console.Write("No\n"); } if (GFG.wordBreak("iiiiiiii")) { Console.Write("Yes\n"); } else { Console.Write("No\n"); } if (GFG.wordBreak("")) { Console.Write("Yes\n"); } else { Console.Write("No\n"); } if (GFG.wordBreak("ilikelikeimangoiii")) { Console.Write("Yes\n"); } else { Console.Write("No\n"); } if (GFG.wordBreak("samsungandmango")) { Console.Write("Yes\n"); } else { Console.Write("No\n"); } if (GFG.wordBreak("samsungandmangok")) { Console.Write("Yes\n"); } else { Console.Write("No\n"); } }}// This code is contributed by mukulsomukesh |
Javascript
<script>// A Dynamic Programming based program to test whether a given String can// be segmented into space separated words in dictionary /* * 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 */ function dictionaryContains( word) { var dictionary = [ "mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i", "like", "ice", "cream" ]; var size = dictionary.length; for (var i = 0; i < size; i++) if (dictionary[i]===(word)) return true; return false; } // Returns true if String can be segmented into space separated // words, otherwise returns false function wordBreak( str) { var size = str.length; if (size == 0) return true; // Create the DP table to store results of subproblems. The value wb[i] // will be true if str[0..i-1] can be segmented into dictionary words, // otherwise false. var wb = Array(size + 1).fill(false); for (var i = 1; i <= size; i++) { // if wb[i] is false, then check if current prefix can make it true. // Current prefix is "str.substring(0, i)" if (wb[i] == false && dictionaryContains(str.substring(0, i))) wb[i] = true; // wb[i] is true, then check for all subStrings starting from // (i+1)th character and store their results. if (wb[i] == true) { // If we reached the last prefix if (i == size) return true; for (j = i + 1; j <= size; j++) { // Update wb[j] if it is false and can be updated // Note the parameter passed to dictionaryContains() is // subString starting from index 'i' and length 'j-i' if (wb[j] == false && dictionaryContains(str.substring(i, j))) wb[j] = true; // If we reached the last character if (j == size && wb[j] == true) return true; } } } /* * Uncomment these lines to print DP table "wb" for (i = 1; i <= size; * i++) document.write(" " + wb[i]; */ // If we have tried all prefixes and none of them worked return false; } // Driver program to test above functions if (wordBreak("ilikesamsung")) document.write("Yes<br/>"); else document.write("No<br/>"); if (wordBreak("iiiiiiii")) document.write("Yes<br/>"); else document.write("No<br/>"); if (wordBreak("")) document.write("Yes<br/>"); else document.write("No<br/>"); if (wordBreak("ilikelikeimangoiii")) document.write("Yes<br/>"); else document.write("No<br/>"); if (wordBreak("samsungandmango")) document.write("Yes<br/>"); else document.write("No<br/>"); if (wordBreak("samsungandmangok")) document.write("Yes<br/>"); else document.write("No<br/>");// This code contributed by Rajput-Ji </script> |
Output
Yes Yes Yes Yes Yes No
The time complexity of the given implementation of the wordBreak function is O(n^3), where n is the length of the input string.
The space complexity of the implementation is O(n), as an extra boolean array of size n+1 is created. Additionally, the dictionary array occupies a space of O(kL), where k is the number of words in the dictionary and L is the maximum length of a word in the dictionary. However, since the dictionary is a constant and small-sized array, its space complexity can be considered negligible. Therefore, the overall space complexity of the implementation is O(n).
Optimized Dynamic Programming:
In this approach, apart from the dp table, we also maintain all the indexes which have matched earlier. Then we will check the substrings from those indexes to the current index. If anyone of that matches then we can divide the string up to that index.
In this program, we are using some extra space. However, its time complexity is O(n*n) if n>s or O(n*s) if s>n where s is the length of the largest string in the dictionary and n is the length of the given string.
