139. Word Break - Explanation

Description

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of dictionary words.

You are allowed to reuse words in the dictionary an unlimited number of times. You may assume all dictionary words are unique.

Example 1:

Input: s = "neetcode", wordDict = ["neet","code"]

Output: true

Explanation: Return true because "neetcode" can be split into "neet" and "code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple","pen","ape"]

Output: true

Explanation: Return true because "applepenapple" can be split into "apple", "pen" and "apple". Notice that we can reuse words and also not use all the words.

Example 3:

Input: s = "catsincars", wordDict = ["cats","cat","sin","in","car"]

Output: false

Constraints:

1 <= s.length <= 200
1 <= wordDict.length <= 100
1 <= wordDict[i].length <= 20
s and wordDict[i] consist of only lowercase English letters.

Topics

Recommended Time & Space Complexity

You should aim for a solution as good or better than O(n * m * t) time and O(n) space, where n is the length of the string s, m is the number of words in wordDict, and t is the maximum length of any word in wordDict.

Hint 1

Try to think of this problem in terms of recursion, where we explore all possibilities. We iterate through the given string s, attempting to pick a word from wordDict that matches a portion of s, and then recursively continue processing the remaining string. Can you determine the recurrence relation and base condition?

Hint 2

The base condition is to return true if we reach the end of the string s. At each recursive call with index i iterating through s, we check all words in wordDict and recursively process the remaining string by incrementing i by the length of the matched word. If any recursive path returns true, we immediately return true. However, this solution is exponential. Can you think of an optimization? Maybe you should consider an approach that avoids repeated work.

Hint 3

We can avoid recalculating results for recursive calls by using memoization. Since we iterate with index i, we can use a hash map or an array of the same length as s to cache the results of recursive calls and prevent redundant computations.

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Prerequisites

Before attempting this problem, you should be comfortable with:

Recursion - Understanding how to break problems into smaller subproblems and handle base cases
Dynamic Programming - Both memoization (top-down) and tabulation (bottom-up) approaches
Hash Set - Using sets for O(1) lookup to efficiently check if a word exists in the dictionary
String Manipulation - Substring operations and string comparison
Trie (Optional) - A tree-based data structure for efficient prefix matching and word lookup

1. Recursion

Intuition

At every index i in the string, we want to decide:

Can the suffix starting at index i be segmented into valid dictionary words?

The recursive idea is:

Try every word in wordDict
If a word matches the string starting at position i
Recursively check whether the remaining substring (starting at i + len(word)) can also be broken successfully

If any path reaches the end of the string, the answer is true.

This is a classic decision-based recursion where:

Each index i represents a subproblem
Base case: reaching the end means a valid segmentation

Algorithm

Define a recursive function dfs(i):
- If i == len(s), return true
For each word w in wordDict:
- Check if w matches s[i : i + len(w)]
- If it matches and dfs(i + len(w)) is true, return true
If no word leads to a valid segmentation, return false
Start recursion from index 0

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:

        def dfs(i):
            if i == len(s):
                return True

            for w in wordDict:
                if ((i + len(w)) <= len(s) and
                     s[i : i + len(w)] == w
                ):
                    if dfs(i + len(w)):
                        return True
            return False

        return dfs(0)

public class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        return dfs(s, wordDict, 0);
    }

    private boolean dfs(String s, List<String> wordDict, int i) {
        if (i == s.length()) {
            return true;
        }

        for (String w : wordDict) {
            if (i + w.length() <= s.length() &&
                s.substring(i, i + w.length()).equals(w)) {
                if (dfs(s, wordDict, i + w.length())) {
                    return true;
                }
            }
        }
        return false;
    }
}

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        return dfs(s, wordDict, 0);
    }

private:
    bool dfs(const string& s, const vector<string>& wordDict, int i) {
        if (i == s.length()) {
            return true;
        }

        for (const string& w : wordDict) {
            if (i + w.length() <= s.length() &&
                s.substr(i, w.length()) == w) {
                if (dfs(s, wordDict, i + w.length())) {
                    return true;
                }
            }
        }
        return false;
    }
};

class Solution {
    /**
     * @param {string} s
     * @param {string[]} wordDict
     * @return {boolean}
     */
    wordBreak(s, wordDict) {
        return this.dfs(s, wordDict, 0);
    }

    /**
     * @param {string} s
     * @param {string[]} wordDict
     * @param {number} i
     * @return {boolean}
     */
    dfs(s, wordDict, i) {
        if (i === s.length) {
            return true;
        }

        for (let w of wordDict) {
            if (
                i + w.length <= s.length &&
                s.substring(i, i + w.length) === w
            ) {
                if (this.dfs(s, wordDict, i + w.length)) {
                    return true;
                }
            }
        }
        return false;
    }
}

public class Solution {
    public bool WordBreak(string s, List<string> wordDict) {
        return Dfs(s, wordDict, 0);
    }

    private bool Dfs(string s, List<string> wordDict, int i) {
        if (i == s.Length) {
            return true;
        }

        foreach (string w in wordDict) {
            if (i + w.Length <= s.Length &&
                s.Substring(i, w.Length) == w) {
                if (Dfs(s, wordDict, i + w.Length)) {
                    return true;
                }
            }
        }
        return false;
    }
}

func wordBreak(s string, wordDict []string) bool {
    return dfs(s, wordDict, 0)
}

