21. Word Break
mediumAsked at RedisDecide whether a string can be segmented into space-separated dictionary words; Redis uses it as a DP/memoization probe that overlaps with how the engine matches CONFIG GET glob patterns.
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. Words may be reused.
Constraints
1 <= s.length <= 3001 <= wordDict.length <= 10001 <= wordDict[i].length <= 20
Examples
Example 1
s='leetcode', wordDict=['leet','code']trueExample 2
s='catsandog', wordDict=['cats','dog','sand','and','cat']falseApproaches
1. Recursive try every split
Try each prefix; recurse on suffix without memoization.
- Time
- O(2^n)
- Space
- O(n)
function tryBreak(s) {
if (s === '') return true;
for (const w of wordDict)
if (s.startsWith(w) && tryBreak(s.slice(w.length))) return true;
return false;
}Tradeoff:
2. Bottom-up DP
dp[i] is true if s[0..i] is segmentable. Walk i = 1..n and check every j < i where dp[j] && wordSet.has(s[j..i]). Mirrors how Redis caches sub-results to avoid recomputing in nested LUA pcall stacks.
- Time
- O(n^2 * L)
- Space
- O(n)
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:
Redis-specific tips
Redis interviewers like the DP framing and reward you for mentioning how you'd back wordDict with a Redis SET membership lookup for distributed dictionary scenarios.
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