Given a string s and y a dictionary of n words dictionary, check if s can be segmented into a sequence of valid words from the dictionary, separated by spaces.
Examples:
Input: s = "ilike", dictionary[] = ["i", "like", "gfg"]
Output: true
Explanation: The string can be segmented as "i like".
Input: s = "ilikegfg", dictionary[] = ["i", "like", "man", "india", "gfg"]
Output: true
Explanation: The string can be segmented as "i like gfg".
Input: "ilikemangoes", dictionary = ["i", "like", "gfg"]
Output: false
Explanation: The string cannot be segmented.
[Naive Approach] Using Recursion - O(2^n) Time and O(n) Space
The idea is to 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). 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.
C++
// C++ program to implement word break.
#include <bits/stdc++.h>
using namespace std;
// Function to check if the given string can be broken
// down into words from the word list
bool wordBreakRec(int i, string &s, vector<string> &dictionary)
{
// If end of string is reached,
// return true.
if (i == s.length())
return true;
int n = s.length();
string prefix = "";
// Try every prefix
for (int j = i; j < n; j++)
{
prefix += s[j];
// if the prefix s[i..j] is a dictionary word
// and rest of the string can also be broken into
// valid words, return true
if (find(dictionary.begin(), dictionary.end(), prefix) != dictionary.end() &&
wordBreakRec(j + 1, s, dictionary))
{
return true;
}
}
return false;
}
bool wordBreak(string &s, vector<string> &dictionary)
{
return wordBreakRec(0, s, dictionary);
}
int main()
{
string s = "ilike";
vector<string> dictionary = {"i", "like", "gfg"};
cout << (wordBreak(s, dictionary) ? "true" : "false") << endl;
return 0;
}
import java.util.*;
class GfG {
static boolean wordBreakRec(int i, String s,
String[] dictionary)
{
if (i == s.length())
return true;
String prefix = "";
for (int j = i; j < s.length(); j++) {
prefix += s.charAt(j);
// Check if the prefix exists in the dictionary
if (Arrays.asList(dictionary).contains(prefix)
&& wordBreakRec(j + 1, s, dictionary)) {
return true;
}
}
return false;
}
static boolean wordBreak(String s, String[] dictionary)
{
return wordBreakRec(0, s, dictionary);
}
public static void main(String[] args)
{
String s = "ilike";
String[] dictionary = { "i", "like", "gfg" };
System.out.println(
wordBreak(s, dictionary) ? "true" : "false");
}
}
def wordBreakRec(i, s, dictionary):
# If end of string is reached,
# return true.
if i == len(s):
return 1
n = len(s)
prefix = ""
# Try every prefix
for j in range(i, n):
prefix += s[j]
# if the prefix s[i..j] is a dictionary word
# and rest of the string can also be broken into
# valid words, return true
if prefix in dictionary and wordBreakRec(j + 1, s, dictionary) == 1:
return 1
return 0
def wordBreak(s, dictionary):
return wordBreakRec(0, s, dictionary)
if __name__ == "__main__":
s = "ilike"
dictionary = {"i", "like", "gfg"}
print("true" if wordBreak(s, dictionary) else "false")
using System;
class Program
{
static bool WordBreakRec(int index, string s, string[] dictionary)
{
// If end of the string is reached, return true.
if (index == s.Length)
return true;
string prefix = "";
// Try every prefix from the current index
for (int j = index; j < s.Length; j++)
{
prefix += s[j];
// Check if the current prefix exists in the dictionary
if (Array.Exists(dictionary, word => word == prefix))
{
// If prefix is valid, recursively check for the remaining substring
if (WordBreakRec(j + 1, s, dictionary))
return true;
}
}
return false;
}
static bool WordBreak(string s, string[] dictionary)
{
return WordBreakRec(0, s, dictionary);
}
static void Main()
{
string[] dictionary = {"i", "like", "gfg"};
string s = "ilike";
Console.WriteLine(WordBreak(s, dictionary) ? "true" : "false");
}
}
// JavaScript program to implement word break.
// Function to check if the given string can be broken
// down into words from the word list.
