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28. Word Break

mediumAsked at Spotify

Determine if a string can be segmented into dictionary words using DP — a tokenisation technique Spotify applies in NLP preprocessing of lyrics and podcast transcripts before feeding text to recommendation models.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words. The same dictionary word may be reused multiple times.

Constraints

  • 1 <= s.length <= 300
  • 1 <= wordDict.length <= 1000
  • 1 <= wordDict[i].length <= 20
  • All strings consist of lowercase English letters
  • All words in wordDict are unique

Examples

Example 1

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

Example 2

Input
s = "applepenapple", wordDict = ["apple","pen"]
Output
true

Explanation: Segmented as 'apple pen apple'.

Example 3

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

Approaches

1. Recursive with memoisation

Try every prefix at each position; if the prefix is in the dict, recurse on the remainder. Cache results to avoid re-solving the same suffix.

Time
O(n^2 * m) where m = avg word length
Space
O(n)
function wordBreak(s, wordDict) {
  const dict = new Set(wordDict);
  const memo = new Map();
  const dp = (start) => {
    if (start === s.length) return true;
    if (memo.has(start)) return memo.get(start);
    for (let end = start + 1; end <= s.length; end++) {
      if (dict.has(s.slice(start, end)) && dp(end)) {
        memo.set(start, true);
        return true;
      }
    }
    memo.set(start, false);
    return false;
  };
  return dp(0);
}

Tradeoff:

2. Bottom-up DP

dp[i] is true if s[0..i-1] can be segmented. For each position, check all previous true positions and see if the slice from there to i is a valid word.

Time
O(n^2)
Space
O(n + dict size)
function wordBreak(s, wordDict) {
  const dict = new Set(wordDict);
  const dp = new Array(s.length + 1).fill(false);
  dp[0] = true;
  for (let i = 1; i <= s.length; i++) {
    for (let j = 0; j < i; j++) {
      if (dp[j] && dict.has(s.slice(j, i))) {
        dp[i] = true;
        break;
      }
    }
  }
  return dp[s.length];
}

Tradeoff:

Spotify-specific tips

Spotify's NLP team tokenises podcast captions and lyric lines before building word embeddings. The word-break DP is the canonical sub-problem. Interviewers here want to see you set a Set for O(1) word lookup first — iterating the array for every slice check is the most common performance miss.

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Output

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