func dfs(s string, wordDict []string, i int) bool {
    if i == len(s) {
        return true
    }

    for _, w := range wordDict {
        if len(s[i:]) >= len(w) && s[i:i+len(w)] == w {
            if dfs(s, wordDict, i+len(w)) {
                return true
            }
        }
    }

    return false
}

class Solution {
    fun wordBreak(s: String, wordDict: List<String>): Boolean {
        return dfs(s, wordDict, 0)
    }

    private fun dfs(s: String, wordDict: List<String>, i: Int): Boolean {
        if (i == s.length) {
            return true
        }

        for (w in wordDict) {
            if (s.length - i >= w.length && s.substring(i, i + w.length) == w) {
                if (dfs(s, wordDict, i + w.length)) {
                    return true
                }
            }
        }

        return false
    }
}

class Solution {
    func wordBreak(_ s: String, _ wordDict: [String]) -> Bool {
        let chars = Array(s)

        func dfs(_ i: Int) -> Bool {
            if i == chars.count {
                return true
            }

            for w in wordDict {
                if i + w.count <= chars.count,
                   String(chars[i..<i + w.count]) == w {
                    if dfs(i + w.count) {
                        return true
                    }
                }
            }
            return false
        }

        return dfs(0)
    }
}

Time & Space Complexity

Time complexity: $O(t * m ^ n)$
Space complexity: $O(n)$

Where $n$ is the length of the string $s$ , $m$ is the number of words in $wordDict$ and $t$ is the maximum length of any word in $wordDict$ .

2. Recursion (Hash Set)

Intuition

This version improves the brute-force recursion by optimizing word lookup.

Instead of trying every word from wordDict at each index, we:

Fix a starting index i
Try all possible substrings s[i : j+1]
Check if the substring exists in a Hash Set (O(1) lookup)

If a valid word is found:

Recursively check whether the remaining suffix starting at j + 1 can be segmented

The key idea:

If we can split the string at any valid word boundary and the rest is solvable, then the whole string is solvable.

Algorithm

Convert wordDict into a hash set wordSet for fast lookup
Define a recursive function dfs(i):
- If i == len(s), return true
For every j from i to len(s) - 1:
- If s[i : j + 1] is in wordSet
  - If dfs(j + 1) is true, return true
If no split works, return false
Start recursion from index 0

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        wordSet = set(wordDict)

        def dfs(i):
            if i == len(s):
                return True

            for j in range(i, len(s)):
                if s[i : j + 1] in wordSet:
                    if dfs(j + 1):
                        return True
            return False

        return dfs(0)

public class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        HashSet<String> wordSet = new HashSet<>(wordDict);

        return dfs(s, wordSet, 0);
    }

    private boolean dfs(String s, HashSet<String> wordSet, int i) {
        if (i == s.length()) {
            return true;
        }

        for (int j = i; j < s.length(); j++) {
            if (wordSet.contains(s.substring(i, j + 1))) {
                if (dfs(s, wordSet, j + 1)) {
                    return true;
                }
            }
        }
        return false;
    }
}

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> wordSet(wordDict.begin(), wordDict.end());
        return dfs(s, wordSet, 0);
    }

    bool dfs(const string& s, const unordered_set<string>& wordSet, int i) {
        if (i == s.size()) {
            return true;
        }

        for (int j = i; j < s.size(); j++) {
            if (wordSet.find(s.substr(i, j - i + 1)) != wordSet.end()) {
                if (dfs(s, wordSet, j + 1)) {
                    return true;
                }
            }
        }
        return false;
    }
};

class Solution {
    /**
     * @param {string} s
     * @param {string[]} wordDict
     * @return {boolean}
     */
    wordBreak(s, wordDict) {
        const wordSet = new Set(wordDict);
        const dfs = (i) => {
            if (i === s.length) {
                return true;
            }

            for (let j = i; j < s.length; j++) {
                if (wordSet.has(s.substring(i, j + 1))) {
                    if (dfs(j + 1)) {
                        return true;
                    }
                }
            }
            return false;
        };

        return dfs(0);
    }
}

public class Solution {
    public bool WordBreak(string s, List<string> wordDict) {
        HashSet<string> wordSet = new HashSet<string>(wordDict);
        return Dfs(s, wordSet, 0);
    }

    private bool Dfs(string s, HashSet<string> wordSet, int i) {
        if (i == s.Length) {
            return true;
        }

        for (int j = i; j < s.Length; j++) {
            if (wordSet.Contains(s.Substring(i, j - i + 1))) {
                if (Dfs(s, wordSet, j + 1)) {
                    return true;
                }
            }
        }
        return false;
    }
}

func wordBreak(s string, wordDict []string) bool {
    wordSet := make(map[string]bool)
    for _, w := range wordDict {
        wordSet[w] = true
    }
    return dfs(s, wordSet, 0)
}

func dfs(s string, wordSet map[string]bool, i int) bool {
    if i == len(s) {
        return true
    }

    for j := i; j < len(s); j++ {
        if wordSet[s[i:j+1]] {
            if dfs(s, wordSet, j+1) {
                return true
            }
        }
    }

    return false
}

class Solution {
    fun wordBreak(s: String, wordDict: List<String>): Boolean {
        val wordSet = wordDict.toHashSet()
        return dfs(s, wordSet, 0)
    }

