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20. Word Break

mediumAsked at Instacart

Determine if a string can be segmented using a dictionary — Instacart applies this DP pattern when parsing user-typed grocery queries that contain concatenated product tokens against the catalog word list.

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 word in the dictionary may be reused multiple times.

Constraints

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

Examples

Example 1

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

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

Example 2

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

Approaches

1. Recursive with memoization

Try all prefix splits from the current index. If a prefix is in the dictionary and the suffix can also be segmented (memoized), return true.

Time
O(n^2 * m)
Space
O(n)
function wordBreak(s, wordDict) {
  const wordSet = new Set(wordDict);
  const memo = new Map();

  function 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 (wordSet.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] = true if s[0..i) can be segmented. For each position i, check all j < i where dp[j] is true and s[j..i) is in the dictionary.

Time
O(n^2)
Space
O(n)
function wordBreak(s, wordDict) {
  const wordSet = 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] && wordSet.has(s.slice(j, i))) {
        dp[i] = true;
        break;
      }
    }
  }
  return dp[s.length];
}

Tradeoff:

Instacart-specific tips

Instacart's search team parses concatenated product strings against a large SKU dictionary — the interviewer wants to see you build the DP table iteratively rather than recurse, and confirm you understand that a Set lookup on the substring is O(word length), not O(1), so the true complexity depends on average word length.

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Output

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