C++
// A Dynamic Programming based program to test// whether a given string can be segmented into// space separated words in dictionary#include <bits/stdc++.h>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", "like", "ice", "cream" }; int size = sizeof(dictionary) / sizeof(dictionary[0]); for (int i = 0; i < size; i++) if (dictionary[i].compare(word) == 0) return true; return false;}// Returns true if string can be segmented into space// separated words, otherwise returns falsebool wordBreak(string s){ int n = s.size(); if (n == 0) return true; // Create the DP table to store results of subproblems. // The value dp[i] will be true if str[0..i] can be // segmented into dictionary words, otherwise false. vector<bool> dp(n + 1, 0); // Initialize all values // as false. // matched_index array represents the indexes for which // dp[i] is true. Initially only -1 element is present // in this array. vector<int> matched_index; matched_index.push_back(-1); for (int i = 0; i < n; i++) { int msize = matched_index.size(); // Flag value which tells that a substring matches // with given words or not. int f = 0; // Check all the substring from the indexes matched // earlier. If any of that substring matches than // make flag value = 1; for (int j = msize - 1; j >= 0; j--) { // sb is substring starting from // matched_index[j] // + 1 and of length i - matched_index[j] string sb = s.substr(matched_index[j] + 1, i - matched_index[j]); if (dictionaryContains(sb)) { f = 1; break; } } // If substring matches than do dp[i] = 1 and // push that index in matched_index array. if (f == 1) { dp[i] = 1; matched_index.push_back(i); } } return dp[n - 1];}// Driver codeint main(){ wordBreak("ilikesamsung") ? cout << "Yes\n" : cout << "No\n"; wordBreak("iiiiiiii") ? cout << "Yes\n" : cout << "No\n"; wordBreak("") ? cout << "Yes\n" : cout << "No\n"; wordBreak("ilikelikeimangoiii") ? cout << "Yes\n" : cout << "No\n"; wordBreak("samsungandmango") ? cout << "Yes\n" : cout << "No\n"; wordBreak("samsungandmangok") ? cout << "Yes\n" : cout << "No\n"; return 0;} |
Java
import java.io.*;import java.util.*;class GFG { public static boolean wordBreak(String s, List<String> dictionary) { // create a dp table to store results of subproblems // value of dp[i] will be true if string s can be segmented // into dictionary words from 0 to i. boolean[] dp = new boolean[s.length() + 1]; // dp[0] is true because an empty string can always be segmented. dp[0] = true; for(int i = 0; i <= s.length(); i++){ for(int j = 0; j < i; j++){ if(dp[j] && dictionary.contains(s.substring(j, i))){ dp[i] = true; break; } } } return dp[s.length()]; } public static void main (String[] args) { String[] dictionary = { "mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i", "like", "ice", "cream" }; List<String> dict = new ArrayList<>(); for(String s : dictionary){ dict.add(s); } if (wordBreak("ilikesamsung", dict)) { System.out.println("Yes"); } else { System.out.println("No"); } if (wordBreak("iiiiiiii", dict)) { System.out.println("Yes"); } else { System.out.println("No"); } if (wordBreak("", dict)) { System.out.println("Yes"); } else { System.out.println("No"); } if (wordBreak("samsungandmango", dict)) { System.out.println("Yes"); } else { System.out.println("No"); } if (wordBreak("ilikesamsung", dict)) { System.out.println("Yes"); } else { System.out.println("No"); } if (wordBreak("samsungandmangok", dict)) { System.out.println("Yes"); } else { System.out.println("No"); } }} |
Python3
def wordBreak(s, dictionary): # create a dp table to store results of subproblems # value of dp[i] will be true if string s can be segmented # into dictionary words from 0 to i. dp = [False for i in range(len(s) + 1)] # dp[0] is true because an empty string can always be segmented. dp[0] = True for i in range(len(s) + 1): for j in range(i): if dp[j] and s[j: i] in dictionary: dp[i] = True break return dp[len(s)] # driver codedictionary = [ "mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i", "like", "ice", "cream" ]dict = set()for s in dictionary: dict.add(s)if (wordBreak("ilikesamsung", dict)): print("Yes")else : print("No")if (wordBreak("iiiiiiii", dict)): print("Yes")else: print("No")if (wordBreak("", dict)): print("Yes")else: print("No")if (wordBreak("samsungandmango", dict)): print("Yes")else: print("No")if (wordBreak("ilikesamsung", dict)): print("Yes")else: print("No")if (wordBreak("samsungandmangok", dict)): print("Yes")else: print("No")# This code is contributed by shinjanpatra |
