Word Break
A medium Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 2
- Core for
- 9 roles
The problem
You get a string and a list of dictionary words. Decide whether the string can be cut into pieces so that every piece is a word from the list. A word may be used as many times as you need.
Example 1
- Input
- s = "rainbow", wordDict = ["rain","bow"]
- Output
- true
- Why
- "rain" + "bow" covers the whole string.
Example 2
- Input
- s = "bugbitebug", wordDict = ["bug","bite"]
- Output
- true
- Why
- "bug" + "bite" + "bug" works; words may be reused.
Example 3
- Input
- s = "ratraces", wordDict = ["rat","race","aces"]
- Output
- false
- Why
- After "rat" the rest is "races", and neither "race" nor "aces" finishes it cleanly.
Constraints
- 1 <= s.length <= 300
- 1 <= wordDict.length <= 1000
- 1 <= wordDict[i].length <= 20
- s and wordDict[i] consist of lowercase English letters
How to think about it
Updated 2026-09-29A prefix is valid if and only if it ends with a dictionary word that immediately follows another valid prefix. Instead of searching forward through every combination of words, mark each index as reachable and look back only at word-length intervals to check connectivity.
Approaches, worst first
Recursive prefix matching
time O(2^n) · space O(n)
From the start of the string, find all dictionary words that match the prefix and recurse on the remainder. Without memoization, repeating failed suffixes causes combinatorial explosion on inputs like 'aaaaab' with words ['a', 'aa', 'aaa'].
Boolean reachability array
time O(n^3) · space O(n + m * k)
Define dp[i] as true if s[0..i-1] can be segmented. Set dp[0] = true. For each i from 1 to n, iterate j from 0 to i - 1; if dp[j] is true and s[j..i-1] exists in the word hash set, set dp[i] = true and break early.
Length-bounded lookupWrite this one
time O(n * L^2) · space O(n + m * k)
Store words in a hash set and find the maximum word length L. For each index i, only scan cut points j between i - L and i - 1 where dp[j] is true. This caps substring extractions to the maximum dictionary word length.
Where people lose marks · 3
- Looking up words in a raw array instead of a hash set. Array lookups turn every subsegment check into an O(wordDict.length) scan.
- Failing to break early once dp[i] becomes true. Continuing the inner loop wastes string slicing and hashing operations for an already reachable prefix.
- Forgetting to set dp[0] = true. If the empty prefix base case is false, no transition can ever trigger.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
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