
Problem Statement
In this task, we are provided with two main inputs: a string s
and a list of strings called wordDict
. The objective is to determine whether the string s
can be completely segmented into a sequence of one or more words that exactly match the words found in wordDict
.
Key rules:
- Words from the
wordDict
can be used multiple times. - The order of segments in
s
must exactly form the original string. - The complete string
s
must be used up by concatenating whole words fromwordDict
.
Return true
if such segmentation is possible and false
otherwise.
Examples
Example 1
Input:
s = "vultrcode", wordDict = ["vultr","code"]
Output:
true
Explanation:
Return true because "vultrcode" can be segmented as "vultr code".
Example 2
Input:
s = "applepenapple", wordDict = ["apple","pen"]
Output:
true
Explanation:
Return true because "applepenapple" can be segmented as "apple pen apple". Note that you are allowed to reuse a dictionary word.
Example 3
Input:
s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
Output:
false
Constraints
1 <= s.length <= 300
1 <= wordDict.length <= 1000
1 <= wordDict[i].length <= 20
s
andwordDict[i]
consist of only lowercase English letters- All the strings of
wordDict
are unique
Approach and Intuition
This problem is a classic case for Dynamic Programming. We'll define a boolean DP array dp
where:
dp[i]
istrue
if the substrings[0:i]
can be formed by concatenating words inwordDict
.dp[0]
is initialized astrue
because an empty string is trivially valid.
Implementation Steps:
Initialize
dp = [False] * (len(s) + 1)
, and setdp[0] = True
.For every index
i
from1
tolen(s)
:For each word in
wordDict
, check:- If
s[i-len(word):i] == word
anddp[i-len(word)] == True
, then setdp[i] = True
.
- If
Return
dp[len(s)]
.
This solution runs in O(n × k) time, where n = len(s)
and k
is the total number of characters across all words in wordDict
.
Example Walkthrough
For s = "vultrcode"
and wordDict = ["vultr", "code"]
:
dp[5] = True because "vultr" ∈ wordDict Then, since dp[5] = True and s[5:9] = "code" ∈ wordDict → dp[9] = True Final result → dp[9] = True
For s = "catsandog"
and wordDict = ["cats", "dog", "sand", "and", "cat"]
:
Multiple segments are valid up to "cats" or "cat", "sand", and "and" However, "og" at the end does not match any wordDict entry after valid prefix segments Final result → False
This strategy ensures optimal performance while respecting the constraints.
Solutions
- C++
- Java
- C
- JavaScript
- Python
class Solution {
public:
bool canSegmentString(string s, vector<string>& dict) {
unordered_set<string> wordSet(dict.begin(), dict.end());
vector<bool> canBreak(s.length() + 1, false);
canBreak[0] = true;
for (int end = 1; end <= s.length(); end++) {
for (int start = 0; start < end; start++) {
if (canBreak[start] && wordSet.find(s.substr(start, end - start)) != wordSet.end()) {
canBreak[end] = true;
break;
}
}
}
return canBreak[s.length()];
}
};
This implementation provides a method canSegmentString
that determines if a given string s
can be segmented into elements that are present in a dictionary dict
using dynamic programming. The dictionary is initially transformed into an unordered set wordSet
for efficient look-up operations.
- The method uses a
vector<bool>
namedcanBreak
which keeps track of the possibility of segmenting the string up to each index. The size ofcanBreak
is one more than the string length to handle segment checks efficiently. canBreak[0]
is initialized totrue
representing that a string of zero length is always segmentable.- The outer loop iterates over the string with an index
end
from 1 to the string's length inclusive, determining if the substring ending at each position can be segmented. - Within this outer loop, another inner loop iterates with a variable
start
from 0 up toend
. This loop checks if the substring fromstart
toend
exists in thewordSet
. - If both the substring can be found in the
wordSet
and the string up tostart
can be segmented (canBreak[start]
is true), then the string up toend
can also be segmented (canBreak[end]
is set to true). - The loop breaks once a valid segmentation is found for a specific
end
.
Finally, the method returns the value of canBreak[s.length()]
, indicating whether the entire string can be segmented based on the provided dictionary. This method efficiently checks segmentations, avoiding unnecessary computation through dynamic programming and leveraging efficient data structures.
class Solution {
public boolean canSegmentString(String str, List<String> dictionary) {
int length = str.length();
Set<String> dictSet = new HashSet<>(dictionary);
boolean[] canSegment = new boolean[length + 1];
canSegment[0] = true;
for (int end = 1; end <= length; end++) {
for (int start = 0; start < end; start++) {
if (canSegment[start] && dictSet.contains(str.substring(start, end))) {
canSegment[end] = true;
break;
}
}
}
return canSegment[length];
}
}
The Word Break problem determines if a string can be segmented into space-separated words found in a given dictionary. The provided Java solution employs dynamic programming to solve this efficiently.
Start by initializing a boolean array
canSegment
wherecanSegment[i]
will be true if the firsti
characters of the string can be segmented using the dictionary.The algorithm constructs a
Set
from the dictionary to allow for O(1) average-time complexity checks for word existence.Initially, set
canSegment[0]
to true, signifying that a zero-length string can be segmented trivially (i.e., do nothing).Use a nested loop where the outer loop goes from 1 to the length of the string. For each position
end
, the inner loop checks all possible start positions from 0 toend
.For each pair
(start, end)
, if the substringstr.substring(start, end)
is in the dictionary andcanSegment[start]
is true, then setcanSegment[end]
to true and break out of the inner loop.The value at
canSegment[str.length()]
gives the final answer; it will be true if the whole string can be segmented according to the dictionary.