    private fun dfs(s: String, wordSet: HashSet<String>, i: Int): Boolean {
        if (i == s.length) {
            return true
        }

        for (j in i until s.length) {
            if (wordSet.contains(s.substring(i, j + 1))) {
                if (dfs(s, wordSet, j + 1)) {
                    return true
                }
            }
        }

        return false
    }
}

class Solution {
    func wordBreak(_ s: String, _ wordDict: [String]) -> Bool {
        let wordSet = Set(wordDict)
        let chars = Array(s)

        func dfs(_ i: Int) -> Bool {
            if i == chars.count {
                return true
            }

            for j in i..<chars.count {
                if wordSet.contains(String(chars[i...j])) {
                    if dfs(j + 1) {
                        return true
                    }
                }
            }
            return false
        }

        return dfs(0)
    }
}

Time & Space Complexity

Time complexity: $O( (n * 2 ^ n) + m)$
Space complexity: $O(n + (m * t))$

Where $n$ is the length of the string $s$ and $m$ is the number of words in $wordDict$ .

3. Dynamic Programming (Top-Down)

Intuition

This is an optimized version of recursion using memoization.

The key observation:

While recursively checking splits, the same index i is reached many times
The result of dfs(i) (can suffix s[i:] be segmented?) never changes

So we cache the result for each index:

If dfs(i) was already computed, reuse it
This avoids recomputing exponential subtrees

In short:

Convert exponential recursion into linear states using memoization.

Algorithm

Use a hash map memo where:
- memo[i] = true/false means whether s[i:] can be segmented
- Base case: memo[len(s)] = true
Define dfs(i):
- If i is in memo, return memo[i]
For each word w in wordDict:
- If s[i : i + len(w)] == w
  - Recursively call dfs(i + len(w))
  - If it returns true, store memo[i] = true and return true
If no word leads to a valid split:
- Store memo[i] = false
Return dfs(0)

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        memo = {len(s) : True}
        def dfs(i):
            if i in memo:
                return memo[i]

            for w in wordDict:
                if ((i + len(w)) <= len(s) and
                     s[i : i + len(w)] == w
                ):
                    if dfs(i + len(w)):
                        memo[i] = True
                        return True
            memo[i] = False
            return False

        return dfs(0)

public class Solution {
    private Map<Integer, Boolean> memo;

    public boolean wordBreak(String s, List<String> wordDict) {
        memo = new HashMap<>();
        memo.put(s.length(), true);
        return dfs(s, wordDict, 0);
    }

    private boolean dfs(String s, List<String> wordDict, int i) {
        if (memo.containsKey(i)) {
            return memo.get(i);
        }

        for (String w : wordDict) {
            if (i + w.length() <= s.length() &&
                s.substring(i, i + w.length()).equals(w)) {
                if (dfs(s, wordDict, i + w.length())) {
                    memo.put(i, true);
                    return true;
                }
            }
        }
        memo.put(i, false);
        return false;
    }
}

class Solution {
public:
    unordered_map<int, bool> memo;

    bool wordBreak(string s, vector<string>& wordDict) {
        memo[s.length()] = true;
        return dfs(s, wordDict, 0);
    }

    bool dfs(string& s, vector<string>& wordDict, int i) {
        if (memo.find(i) != memo.end()) {
            return memo[i];
        }

        for (const string& w : wordDict) {
            if (i + w.length() <= s.length() &&
                s.substr(i, w.length()) == w) {
                if (dfs(s, wordDict, i + w.length())) {
                    memo[i] = true;
                    return true;
                }
            }
        }
        memo[i] = false;
        return false;
    }
};

class Solution {
    constructor() {
        this.memo = {};
    }

    /**
     * @param {string} s
     * @param {string[]} wordDict
     * @return {boolean}
     */
    wordBreak(s, wordDict) {
        this.memo = { [s.length]: true };
        return this.dfs(s, wordDict, 0);
    }

    /**
     * @param {string} s
     * @param {string[]} wordDict
     * @param {number} i
     * @return {boolean}
     */
    dfs(s, wordDict, i) {
        if (i in this.memo) {
            return this.memo[i];
        }

        for (let w of wordDict) {
            if (
                i + w.length <= s.length &&
                s.substring(i, i + w.length) === w
            ) {
                if (this.dfs(s, wordDict, i + w.length)) {
                    this.memo[i] = true;
                    return true;
                }
            }
        }
        this.memo[i] = false;
        return false;
    }
}

public class Solution {
    private Dictionary<int, bool> memo;

    public bool WordBreak(string s, List<string> wordDict) {
        memo = new Dictionary<int, bool> { { s.Length, true } };
        return Dfs(s, wordDict, 0);
    }

    private bool Dfs(string s, List<string> wordDict, int i) {
        if (memo.ContainsKey(i)) {
            return memo[i];
        }

        foreach (var w in wordDict) {
            if (i + w.Length <= s.Length &&
                s.Substring(i, w.Length) == w) {
                if (Dfs(s, wordDict, i + w.Length)) {
                    memo[i] = true;
                    return true;
                }
            }
        }
        memo[i] = false;
        return false;
    }
}

func wordBreak(s string, wordDict []string) bool {
    memo := make(map[int]bool)
    memo[len(s)] = true