C#
using System;using System.Collections.Generic;public class WordBreakProgram { // A utility function to check whether a word // is present in dictionary or not. A HashSet of // strings is used for dictionary. static bool DictionaryContains(HashSet<string> dictionary, string word) { return dictionary.Contains(word); } // Returns true if string can be segmented into space // separated words, otherwise returns false static bool WordBreak(string s, HashSet<string> dictionary) { int n = s.Length; if (n == 0) return true; // Create the DP table to store results of // subproblems. The value dp[i] will be true if // str[0..i] can be segmented into dictionary words, // otherwise false. bool[] dp = new bool[n + 1]; // matchedIndex array represents the indexes for // which dp[i] is true. Initially only -1 element is // present in this array. List<int> matchedIndex = new List<int>(); matchedIndex.Add(-1); for (int i = 0; i < n; i++) { int msize = matchedIndex.Count; // Flag value which tells that a substring // matches with given words or not. int f = 0; // Check all the substring from the indexes // matched earlier. If any of that substring // matches than make flag value = 1; for (int j = msize - 1; j >= 0; j--) { // sb is substring starting from // matchedIndex[j] // + 1 and of length i - matchedIndex[j] string sb = s.Substring(matchedIndex[j] + 1, i - matchedIndex[j]); if (DictionaryContains(dictionary, sb)) { f = 1; break; } } // If substring matches than do dp[i] = 1 and // push that index in matchedIndex array. if (f == 1) { dp[i] = true; matchedIndex.Add(i); } } return dp[n - 1]; } // Driver code public static void Main() { HashSet<string> dictionary = new HashSet<string>{ "mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i", "like", "ice", "cream" }; Console.WriteLine( WordBreak("ilikesamsung", dictionary) ? "Yes" : "No"); Console.WriteLine(WordBreak("iiiiiiii", dictionary) ? "Yes" : "No"); Console.WriteLine(WordBreak("", dictionary) ? "Yes" : "No"); Console.WriteLine( WordBreak("ilikelikeimangoiii", dictionary) ? "Yes" : "No"); Console.WriteLine( WordBreak("samsungandmango", dictionary) ? "Yes" : "No"); Console.WriteLine( WordBreak("samsungandmangok", dictionary) ? "Yes" : "No"); }}// This code is contributed by Rohit Yadav |
Javascript
<script> function wordBreak( s, dictionary) { // create a dp table to store results of subproblems // value of dp[i] will be true if string s can be segmented // into dictionary words from 0 to i. var dp = Array(s.length + 1).fill(false); // dp[0] is true because an empty string can always be segmented. dp[0] = true; for (var i = 0; i <= s.length; i++) { for (var j = 0; j < i; j++) { if (dp[j] && dictionary.has(s.substring(j, i))) { dp[i] = true; break; } } } return dp[s.length]; } var dictionary = [ "mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i", "like", "ice", "cream" ]; var dict = new Set(); for (var s of dictionary) { dict.add(s); } if (wordBreak("ilikesamsung", dict)) { document.write("<br/>Yes"); } else { document.write("<br/>No"); } if (wordBreak("iiiiiiii", dict)) { document.write("<br/>Yes"); } else { document.write("<br/>No"); } if (wordBreak("", dict)) { document.write("<br/>Yes"); } else { document.write("<br/>No"); } if (wordBreak("samsungandmango", dict)) { document.write("<br/>Yes"); } else { document.write("<br/>No"); } if (wordBreak("ilikesamsung", dict)) { document.write("<br/>Yes"); } else { document.write("<br/>No"); } if (wordBreak("samsungandmangok", dict)) { document.write("<br/>Yes"); } else { document.write("<br/>No"); }// This code is contributed by Rajput-Ji</script> |
Output
Yes Yes Yes Yes Yes No
Word Break Problem | (Trie solution)
Exercise:
The above solutions only find out whether a given string can be segmented or not. Extend the above Dynamic Programming solution to print all possible partitions of input string.
Examples:
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
Input: ilikesamsungmobile
Output:
i like sam sung mobile
i like samsung mobile
Word Break Problem | (Hashmap solution):
In this approach first, we are storing all the words in a Hashmap. after that, we traverse the input string and check if there is a match or not.