By the end of the use of this algorithm, efficiently determine if the input string can be split into dictionary words without redundant checks, leveraging the power of dynamic programming. This approach ensures that each substring is processed only once and only if needed, providing significant performance benefits for larger strings or dictionaries.
#define BUFFER_SIZE 1000
#define DICT_CAPACITY 1000
struct hash_entry {
char key[BUFFER_SIZE + 1];
UT_hash_handle hh;
};
bool canSegmentString(char* s, char** dict, int dictSize) {
struct hash_entry* hashTable = NULL;
for (int idx = 0; idx < dictSize; idx++) {
struct hash_entry* newEntry = malloc(sizeof(struct hash_entry));
strcpy(newEntry->key, dict[idx]);
HASH_ADD_STR(hashTable, key, newEntry);
}
bool dp[BUFFER_SIZE + 1];
memset(dp, false, sizeof(dp));
dp[0] = true;
int stringLength = strlen(s);
for (int i = 1; i <= stringLength; i++) {
for (int j = 0; j < i; j++) {
if (dp[j]) {
struct hash_entry* lookupEntry;
char extractedWord[BUFFER_SIZE + 1];
strncpy(extractedWord, &s[j], i - j);
extractedWord[i - j] = '\0'; // Ensure null termination
HASH_FIND_STR(hashTable, extractedWord, lookupEntry);
if (lookupEntry) {
dp[i] = true;
break;
}
}
}
}
return dp[stringLength];
}
The given C code solves the "Word Break" problem by determining if a string can be segmented into space-separated dictionary words. The implementation utilizes a dynamic programming approach along with a hash table for efficient word look-up. Below is an overview of the code structure:
Define Constants and Structures: Constants like
BUFFER_SIZE
define the maximum length of words andDICT_CAPACITY
for dictionary capacity. Astruct hash_entry
is defined to use with the hash table for storing dictionary words.Initialization of Hash Table: The dictionary words are added into a hash table,
hashTable
, using the uthash library. This allows for efficient look-up times during the segmentation process.Dynamic Programming Array: The boolean array
dp
is utilized to store whether a substring can be segmented up till each index. It starts withdp[0]
initialized totrue
, signifying that a zero-length string is trivially segmentable.Segmentation Logic: The function iterates over the string using two nested loops, checking for each substring if it ends at a dictionary word which starts at a previously segmentable index. If it matches, the current position in
dp
is set totrue
.Result: The value at
dp[stringLength]
determines if the whole string can be segmented, which is returned as the result of thecanSegmentString
function.
The solution is efficient for scenarios where dictionary look-up operations are a bottleneck, as the hash table significantly reduces the look-up time. The dynamic programming component ensures that no substring is redundantly processed, making it efficient in terms of time complexity as well. This implementation provides a robust method to solve segmentation problems in strings.
var canSegment = function (str, dictionary) {
let length = str.length;
let wordSet = new Set(dictionary);
let dpTable = new Array(length + 1).fill(false);
dpTable[0] = true;
for (let i = 1; i <= length; i++) {
for (let j = 0; j < i; j++) {
if (dpTable[j] && wordSet.has(str.substring(j, i))) {
dpTable[i] = true;
break;
}
}
}
return dpTable[length];
};
This JavaScript solution addresses the problem of determining if a string can be segmented into space-separated words that are found in a given dictionary.
- Start by creating a set from the dictionary to facilitate fast lookups.
- Use dynamic programming to solve the problem. Create a boolean array
dpTable
initialized withfalse
values and setdpTable[0]
totrue
to represent an empty substring. - Iterate through the string using two nested loops:
- Outer loop goes from position 1 to the length of the string.
- Inner loop checks each substring from the start of the string up to the current position in the outer loop.
- For each position, check if there exists any breakpoint
j
within the current substring such that the substring fromj
toi
exists in the dictionary and the positionj
indpTable
istrue
. If found, setdpTable[i]
totrue
, breaking the inner loop early as no further checks are needed. - Return the value of
dpTable[length]
which indicates whether the entire string can be segmented into dictionary words.
By implementing this solution, effectively segment strings using dictionary words through a combine use of dynamic programming and hash set lookups. This ensures optimal performance while checking every potential word segment against the dictionary quickly.
class Solution:
def canSegmentString(self, text: str, dictionary: List[str]) -> bool:
length = len(text)
word_set = set(dictionary)
can_segment = [False] * (length + 1)
can_segment[0] = True
for i in range(1, length + 1):
for j in range(i):
if can_segment[j] and text[j:i] in word_set:
can_segment[i] = True
break
return can_segment[length]
The given Python solution is designed to solve the "Word Break" problem. The goal is to determine if a given string (text
) can be segmented into one or more words that exist in a supplied dictionary.
The solution employs a dynamic programming approach where:
length
stores the size of the text.word_set
is a set created from the input dictionary for faster look-ups.can_segment
is a Boolean list, initialized to accommodate each character of the text plus an empty character scenario. The first element is set toTrue
, representing an empty string.
The function iterates through the
text
by:- Dividing
text
into all possible substrings. - Checking each substring against the
word_set
. - For each valid substring (one found in
word_set
), it updates thecan_segment
list to record the positions where valid breaks occur.
- Dividing
It continues segmenting until it either finds a valid sequence of words covering the entire string or exhausts all possibilities.
At the end of execution, the result reflects in can_segment[length]
, which indicates if the entire text
can be segmented based on the dictionary provided. This approach ensures that you capture whether a holistic solution to the provided text exists.
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