    var dfs func(int) bool
    dfs = func(i int) bool {
        if val, found := memo[i]; found {
            return val
        }

        for _, w := range wordDict {
            if i+len(w) <= len(s) && s[i:i+len(w)] == w {
                if dfs(i + len(w)) {
                    memo[i] = true
                    return true
                }
            }
        }

        memo[i] = false
        return false
    }

    return dfs(0)
}

class Solution {
    fun wordBreak(s: String, wordDict: List<String>): Boolean {
        val memo = hashMapOf(s.length to true)

        fun dfs(i: Int): Boolean {
            memo[i]?.let { return it }

            for (w in wordDict) {
                if (i + w.length <= s.length &&
                    s.substring(i, i + w.length) == w) {
                    if (dfs(i + w.length)) {
                        memo[i] = true
                        return true
                    }
                }
            }

            memo[i] = false
            return false
        }

        return dfs(0)
    }
}

class Solution {
    func wordBreak(_ s: String, _ wordDict: [String]) -> Bool {
        var memo = [Int: Bool]()
        memo[s.count] = true
        let chars = Array(s)

        func dfs(_ i: Int) -> Bool {
            if let cached = memo[i] {
                return cached
            }

            for w in wordDict {
                if i + w.count <= chars.count,
                   String(chars[i..<i + w.count]) == w {
                    if dfs(i + w.count) {
                        memo[i] = true
                        return true
                    }
                }
            }
            memo[i] = false
            return false
        }

        return dfs(0)
    }
}

Time & Space Complexity

Time complexity: $O(n * m * t)$
Space complexity: $O(n)$

Where $n$ is the length of the string $s$ , $m$ is the number of words in $wordDict$ and $t$ is the maximum length of any word in $wordDict$ .

4. Dynamic Programming (Hash Set)

Intuition

This approach is a top-down dynamic programming solution with pruning.

Key ideas:

Checking every possible substring is expensive.
A word can only be as long as the maximum word length in wordDict.
Use a Hash Set for O(1) word lookup.
Use memoization so each index in the string is solved only once.

So we:

Limit how far we try to split from each index
Cache results for indices to avoid repeated work

This turns exponential recursion into efficient DP.

Algorithm

Convert wordDict into a hash set wordSet
Compute t = maximum length of any word in wordDict
Use a memo map where memo[i] means:
- Can substring s[i:] be segmented?
Define dfs(i):
- If i is in memo, return memo[i]
- If i == len(s), return true
For j from i to min(len(s), i + t) - 1:
- If s[i : j + 1] is in wordSet
  - If dfs(j + 1) is true, store and return true
If no valid split works:
- Store memo[i] = false
Return dfs(0)

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        wordSet = set(wordDict)
        t = 0
        for w in wordDict:
            t = max(t, len(w))

        memo = {}
        def dfs(i):
            if i in memo:
                return memo[i]
            if i == len(s):
                return True
            for j in range(i, min(len(s), i + t)):
                if s[i : j + 1] in wordSet:
                    if dfs(j + 1):
                        memo[i] = True
                        return True
            memo[i] = False
            return False

        return dfs(0)

public class Solution {
    private HashSet<String> wordSet;
    private Boolean[] memo;
    private int t;

    public boolean wordBreak(String s, List<String> wordDict) {
        wordSet = new HashSet<>(wordDict);
        memo = new Boolean[s.length()];
        t = 0;
        for (int i = 0; i < wordDict.size(); i++) {
            t = Math.max(t, wordDict.get(i).length());
        }
        return dfs(s, 0);
    }

    private boolean dfs(String s, int i) {
        if (i == s.length()) {
            return true;
        }
        if (memo[i] != null) {
            return memo[i];
        }

        for (int j = i; j < Math.min(i + t, s.length()); j++) {
            if (wordSet.contains(s.substring(i, j + 1))) {
                if (dfs(s, j + 1)) {
                    memo[i] = true;
                    return true;
                }
            }
        }
        memo[i] = false;
        return false;
    }
}

class Solution {
public:
    unordered_set<string> wordSet;
    vector<int> memo;
    int t;

    bool wordBreak(string s, vector<string>& wordDict) {
        wordSet.insert(wordDict.begin(), wordDict.end());
        memo.resize(s.size(), -1);
        t = 0;
        for (string& w : wordDict) {
            t = max(t, int(w.length()));
        }
        return dfs(s, 0);
    }

    bool dfs(string& s, int i) {
        if (i == s.size()) {
            return true;
        }
        if (memo[i] != -1) {
            return memo[i] == 1;
        }

        for (int j = i; j < (i + t, s.size()); j++) {
            if (wordSet.count(s.substr(i, j - i + 1))) {
                if (dfs(s, j + 1)) {
                    memo[i] = 1;
                    return true;
                }
            }
        }
        memo[i] = 0;
        return false;
    }
};

class Solution {
    /**
     * @param {string} s
     * @param {string[]} wordDict
     * @return {boolean}
     */
    wordBreak(s, wordDict) {
        this.wordSet = new Set(wordDict);
        this.memo = new Array(s.length).fill(null);
        this.t = 0;
        for (const w of wordDict) {
            this.t = Math.max(this.t, w.length);
        }
        return this.dfs(s, 0);
    }