C++
#include<bits/stdc++.h>using namespace std;bool CanParseUtil(unordered_map<string, bool>mp, string word){ // if the size id zero that means we completed the word. so we can return true int size = word.size(); if(size == 0) { return true; } string temp = ""; for(int i = 0; i < word.length(); i++) { temp += word[i]; // if the temp exist in hashmap and the parsing operation of the remaining word is true, we can return true. if(mp.find(temp) != mp.end() && CanParseUtil(mp, word.substr(i+1))) { return true; } } // if there is a mismatch in the dictionary, we can return false. return false;}string CanParse(vector<string>words, string word){ int start = 0; // store the words in the hashmap unordered_map<string, bool>mp; for(auto it : words) { mp[it] = true; } return CanParseUtil(mp,word ) == true ? "YES" : "NO";}int main() { vector<string>words{"mobile","samsung","sam","sung", "man","mango","icecream","and", "go","i","like","ice","cream"}; string word = "samsungandmangok"; cout << CanParse(words, word) << endl;} |
Java
import java.util.*;class GFG { static boolean CanParseUtil(HashMap<String, Boolean> mp, String word) { // if the size id zero that means we completed the // word. so we can return true int size = word.length(); if (size == 0) { return true; } String temp = ""; for (int i = 0; i < word.length(); i++) { temp += word.charAt(i); // if the temp exist in hashmap and the parsing // operation of the remaining word is true, we // can return true. if (mp.containsKey(temp) && CanParseUtil(mp, word.substring(i + 1))) { return true; } } // if there is a mismatch in the dictionary, we can // return false. return false; } static String CanParse(String[] words, String word) { // store the words in the hashmap HashMap<String, Boolean> mp = new HashMap<>(); for (String it : words) { mp.put(it, true); } return CanParseUtil(mp, word) == true ? "YES" : "NO"; } public static void main(String[] args) { String[] words = { "mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i", "like", "ice", "cream" }; String word = "samsungandmangok"; System.out.println(CanParse(words, word)); }}// This code is contributed by karandeep1234. |
Python3
def CanParseUtil(mp,word): # if the size id zero that means we completed the word. so we can return True size = len(word) if(size == 0): return True temp = "" for i in range(len(word)): temp += word[i] # if the temp exist in hashmap and the parsing operation of the remaining word is True, we can return True. if(temp in mp and CanParseUtil(mp, word[i+1:])): return True # if there is a mismatch in the dictionary, we can return false. return Falsedef CanParse(words,word): start = 0 # store the words in the hashmap mp = {} for it in words: mp[it] = True return "YES" if CanParseUtil(mp,word ) == True else "NO"# driver codewords = ["mobile","samsung","sam","sung", "man","mango","icecream","and", "go","i","like","ice","cream"]word = "samsungandmangok"print(CanParse(words, word))# This code is contributed by shinjanpatra |
C#
using System;using System.Collections.Generic;class Program { // Utility function to recursively check if a given word can be parsed from a dictionary of words static bool CanParseUtil(Dictionary<string, bool> mp, string word) { // If the size of the word is zero, that means we've completed the word, so we can return true int size = word.Length; if(size == 0) { return true; } string temp = ""; // Iterate over each character in the word and check if we can form a valid word by adding the character to the prefix for(int i = 0; i < word.Length; i++) { temp += word[i]; // If the prefix exists in the dictionary and the remaining portion of the word can also be parsed, return true if(mp.ContainsKey(temp) && CanParseUtil(mp, word.Substring(i+1))) { return true; } } // If there is a mismatch in the dictionary, we can return false return false; } // Main function to check if a given word can be parsed from a list of words static string CanParse(List<string> words, string word) { int start = 0; // Store the words in a dictionary for faster lookup Dictionary<string, bool> mp = new Dictionary<string, bool>(); foreach(string it in words) { mp.Add(it, true); } // Call the utility function to check if the given word can be parsed from the dictionary return CanParseUtil(mp, word) == true ? "YES" : "NO"; } static void Main(string[] args) { // Sample input words and word to be parsed List<string> words = new List<string> { "mobile", "samsung", "sam", "sung", "man", "mango", "icecream", "and", "go", "i", "like", "ice", "cream" }; string word = "samsungandmangok"; // Call the CanParse function to check if the given word can be parsed from the list of words Console.WriteLine(CanParse(words, word)); }} |
Javascript
<script>function CanParseUtil(mp,word){ // if the size id zero that means we completed the word. so we can return true let size = word.length; if(size == 0) { return true; } let temp = ""; for(let i = 0; i < word.length; i++) { temp += word[i]; // if the temp exist in hashmap and the parsing operation of the remaining word is true, we can return true. if(mp.has(temp) === true && CanParseUtil(mp, word.substring(i+1))) { return true; } } // if there is a mismatch in the dictionary, we can return false. return false;}function CanParse(words,word){ let start = 0; // store the words in the hashmap let mp = new Map(); for(let it of words) { mp.set(it , true); } return CanParseUtil(mp,word ) == true ? "YES" : "NO";}// driver codelet words = ["mobile","samsung","sam","sung", "man","mango","icecream","and", "go","i","like","ice","cream"];let word = "samsungandmangok";document.write(CanParse(words, word),"</br>");// code is contributed by shinjanpatra</script> |
Output
NO
Time Complexity: The time complexity of the above code will be O(2^n).
Auxiliary Space: The space complexity will be O(n) as we are using recursion and the recursive call stack will take O(n) space.
Refer below post for solution of exercise.
Word Break Problem using Backtracking
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
