22. Word Break
mediumAsked at WiseDecide if a string segments into dictionary words — Wise uses it as a stand-in for whether a multi-currency payment can be decomposed into legal corridor hops.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given a string s and a dictionary of words, 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 <= each word length <= 20
Examples
Example 1
s='leetcode', wordDict=['leet','code']trueExample 2
s='applepenapple', wordDict=['apple','pen']trueApproaches
1. Naive recursion
At each index try every prefix that matches a word and recurse on the suffix.
- Time
- O(2^n)
- Space
- O(n)
const set=new Set(wordDict);
function f(i){
if (i===s.length) return true;
for (let j=i+1;j<=s.length;j++)
if (set.has(s.slice(i,j)) && f(j)) return true;
return false;
}
return f(0);Tradeoff:
2. Bottom-up DP
dp[i] = true if s[0..i) is segmentable; for each i scan j<i and check dp[j] && s.slice(j,i) in dict. Linear table, O(n^2) scan.
- Time
- O(n^2)
- 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:
Wise-specific tips
Wise grades the DP framing — their corridor-decomposition engine uses the same dp[i] reachability shape, and they want to see you reach for it instead of memoized recursion.
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