// Returns 1 if string can be segmented
function wordBreakRec(i, s, dictionary)
{
// If end of string is reached,
// return true.
if (i === s.length)
return 1;
let n = s.length;
let prefix = "";
// Try every prefix
for (let j = i; j < n; j++) {
prefix += s[j];
// if the prefix s[i..j] is a dictionary word
// and rest of the string can also be broken into
// valid words, return true
if (dictionary.find((pre) => pre == prefix)
!== undefined
&& wordBreakRec(j + 1, s, dictionary) === 1) {
return 1;
}
}
return 0;
}
function wordBreak(s, dictionary)
{
return wordBreakRec(0, s, dictionary);
}
let s = "ilike";
let dictionary = [ "i", "like", "gfg" ];
console.log(wordBreak(s, dictionary) ? "true" : "false");
[Expected Approach - 1] Using Top-Down DP - O(n^2) Time and O(n+m) Space
The idea is to use dynamic programming in the recursive solution to avoid recomputing same subproblems. To further improve the time complexity, store the words of the dictionary in a set to improve the time complexity of looking for a word in dictionary from O(m) to O(1).
If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming:
1. Optimal Substructure:
To check if the string can be segmented starting from index i, i.e., wordBreakRec(i), depends on the solutions of the subproblems wordBreakRec(j) where j lies between i and n. Return true if s[i:j] is present in dictionary and wordBreakRec(j) returns true.
2. Overlapping Subproblems:
While applying a recursive approach in this problem, we notice that certain subproblems are computed multiple times. For example, for wordBreakRec(0), wordBreakRec(1) and wordBreakRec(2) is called. wordBreakRec(1) will again call wordBreakRec(2).
- There is only one parameter: i that changes in the recursive solution. So we create a 1D array of size n for memoization.
- We initialize this array as -1 to indicate nothing is computed initially.
- Now we modify our recursive solution to first check if the value is -1, then only make recursive calls. This way, we avoid re-computations of the same subproblems.
C++
#include <bits/stdc++.h>
using namespace std;
bool wordBreakRec(int ind, string &s, vector<string> &dictionary, vector<int> &dp)
{
if (ind >= s.size())
{
return true;
}
if (dp[ind] != -1)
return dp[ind];
bool possible = false;
for (int i = 0; i < dictionary.size(); i++)
{
string temp = dictionary[i];
if (temp.size() > s.size() - ind)
continue;
bool ok = true;
int k = ind;
for (int j = 0; j < temp.size(); j++)
{
if (temp[j] != s[k])
{
ok = false;
break;
}
else
k++;
}
if (ok)
{
possible |= wordBreakRec(ind + temp.size(), s, dictionary, dp);
}
}
return dp[ind] = possible;
}
bool wordBreak(string s, vector<string> &dictionary)
{
int n = s.size();
vector<int> dp(n + 1, -1);
string temp = "";
return wordBreakRec(0, s, dictionary, dp);
}
int main()
{
string s = "ilike";
vector<string> dictionary = {"i", "like", "gfg"};
cout << (wordBreak(s, dictionary) ? "true" : "false") << endl;
return 0;
}
import java.util.*;
class GfG {
static boolean wordBreakRec(int ind, String s, String[] dict, int[] dp) {
if (ind >= s.length()) {
return true;
}
if (dp[ind] != -1) {
return dp[ind] == 1;
}
boolean possible = false;
for (String temp : dict) {
if (temp.length() > s.length() - ind) {
continue;
}
boolean ok = true;
int k = ind;
for (int j = 0; j < temp.length(); j++) {
if (temp.charAt(j) != s.charAt(k)) {
ok = false;
break;
}
k++;
}
if (ok) {
possible |= wordBreakRec(ind + temp.length(), s, dict, dp);
}
}
dp[ind] = possible ? 1 : 0;
return possible;
}
public static boolean wordBreak(String s, String[] dict) {
int n = s.length();
int[] dp = new int[n + 1];
Arrays.fill(dp, -1);
return wordBreakRec(0, s, dict, dp);
}
public static void main(String[] args) {
String s = "ilike";
String[] dict = {"i", "like", "gfg"};
System.out.println(wordBreak(s, dict) ? "true" : "false");
}
}
def wordBreakRec(ind, s, dict, dp):
if ind >= len(s):
return True
if dp[ind] != -1:
return dp[ind] == 1
possible = False
for temp in dict:
if len(temp) > len(s) - ind:
continue
if s[ind:ind+len(temp)] == temp:
possible |= wordBreakRec(ind + len(temp), s, dict, dp)
