Input:  s = "leetcode", wordDict = ["leet", "code"]
Output: true

Explanation: "leetcode" can be segmented as "leet code".

Input:  s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false

Explanation: No valid segmentation exists.

dp[i] = true if s[0..i-1] (prefix of length i) can be segmented into valid words

dp[i] = true if there exists j < i such that:
  - dp[j] is true (prefix s[0..j-1] is segmentable), AND
  - s[j..i-1] is a valid word in the dictionary

dp[0] = true    (empty string is always segmentable — vacuously true)

function wordBreak(s, wordDict) {
    const n = s.length;
    const wordSet = new Set(wordDict);
    const dp = new Array(n + 1).fill(false);
    dp[0] = true;  // Empty string is segmentable

    for (let i = 1; i <= n; i++) {
        for (let j = 0; j < i; j++) {
            if (dp[j] && wordSet.has(s.substring(j, i))) {
                dp[i] = true;
                break;  // Found a valid segmentation — no need to check further
            }
        }
    }

    return dp[n];
}

console.log(wordBreak("leetcode", ["leet", "code"]));           // true
console.log(wordBreak("applepenapple", ["apple", "pen"]));      // true
console.log(wordBreak("catsandog", ["cats", "dog", "sand", "and", "cat"])); // false

s = "leetcode"
wordSet = {"leet", "code"}

dp[0] = true

i=1: "l" → not in dict → dp[1] = false
i=2: "le" → not in dict → dp[2] = false
i=3: "lee" → not in dict → dp[3] = false
i=4: j=0: dp[0]=T, "leet" in dict → dp[4] = true ✅

i=5: j=4: dp[4]=T, "c" not in dict
     others: no match → dp[5] = false

i=6: j=4: dp[4]=T, "co" not in dict
     others: no match → dp[6] = false

i=7: j=4: dp[4]=T, "cod" not in dict
     others: no match → dp[7] = false

i=8: j=4: dp[4]=T, "code" in dict → dp[8] = true ✅

Answer: true

function wordBreak(s, wordDict) {
    const n = s.length;
    const wordSet = new Set(wordDict);
    const dp = new Array(n + 1).fill(false);
    dp[0] = true;

    for (let i = 1; i <= n; i++) {
        for (const word of wordDict) {
            const len = word.length;
            if (i >= len && dp[i - len] && s.substring(i - len, i) === word) {
                dp[i] = true;
                break;
            }
        }
    }

    return dp[n];
}

function wordBreakBFS(s, wordDict) {
    const wordSet = new Set(wordDict);
    const n = s.length;
    const visited = new Array(n).fill(false);
    const queue = [0];

    while (queue.length > 0) {
        const start = queue.shift();

        if (start === n) return true;  // Reached the end

        if (visited[start]) continue;
        visited[start] = true;

        for (let end = start + 1; end <= n; end++) {
            if (wordSet.has(s.substring(start, end))) {
                queue.push(end);  // Found a valid word ending at 'end'
            }
        }
    }

    return false;
}

console.log(wordBreakBFS("leetcode", ["leet", "code"]));  // true

function wordBreak(s, wordDict) {
    const wordSet = new Set(wordDict);
    const memo = new Map();  // key: start index → [sentences]

    function dfs(start) {
        if (start === s.length) return [""];  // Base: empty suffix
        if (memo.has(start)) return memo.get(start);

        const sentences = [];

        for (let end = start + 1; end <= s.length; end++) {
            const word = s.substring(start, end);
            if (wordSet.has(word)) {
                const subSentences = dfs(end);
                for (const sub of subSentences) {
                    sentences.push(sub ? word + " " + sub : word);
                }
            }
        }

        memo.set(start, sentences);
        return sentences;
    }

    return dfs(0);
}

console.log(wordBreak("catsanddog", ["cat", "cats", "and", "sand", "dog"]));
// ["cat sand dog", "cats and dog"]

console.log(wordBreak("pineapplepenapple", ["apple", "pen", "applepen", "pine", "pineapple"]));
// ["pine apple pen apple", "pineapple pen apple", "pine applepen apple"]

dfs(0) → tries:
  "cat" at (0,3): dfs(3) → returns ["sand dog"]
    → "cat sand dog"
  "cats" at (0,4): dfs(4) → returns ["and dog"]
    → "cats and dog"
  → returns ["cat sand dog", "cats and dog"]

dfs(3) → tries:
  "sand" at (3,7): dfs(7) → returns ["dog"]
    → "sand dog"
  → returns ["sand dog"]

dfs(4) → tries:
  "and" at (4,7): dfs(7) → returns ["dog"]
    → "and dog"
  → returns ["and dog"]

dfs(7) → tries:
  "dog" at (7,10): dfs(10) → returns [""]
    → "dog"
  → returns ["dog"]

dfs(10) → returns [""] (base case)