    /**
     * @param {string} s
     * @param {number} i
     * @return {boolean}
     */
    dfs(s, i) {
        if (i === s.length) {
            return true;
        }
        if (this.memo[i] !== null) {
            return this.memo[i];
        }

        for (let j = i; j < Math.min(i + this.t, s.length); j++) {
            if (this.wordSet.has(s.substring(i, j + 1))) {
                if (this.dfs(s, j + 1)) {
                    this.memo[i] = true;
                    return true;
                }
            }
        }
        this.memo[i] = false;
        return false;
    }
}

public class Solution {
    private HashSet<string> wordSet;
    private Dictionary<int, bool> memo;
    private int t;

    public bool WordBreak(string s, List<string> wordDict) {
        wordSet = new HashSet<string>(wordDict);
        memo = new Dictionary<int, bool>();
        t = 0;
        foreach (var w in wordDict) {
            t = Math.Max(t, w.Length);
        }
        return Dfs(s, 0);
    }

    private bool Dfs(string s, int i) {
        if (i == s.Length) {
            return true;
        }
        if (memo.ContainsKey(i)) {
            return memo[i];
        }

        for (int j = i; j < Math.Min(i + t, s.Length); j++) {
            if (wordSet.Contains(s.Substring(i, j - i + 1))) {
                if (Dfs(s, j + 1)) {
                    memo[i] = true;
                    return true;
                }
            }
        }
        memo[i] = false;
        return false;
    }
}

func wordBreak(s string, wordDict []string) bool {
    wordSet := make(map[string]bool)
    maxLen := 0

    for _, w := range wordDict {
        wordSet[w] = true
        if len(w) > maxLen {
            maxLen = len(w)
        }
    }

    memo := make(map[int]bool)

    var dfs func(int) bool
    dfs = func(i int) bool {
        if val, found := memo[i]; found {
            return val
        }
        if i == len(s) {
            return true
        }
        for j := i; j < len(s) && j < i+maxLen; j++ {
            if wordSet[s[i:j+1]] {
                if dfs(j + 1) {
                    memo[i] = true
                    return true
                }
            }
        }
        memo[i] = false
        return false
    }

    return dfs(0)
}

class Solution {
    fun wordBreak(s: String, wordDict: List<String>): Boolean {
        val wordSet = wordDict.toSet()
        val maxLen = wordDict.maxOfOrNull { it.length } ?: 0
        val memo = HashMap<Int, Boolean>()

        fun dfs(i: Int): Boolean {
            memo[i]?.let { return it }
            if (i == s.length) return true

            for (j in i until minOf(s.length, i + maxLen)) {
                if (s.substring(i, j + 1) in wordSet) {
                    if (dfs(j + 1)) {
                        memo[i] = true
                        return true
                    }
                }
            }

            memo[i] = false
            return false
        }

        return dfs(0)
    }
}

class Solution {
    func wordBreak(_ s: String, _ wordDict: [String]) -> Bool {
        let wordSet = Set(wordDict)
        var maxLen = 0
        for w in wordDict {
            maxLen = max(maxLen, w.count)
        }

        var memo = [Int: Bool]()
        let chars = Array(s)

        func dfs(_ i: Int) -> Bool {
            if let cached = memo[i] {
                return cached
            }
            if i == chars.count {
                return true
            }
            for j in i..<min(chars.count, i + maxLen) {
                if wordSet.contains(String(chars[i...j])) {
                    if dfs(j + 1) {
                        memo[i] = true
                        return true
                    }
                }
            }
            memo[i] = false
            return false
        }

        return dfs(0)
    }
}

Time & Space Complexity

Time complexity: $O((t ^ 2 * n) + m)$
Space complexity: $O(n + (m * t))$

Where $n$ is the length of the string $s$ , $m$ is the number of words in $wordDict$ and $t$ is the maximum length of any word in $wordDict$ .

5. Dynamic Programming (Bottom-Up)

Intuition

This is a bottom-up dynamic programming approach.

Instead of trying to split the string recursively, we solve the problem from the end of the string toward the start.

Key idea:

dp[i] means whether the substring s[i:] can be segmented
If we know the answer for future positions, we can decide the current one
We reuse already computed results → no recursion, no stack overhead

Algorithm

Create a boolean array dp of size len(s) + 1
- dp[i] = true if s[i:] can be segmented
Base case:
- dp[len(s)] = true (empty string is valid)
Iterate i from len(s) - 1 down to 0:
- For each word w in wordDict:
  - If s[i : i + len(w)] == w
    - Set dp[i] = dp[i + len(w)]
  - If dp[i] becomes true, break early
Return dp[0]

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        dp = [False] * (len(s) + 1)
        dp[len(s)] = True

        for i in range(len(s) - 1, -1, -1):
            for w in wordDict:
                if (i + len(w)) <= len(s) and s[i : i + len(w)] == w:
                    dp[i] = dp[i + len(w)]
                if dp[i]:
                    break

        return dp[0]

public class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        boolean[] dp = new boolean[s.length() + 1];
        dp[s.length()] = true;

        for (int i = s.length() - 1; i >= 0; i--) {
            for (String w : wordDict) {
                if ((i + w.length()) <= s.length() &&
                     s.substring(i, i + w.length()).equals(w)) {
                    dp[i] = dp[i + w.length()];
                }
                if (dp[i]) {
                    break;
                }
            }
        }