dp[ind] = 1 if possible else 0
return possible
def word_break(s, dict):
n = len(s)
dp = [-1] * (n + 1)
return wordBreakRec(0, s, dict, dp)
s = "ilike"
dict = ["i", "like", "gfg"]
print("true" if word_break(s, dict) else "false")
function wordBreakRec(ind, s, dict, dp) {
if (ind >= s.length) {
return true;
}
if (dp[ind] !== -1) {
return dp[ind] === 1;
}
let possible = false;
for (let temp of dict) {
if (temp.length > s.length - ind) {
continue;
}
if (s.substring(ind, ind + temp.length) === temp) {
possible ||= wordBreakRec(ind + temp.length, s, dict, dp);
}
}
dp[ind] = possible ? 1 : 0;
return possible;
}
function wordBreak(s, dict) {
let n = s.length;
let dp = new Array(n + 1).fill(-1);
return wordBreakRec(0, s, dict, dp);
}
let s = "ilike";
let dict = ["i", "like", "gfg"];
console.log(wordBreak(s, dict) ? "true" : "false");
[Expected Approach - 2] Using Bottom Up DP - O(n*m*k) time and O(n) space
The idea is to use bottom-up dynamic programming to determine if a string can be segmented into dictionary words. Create a boolean array d[] where each position dp[i] represents whether the substring from 0 to that position can be broken into dictionary words.
Step by step approach:
- Start from the beginning of the string and mark it as valid (base case). i.e., dp[0] = true
- For each position, check if any dictionary word ends at that position and leads to an already valid position.
- If such a word exists, mark the current position as valid, i.e., dp[i] = true
- At the end return the last entry of dp[]
C++
// C++ program to implement word break.
#include <bits/stdc++.h>
using namespace std;
bool wordBreak(string &s, vector<string> &dictionary)
{
int n = s.size();
vector<bool> dp(n + 1, 0);
dp[0] = 1;
// Traverse through the given string
for (int i = 1; i <= n; i++)
{
// Traverse through the dictionary words
for (string &w : dictionary)
{
// Check if current word is present
// the prefix before the word is also
// breakable
int start = i - w.size();
if (start >= 0 && dp[start] && s.substr(start, w.size()) == w)
{
dp[i] = 1;
break;
}
}
}
return dp[n];
}
int main()
{
string s = "ilike";
vector<string> dictionary = {"i", "like", "gfg"};
cout << (wordBreak(s, dictionary) ? "true" : "false") << endl;
return 0;
}
import java.util.*;
class GfG {
static boolean wordBreak(String s, String[] dictionary)
{
int n = s.length();
boolean[] dp = new boolean[n + 1];
dp[0] = true;
// Traverse through the given string
for (int i = 1; i <= n; i++) {
// Traverse through the dictionary words
for (String w : dictionary) {
// Check if the current word is present and
// the prefix before the word is also
// breakable
int start = i - w.length();
if (start >= 0 && dp[start]
&& s.substring(start,
start + w.length())
.equals(w)) {
dp[i] = true;
break;
}
}
}
return dp[n]; // Returning true or false
}
public static void main(String[] args)
{
String s = "ilike";
String[] dictionary
= { "i", "like", "gfg" }; // Using String array
System.out.println(
wordBreak(s, dictionary) ? "true" : "false");
}
}
# Python program to implement word break
def wordBreak(s, dictionary):
n = len(s)
dp = [False] * (n + 1)
dp[0] = True
# Traverse through the given string
for i in range(1, n + 1):
# Traverse through the dictionary words
for w in dictionary:
# Check if current word is present
# the prefix before the word is also
# breakable
start = i - len(w)
if start >= 0 and dp[start] and s[start:start + len(w)] == w:
dp[i] = True
break
return 1 if dp[n] else 0
if __name__ == '__main__':
s = "ilike"
dictionary = ["i", "like", "gfg"]
print("true" if wordBreak(s, dictionary) else "false")
using System;
using System.Collections.Generic;
class GfG {
static bool wordBreak(string s, string[] dictionary)
{
int n = s.Length;
bool[] dp = new bool[n + 1];
dp[0] = true;
// Traverse through the given string
for (int i = 1; i <= n; i++) {
// Traverse through the dictionary words
foreach(string w in dictionary)
{
// Check if current word is present and the
// prefix before the word is also breakable
int start = i - w.Length;
if (start >= 0 && dp[start]
&& s.Substring(start, w.Length) == w) {
dp[i] = true;
