“Longest String Chain” is a popular interview problem because it looks like a simple word puzzle, but it quietly tests whether you can combine sorting, dynamic programming, and careful comparisons. You’re given a list of words, and your task is to determine the length of the longest possible chain where each word is formed by adding exactly one character to the previous word.
A word A is considered a predecessor of word B if you can insert one letter anywhere in A, without changing the order of existing letters, to create B. This rule preserves the original sequence of characters while allowing exactly one additional character to appear.
For example, the sequence "a" → "ab" → "abc" forms a valid chain because each step adds one character while keeping the original letters in the same order. Similarly, "cat" → "coat" is valid because you can insert the letter 'o' without rearranging the other characters. However, "ab" → "ba" is not valid because the characters change order rather than simply adding a new one.
Your goal is to examine all possible chains that can be formed from the list of words and return the maximum chain length. The challenge lies in efficiently identifying valid predecessor relationships without checking every possible combination.
Why brute force fails quickly
A naive solution might attempt to construct chains starting from every word in the list. For each starting point, it would try to extend the chain by checking whether other words can follow it as valid successors.
This approach often involves testing every pair of words to determine whether one can be formed by adding a character to the other. While that might sound manageable at first, the number of comparisons grows quickly as the list of words increases.
Some brute-force implementations also use recursion to explore every possible chain path. Unfortunately, this leads to repeated checks of the same predecessor relationships and quickly becomes computationally expensive.
The real insight is recognizing that the problem behaves like a longest path problem in a directed acyclic graph (DAG). Each edge points from a shorter word to a longer word because each step adds exactly one character. Once you recognize the “shorter → longer” structure, dynamic programming becomes the natural and efficient solution.
Want to explore more coding problem solutions? Check out the Longest Univalue Path and Largest Divisible Subset coding problem solutions.
How the solution works
The efficient solution relies on three key ideas: sorting, dynamic programming, and generating predecessors directly. Together, these steps allow you to build the longest chains incrementally without redundant work.
1) Sort words by length
The first step is to sort the words in ascending order based on their length. This ordering guarantees that when you process a word, all possible predecessors—words that are exactly one character shorter—have already been considered.
Sorting creates a natural progression for dynamic programming. You build solutions for short words first and then extend those solutions when processing longer words.
2) Use dynamic programming to track chain lengths
For each word, you maintain a value representing the length of the longest chain that ends with that word. This can be stored in a dictionary or hash map where the key is the word and the value is the chain length.
At minimum, every word can form a chain of length 1 by itself. The goal is to determine whether the word can extend an existing chain that ends with one of its predecessors.
3) Generate predecessors by deleting characters
Instead of comparing the current word against every other word in the list, you can generate all possible predecessors directly. You do this by deleting one character at a time from the word and checking whether the resulting shorter word exists in the dynamic programming table.
For example, if the current word is "abcd", removing one character at each position produces "bcd", "acd", "abd", and "abc". Each of these strings is a possible predecessor candidate.
If any generated predecessor already appears in the DP map, you can extend its chain by one. In that case, you update the value for the current word by taking the maximum between its existing chain length and the predecessor’s chain length plus one.
This small optimization eliminates expensive pairwise comparisons and replaces them with a predictable number of checks per word.
Why this solution is efficient
For a word of length L, there are exactly L possible predecessors that can be generated by deleting one character at a time. That means each word requires only a small number of checks rather than comparisons with every other word.
Because the algorithm processes words in order of increasing length, the dynamic programming table already contains all relevant shorter words when needed. This avoids repeated computations and keeps the algorithm efficient even for larger input sets.
In conceptual terms, the time complexity is often described as O(n · L²). This accounts for generating predecessor strings and handling string operations, though in practice the approach performs very well under typical interview constraints.
The space complexity is O(n) because the dynamic programming structure stores a single value for each word. This makes the algorithm both time-efficient and memory-efficient for practical inputs.
Common pitfalls interviewers watch for
One common mistake is forgetting to sort the words by length before applying dynamic programming. Without sorting, the algorithm may try to extend chains using predecessors that have not yet been processed, which breaks the DP logic.
Another frequent issue is comparing words of the same length when looking for predecessors. Words must differ in length by exactly one character for a valid predecessor relationship, so equal-length comparisons waste time and lead to incorrect reasoning.
Some candidates also attempt to verify predecessor relationships by rearranging characters or sorting strings. This approach violates the problem constraint because the original order of characters must remain unchanged.
A final pitfall is relying on pairwise comparisons across the entire list of words. That strategy dramatically increases the number of checks required and makes the solution slower than necessary.
If you clearly explain the idea of generating predecessors by deleting one character and combining it with dynamic programming, interviewers usually recognize that you understand the optimal strategy.
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