        return dp[0];
    }
}

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        vector<bool> dp(s.size() + 1, false);
        dp[s.size()] = true;

        for (int i = s.size() - 1; i >= 0; i--) {
            for (const auto& w : wordDict) {
                if ((i + w.size()) <= s.size() &&
                     s.substr(i, w.size()) == w) {
                    dp[i] = dp[i + w.size()];
                }
                if (dp[i]) {
                    break;
                }
            }
        }

        return dp[0];
    }
};

class Solution {
    /**
     * @param {string} s
     * @param {string[]} wordDict
     * @return {boolean}
     */
    wordBreak(s, wordDict) {
        const dp = new Array(s.length + 1).fill(false);
        dp[s.length] = true;

        for (let i = s.length - 1; i >= 0; i--) {
            for (const w of wordDict) {
                if (
                    i + w.length <= s.length &&
                    s.slice(i, i + w.length) === w
                ) {
                    dp[i] = dp[i + w.length];
                }
                if (dp[i]) {
                    break;
                }
            }
        }

        return dp[0];
    }
}

public class Solution {
    public bool WordBreak(string s, List<string> wordDict) {
        bool[] dp = new bool[s.Length + 1];
        dp[s.Length] = true;

        for (int i = s.Length - 1; i >= 0; i--) {
            foreach (string w in wordDict) {
                if ((i + w.Length) <= s.Length &&
                     s.Substring(i, w.Length) == w) {
                    dp[i] = dp[i + w.Length];
                }
                if (dp[i]) {
                    break;
                }
            }
        }

        return dp[0];
    }
}

class Solution {
    fun wordBreak(s: String, wordDict: List<String>): Boolean {
        val dp = BooleanArray(s.length + 1)
        dp[s.length] = true

        for (i in s.length - 1 downTo 0) {
            for (w in wordDict) {
                if (i + w.length <= s.length &&
                    s.substring(i, i + w.length) == w) {
                    dp[i] = dp[i + w.length]
                }
                if (dp[i]) break
            }
        }

        return dp[0]
    }
}

class Solution {
    func wordBreak(_ s: String, _ wordDict: [String]) -> Bool {
        var dp = Array(repeating: false, count: s.count + 1)
        dp[s.count] = true
        let chars = Array(s)

        for i in stride(from: s.count - 1, through: 0, by: -1) {
            for w in wordDict {
                if i + w.count <= chars.count, String(chars[i..<i + w.count]) == w {
                    dp[i] = dp[i + w.count]
                }
                if dp[i] {
                    break
                }
            }
        }

        return dp[0]
    }
}

Time & Space Complexity

Time complexity: $O(n * m * t)$
Space complexity: $O(n)$

Where $n$ is the length of the string $s$ , $m$ is the number of words in $wordDict$ and $t$ is the maximum length of any word in $wordDict$ .

6. Dynamic Programming (Trie)

Intuition

The normal DP checks every word at every index, which can waste time comparing strings again and again.

A Trie stores all dictionary words like a prefix tree, so from any starting index i in s, we can walk forward character-by-character and quickly know:

whether the current prefix matches some dictionary word path
and when we hit a complete word (is_word = true)

We still use DP:

dp[i] = can we break the suffix s[i:] into dictionary words?
If from i we can reach some j where s[i..j] is a word, then dp[i] = dp[j+1]

Trie helps us find valid words starting at i efficiently.

Algorithm

Build a Trie from all words in wordDict.
Create a boolean DP array dp of size n + 1 where n = len(s).
- dp[n] = true (empty suffix is always valid)
Let t be the maximum word length in the dictionary (use it as a bound).
Fill DP from right to left:
- For each index i from n down to 0:
  - Try all end positions j from i to min(n-1, i+t-1):
    - If s[i..j] is a word in the Trie, set dp[i] = dp[j+1]
    - If dp[i] becomes true, stop early for this i
Return dp[0].

class TrieNode:
    def __init__(self):
        self.children = {}
        self.is_word = False

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, word):
        node = self.root
        for char in word:
            if char not in node.children:
                node.children[char] = TrieNode()
            node = node.children[char]
        node.is_word = True

    def search(self, s, i, j):
        node = self.root
        for idx in range(i, j + 1):
            if s[idx] not in node.children:
                return False
            node = node.children[s[idx]]
        return node.is_word

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        trie = Trie()
        for word in wordDict:
            trie.insert(word)

        dp = [False] * (len(s) + 1)
        dp[len(s)] = True

        t = 0
        for w in wordDict:
            t = max(t, len(w))

        for i in range(len(s), -1, -1):
            for j in range(i, min(len(s), i + t)):
                if trie.search(s, i, j):
                    dp[i] = dp[j + 1]
                    if dp[i]:
                        break

        return dp[0]

class TrieNode {
    HashMap<Character, TrieNode> children;
    boolean isWord;

    public TrieNode() {
        children = new HashMap<>();
        isWord = false;
    }
}

class Trie {
    TrieNode root;

    public Trie() {
        root = new TrieNode();
    }

    public void insert(String word) {
        TrieNode node = root;
        for (char c : word.toCharArray()) {
            node.children.putIfAbsent(c, new TrieNode());
            node = node.children.get(c);
        }
        node.isWord = true;
    }