break;
}
}
}
return dp[n]; // Return true if word break is
// possible, else false
}
public static void Main()
{
string s = "ilike";
string[] dictionary
= { "i", "like", "gfg" }; // Using string array
Console.WriteLine(
wordBreak(s, dictionary) ? "true" : "false");
}
}
// JavaScript program to implement word break
function wordBreak(s, dictionary)
{
const n = s.length;
const dp = new Array(n + 1).fill(false);
dp[0] = true;
// Traverse through the given string
for (let i = 1; i <= n; i++) {
// Traverse through the dictionary words
for (const w of dictionary) {
// Check if current word is present
// the prefix before the word is also
// breakable
const start = i - w.length;
if (start >= 0 && dp[start]
&& s.substring(start, start + w.length)
=== w) {
dp[i] = true;
break;
}
}
}
return (dp[n]) ? 1 : 0;
}
const s = "ilike";
const dictionary = [ "i", "like", "gfg" ];
console.log(wordBreak(s, dictionary) ? "true" : "false");
Time Complexity: O(n * m * k), where n is the length of string and m is the number of dictionary words and k is the length of maximum sized string in dictionary.
Space Complexity: O(n)
Related Articles:
Word Break Problem | (Trie solution)
Word Break Problem using Backtracking
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Longest Palindromic Subsequence (LPS)Given a string s, find the length of the Longest Palindromic Subsequence in it. Note: The Longest Palindromic Subsequence (LPS) is the maximum-length subsequence of a given string that is also a Palindrome. Longest Palindromic SubsequenceExamples:Input: s = "bbabcbcab"Output: 7Explanation: Subsequen
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Longest Common Increasing Subsequence (LCS + LIS)Given two arrays, a[] and b[], find the length of the longest common increasing subsequence(LCIS). LCIS refers to a subsequence that is present in both arrays and strictly increases.Prerequisites: LCS, LIS.Examples:Input: a[] = [3, 4, 9, 1], b[] = [5, 3, 8, 9, 10, 2, 1]Output: 2Explanation: The long
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Find all distinct subset (or subsequence) sums of an arrayGiven an array arr[] of size n, the task is to find a distinct sum that can be generated from the subsets of the given sets and return them in increasing order. It is given that the sum of array elements is small.Examples: Input: arr[] = [1, 2]Output: [0, 1, 2, 3]Explanation: Four distinct sums can
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Weighted Job SchedulingGiven a 2D array jobs[][] of order n*3, where each element jobs[i] defines start time, end time, and the profit associated with the job. The task is to find the maximum profit you can take such that there are no two jobs with overlapping time ranges.Note: If the job ends at time X, it is allowed to
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Count Derangements (Permutation such that no element appears in its original position)A Derangement is a permutation of n elements, such that no element appears in its original position. For example, a derangement of [0, 1, 2, 3] is [2, 3, 1, 0].Given a number n, find the total number of Derangements of a set of n elements.Examples : Input: n = 2Output: 1Explanation: For two balls [1
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Minimum insertions to form a palindromeGiven a string s, the task is to find the minimum number of characters to be inserted to convert it to a palindrome.Examples:Input: s = "geeks"Output: 3Explanation: "skgeegks" is a palindromic string, which requires 3 insertions.Input: s= "abcd"Output: 3Explanation: "abcdcba" is a palindromic string
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Ways to arrange Balls such that adjacent balls are of different typesThere are 'p' balls of type P, 'q' balls of type Q and 'r' balls of type R. Using the balls we want to create a straight line such that no two balls of the same type are adjacent.Examples : Input: p = 1, q = 1, r = 0Output: 2Explanation: There are only two arrangements PQ and QPInput: p = 1, q = 1,
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