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21. Word Break

mediumAsked at Brex

Determine whether a string can be segmented into dictionary words — a DP string problem Brex applies to category-keyword parsing in its rules engine.

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.

Constraints

  • 1 <= s.length <= 300
  • 1 <= wordDict.length <= 1000
  • 1 <= wordDict[i].length <= 20
  • wordDict contains distinct strings

Examples

Example 1

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

Example 2

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

Approaches

1. Naive recursion

Try every prefix; if it is in the dict recurse on the suffix. Exponential without memoization.

Time
O(2^n)
Space
O(n)
function wb(s, dict) {
  if (!s.length) return true;
  for (let i = 1; i <= s.length; i++)
    if (dict.includes(s.slice(0,i)) && wb(s.slice(i), dict)) return true;
  return false;
}

Tradeoff:

2. Bottom-up DP with set lookup

dp[i] is true if s[0..i-1] can be segmented. For each position j <= i check if dp[j] is true and s[j..i-1] is in the word set. Runs in O(n^2) with O(1) set lookups.

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

Tradeoff:

Brex-specific tips

Brex asks about fintech infrastructure, multi-currency handling, and spend management algorithms. Expect LeetCode-style DSA focused on hash maps, sorting, and dynamic programming.

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Output

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