    public boolean search(String s, int i, int j) {
        TrieNode node = root;
        for (int idx = i; idx <= j; idx++) {
            if (!node.children.containsKey(s.charAt(idx))) {
                return false;
            }
            node = node.children.get(s.charAt(idx));
        }
        return node.isWord;
    }
}

public class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        Trie trie = new Trie();
        for (String word : wordDict) {
            trie.insert(word);
        }

        int n = s.length();
        boolean[] dp = new boolean[n + 1];
        dp[n] = true;

        int maxLen = 0;
        for (String word : wordDict) {
            maxLen = Math.max(maxLen, word.length());
        }

        for (int i = n - 1; i >= 0; i--) {
            for (int j = i; j < Math.min(n, i + maxLen); j++) {
                if (trie.search(s, i, j)) {
                    dp[i] = dp[j + 1];
                    if (dp[i]) break;
                }
            }
        }

        return dp[0];
    }
}

class TrieNode {
public:
    unordered_map<char, TrieNode*> children;
    bool is_word = false;
};

class Trie {
public:
    TrieNode* root;

    Trie() {
        root = new TrieNode();
    }

    void insert(string word) {
        TrieNode* node = root;
        for (char c : word) {
            if (!node->children.count(c)) {
                node->children[c] = new TrieNode();
            }
            node = node->children[c];
        }
        node->is_word = true;
    }

    bool search(string& s, int i, int j) {
        TrieNode* node = root;
        for (int idx = i; idx <= j; ++idx) {
            if (!node->children.count(s[idx])) {
                return false;
            }
            node = node->children[s[idx]];
        }
        return node->is_word;
    }
};

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        Trie trie;
        for (string word : wordDict) {
            trie.insert(word);
        }

        int n = s.length();
        vector<bool> dp(n + 1, false);
        dp[n] = true;

        int maxLen = 0;
        for (string w : wordDict) {
            maxLen = max(maxLen, (int)w.size());
        }

        for (int i = n - 1; i >= 0; --i) {
            for (int j = i; j < min(n, i + maxLen); ++j) {
                if (trie.search(s, i, j)) {
                    dp[i] = dp[j + 1];
                    if (dp[i]) break;
                }
            }
        }

        return dp[0];
    }
};

class TrieNode {
    constructor() {
        this.children = {};
        this.isWord = false;
    }
}

class Trie {
    constructor() {
        this.root = new TrieNode();
    }

    /**
     * @param {string} word
     * @return {void}
     */
    insert(word) {
        let node = this.root;
        for (let char of word) {
            if (!node.children[char]) {
                node.children[char] = new TrieNode();
            }
            node = node.children[char];
        }
        node.isWord = true;
    }

    /**
     * @param {string} s
     * @param {number} i
     * @param {number} j
     * @return {boolean}
     */
    search(s, i, j) {
        let node = this.root;
        for (let idx = i; idx <= j; idx++) {
            if (!node.children[s[idx]]) {
                return false;
            }
            node = node.children[s[idx]];
        }
        return node.isWord;
    }
}

class Solution {
    /**
     * @param {string} s
     * @param {string[]} wordDict
     * @return {boolean}
     */
    wordBreak(s, wordDict) {
        const trie = new Trie();
        for (let word of wordDict) {
            trie.insert(word);
        }

        const dp = new Array(s.length + 1).fill(false);
        dp[s.length] = true;

        let maxLen = 0;
        for (let w of wordDict) {
            maxLen = Math.max(maxLen, w.length);
        }

        for (let i = s.length - 1; i >= 0; i--) {
            for (let j = i; j < Math.min(s.length, i + maxLen); j++) {
                if (trie.search(s, i, j)) {
                    dp[i] = dp[j + 1];
                    if (dp[i]) break;
                }
            }
        }

        return dp[0];
    }
}

public class TrieNode {
    public Dictionary<char, TrieNode> Children;
    public bool IsWord;

    public TrieNode() {
        Children = new Dictionary<char, TrieNode>();
        IsWord = false;
    }
}

public class Trie {
    public TrieNode Root;

    public Trie() {
        Root = new TrieNode();
    }

    public void Insert(string word) {
        TrieNode node = Root;
        foreach (char c in word) {
            if (!node.Children.ContainsKey(c)) {
                node.Children[c] = new TrieNode();
            }
            node = node.Children[c];
        }
        node.IsWord = true;
    }

    public bool Search(string s, int i, int j) {
        TrieNode node = Root;
        for (int idx = i; idx <= j; idx++) {
            if (!node.Children.ContainsKey(s[idx])) {
                return false;
            }
            node = node.Children[s[idx]];
        }
        return node.IsWord;
    }
}

public class Solution {
    public bool WordBreak(string s, IList<string> wordDict) {
        Trie trie = new Trie();
        foreach (string word in wordDict) {
            trie.Insert(word);
        }

        int n = s.Length;
        bool[] dp = new bool[n + 1];
        dp[n] = true;

        int maxLen = 0;
        foreach (string word in wordDict) {
            maxLen = Math.Max(maxLen, word.Length);
        }

        for (int i = n - 1; i >= 0; i--) {
            for (int j = i; j < Math.Min(n, i + maxLen); j++) {
                if (trie.Search(s, i, j)) {
                    dp[i] = dp[j + 1];
                    if (dp[i]) break;
                }
            }
        }

