Longest String Chain
A medium Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 1
- Core for
- 9 roles
The problem
Given a list of words, find the length of the longest string chain where each word is a predecessor of the next. A word A is a predecessor of word B if you can insert exactly one letter into A to make it equal to B.
Example 1
- Input
- words = ["x","xy","xyz","wxy","wxyz","vwxyz"]
- Output
- 5
- Why
- Chain: x -> xy -> xyz -> wxyz -> vwxyz (each adds one letter at front), length 5.
Example 2
- Input
- words = ["p","pq","pqr","qp","qpq","qpqr"]
- Output
- 4
- Why
- Chain: p -> pq -> pqr -> qpqr (each step adds one letter), length 4.
Example 3
- Input
- words = ["abcd","dbqca"]
- Output
- 1
- Why
- "abcd" cannot be a predecessor of "dbqca" (not just one insertion), so the longest chain is length 1.
Constraints
- 1 <= words.length <= 1000
- 1 <= words[i].length <= 16
- words[i] consists of lowercase English letters
How to think about it
Updated 2026-09-29Instead of searching forward by trying to insert all 26 letters at every position of word A, search backward from word B by deleting one character at each of its L positions. If that smaller word exists in your dictionary, it extends whatever chain ended there.
Approaches, worst first
Graph construction with DFS
time O(n^2 * L) · space O(n^2)
Build a directed acyclic graph by comparing all pairs of words (u, v) where len(v) == len(u) + 1. Then run DFS to find the longest path.
Length-sorted map with single-character deletionsWrite this one
time O(n log n + n * L^2) · space O(n * L)
Sort words by length ascending. Store chain lengths in a hash map. For each word of length L, generate all L predecessor words by removing one character at a time. The word's chain length is 1 + max(map[predecessor]).
Where people lose marks · 3
- Processing words out of length order. Shorter words must be completely solved before longer words attempt to query their chain lengths.
- Generating successors (inserting letters) instead of predecessors (deleting letters). Successor generation tests 26 * (L + 1) variations rather than just L deletions.
- A single word always forms a chain of length 1, so the initial answer must be at least 1.
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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