        return dp[0];
    }
}

type TrieNode struct {
    children map[rune]*TrieNode
    isWord   bool
}

func NewTrieNode() *TrieNode {
    return &TrieNode{children: make(map[rune]*TrieNode)}
}

type Trie struct {
    root *TrieNode
}

func NewTrie() *Trie {
    return &Trie{root: NewTrieNode()}
}

func (t *Trie) Insert(word string) {
    node := t.root
    for _, char := range word {
        if _, found := node.children[char]; !found {
            node.children[char] = NewTrieNode()
        }
        node = node.children[char]
    }
    node.isWord = true
}

func (t *Trie) Search(s string, i, j int) bool {
    node := t.root
    for idx := i; idx <= j; idx++ {
        char := rune(s[idx])
        if _, found := node.children[char]; !found {
            return false
        }
        node = node.children[char]
    }
    return node.isWord
}

func wordBreak(s string, wordDict []string) bool {
    trie := NewTrie()
    for _, word := range wordDict {
        trie.Insert(word)
    }

    dp := make([]bool, len(s)+1)
    dp[len(s)] = true

    maxLength := 0
    for _, word := range wordDict {
        if len(word) > maxLength {
            maxLength = len(word)
        }
    }

    for i := len(s) - 1; i >= 0; i-- {
        for j := i; j < len(s) && j < i+maxLength; j++ {
            if trie.Search(s, i, j) {
                dp[i] = dp[j+1]
                if dp[i] {
                    break
                }
            }
        }
    }

    return dp[0]
}

class TrieNode {
    val children = mutableMapOf<Char, TrieNode>()
    var isWord = false
}

class Trie {
    private val root = TrieNode()

    fun insert(word: String) {
        var node = root
        for (char in word) {
            node = node.children.computeIfAbsent(char) { TrieNode() }
        }
        node.isWord = true
    }

    fun search(s: String, i: Int, j: Int): Boolean {
        var node = root
        for (idx in i..j) {
            val char = s[idx]
            node = node.children[char] ?: return false
        }
        return node.isWord
    }
}

class Solution {
    fun wordBreak(s: String, wordDict: List<String>): Boolean {
        val trie = Trie()
        wordDict.forEach { trie.insert(it) }

        val dp = BooleanArray(s.length + 1)
        dp[s.length] = true

        val maxLength = wordDict.maxOfOrNull { it.length } ?: 0

        for (i in s.length - 1 downTo 0) {
            for (j in i until minOf(s.length, i + maxLength)) {
                if (trie.search(s, i, j)) {
                    dp[i] = dp[j + 1]
                    if (dp[i]) break
                }
            }
        }

        return dp[0]
    }
}

class TrieNode {
    var children = [Character: TrieNode]()
    var isWord = false
}

class Trie {
    private let root = TrieNode()

    func insert(_ word: String) {
        var node = root
        for char in word {
            if node.children[char] == nil {
                node.children[char] = TrieNode()
            }
            node = node.children[char]!
        }
        node.isWord = true
    }

    func search(_ s: [Character], _ i: Int, _ j: Int) -> Bool {
        var node = root
        for idx in i...j {
            if node.children[s[idx]] == nil {
                return false
            }
            node = node.children[s[idx]]!
        }
        return node.isWord
    }
}

class Solution {
    func wordBreak(_ s: String, _ wordDict: [String]) -> Bool {
        let trie = Trie()
        for word in wordDict {
            trie.insert(word)
        }

        var dp = Array(repeating: false, count: s.count + 1)
        dp[s.count] = true
        let chars = Array(s)

        var maxLen = 0
        for word in wordDict {
            maxLen = max(maxLen, word.count)
        }

        for i in stride(from: s.count, through: 0, by: -1) {
            for j in i..<min(s.count, i + maxLen) {
                if trie.search(chars, i, j) {
                    dp[i] = dp[j + 1]
                    if dp[i] {
                        break
                    }
                }
            }
        }

        return dp[0]
    }
}

Time & Space Complexity

Time complexity: $O((n * t ^ 2) + m)$
Space complexity: $O(n + (m * t))$

Where $n$ is the length of the string $s$ , $m$ is the number of words in $wordDict$ and $t$ is the maximum length of any word in $wordDict$ .

Common Pitfalls

Off-by-One Error in DP Array Initialization

The DP array needs size n + 1 to represent the state after processing all characters. Using size n causes index-out-of-bounds when checking dp[n] as the base case.

# Wrong: array too small
dp = [False] * len(s)
dp[len(s)] = True  # IndexError!

# Correct: need n+1 elements
dp = [False] * (len(s) + 1)
dp[len(s)] = True

Checking Substring Beyond String Length

When iterating through possible word matches, failing to check if the word extends beyond the string causes substring errors or incorrect matches.

# Wrong: may go out of bounds
for w in wordDict:
    if s[i:i + len(w)] == w:  # no length check

# Correct: verify word fits in remaining string
for w in wordDict:
    if i + len(w) <= len(s) and s[i:i + len(w)] == w:

Not Converting wordDict to a Set for Efficient Lookup

Using a list for the word dictionary when checking substrings against it results in O(m) lookup time per check, causing TLE on large inputs.

# Wrong: O(m) per lookup
if s[i:j+1] in wordDict:  # wordDict is a list

# Correct: O(1) average lookup
wordSet = set(wordDict)
if s[i:j+1] in wordSet: