LeetCode #1048 — MEDIUM

Longest String Chain

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given an array of words where each word consists of lowercase English letters.

word_A is a predecessor of word_B if and only if we can insert exactly one letter anywhere in word_A without changing the order of the other characters to make it equal to word_B.

For example, "abc" is a predecessor of "abac", while "cba" is not a predecessor of "bcad".

A word chain is a sequence of words [word₁, word₂, ..., word_k] with k >= 1, where word₁ is a predecessor of word₂, word₂ is a predecessor of word₃, and so on. A single word is trivially a word chain with k == 1.

Return the length of the longest possible word chain with words chosen from the given list of words.

Example 1:

Input: words = ["a","b","ba","bca","bda","bdca"]
Output: 4
Explanation: One of the longest word chains is ["a","ba","bda","bdca"].

Example 2:

Input: words = ["xbc","pcxbcf","xb","cxbc","pcxbc"]
Output: 5
Explanation: All the words can be put in a word chain ["xb", "xbc", "cxbc", "pcxbc", "pcxbcf"].

Example 3:

Input: words = ["abcd","dbqca"]
Output: 1
Explanation: The trivial word chain ["abcd"] is one of the longest word chains.
["abcd","dbqca"] is not a valid word chain because the ordering of the letters is changed.

Constraints:

1 <= words.length <= 1000
1 <= words[i].length <= 16
words[i] only consists of lowercase English letters.

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: You are given an array of words where each word consists of lowercase English letters. wordA is a predecessor of wordB if and only if we can insert exactly one letter anywhere in wordA without changing the order of the other characters to make it equal to wordB. For example, "abc" is a predecessor of "abac", while "cba" is not a predecessor of "bcad". A word chain is a sequence of words [word1, word2, ..., wordk] with k >= 1, where word1 is a predecessor of word2, word2 is a predecessor of word3, and so on. A single word is trivially a word chain with k == 1. Return the length of the longest possible word chain with words chosen from the given list of words.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Two Pointers · Dynamic Programming

Example 1

["a","b","ba","bca","bda","bdca"]

Example 2

["xbc","pcxbcf","xb","cxbc","pcxbc"]

Example 3

["abcd","dbqca"]

Step 02

Core Insight

What unlocks the optimal approach

Instead of adding a character, try deleting a character to form a chain in reverse.
For each word in order of length, for each word2 which is word with one character removed, length[word2] = max(length[word2], length[word] + 1).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1048: Longest String Chain
class Solution {
    public int longestStrChain(String[] words) {
        Arrays.sort(words, Comparator.comparingInt(String::length));
        int res = 0;
        Map<String, Integer> map = new HashMap<>();
        for (String word : words) {
            int x = 1;
            for (int i = 0; i < word.length(); ++i) {
                String pre = word.substring(0, i) + word.substring(i + 1);
                x = Math.max(x, map.getOrDefault(pre, 0) + 1);
            }
            map.put(word, x);
            res = Math.max(res, x);
        }
        return res;
    }
}

// Accepted solution for LeetCode #1048: Longest String Chain
func longestStrChain(words []string) int {
	sort.Slice(words, func(i, j int) bool { return len(words[i]) < len(words[j]) })
	res := 0
	mp := make(map[string]int)
	for _, word := range words {
		x := 1
		for i := 0; i < len(word); i++ {
			pre := word[0:i] + word[i+1:len(word)]
			x = max(x, mp[pre]+1)
		}
		mp[word] = x
		res = max(res, x)
	}
	return res
}

# Accepted solution for LeetCode #1048: Longest String Chain
class Solution:
    def longestStrChain(self, words: List[str]) -> int:
        def check(w1, w2):
            if len(w2) - len(w1) != 1:
                return False
            i = j = cnt = 0
            while i < len(w1) and j < len(w2):
                if w1[i] != w2[j]:
                    cnt += 1
                else:
                    i += 1
                j += 1
            return cnt < 2 and i == len(w1)

        n = len(words)
        dp = [1] * (n + 1)
        words.sort(key=lambda x: len(x))
        res = 1
        for i in range(1, n):
            for j in range(i):
                if check(words[j], words[i]):
                    dp[i] = max(dp[i], dp[j] + 1)
            res = max(res, dp[i])
        return res

// Accepted solution for LeetCode #1048: Longest String Chain
use std::collections::HashMap;

impl Solution {
    #[allow(dead_code)]
    pub fn longest_str_chain(words: Vec<String>) -> i32 {
        let mut words = words;
        let mut ret = 0;
        let mut map: HashMap<String, i32> = HashMap::new();

        // Sort the words vector first
        words.sort_by(|lhs, rhs| lhs.len().cmp(&rhs.len()));

        // Begin the "dp" process
        for w in words.iter() {
            let n = w.len();
            let mut x = 1;

            for i in 0..n {
                let s = w[..i].to_string() + &w[i + 1..];
                let v = map.entry(s.clone()).or_default();
                x = std::cmp::max(x, *v + 1);
            }

            map.insert(w.clone(), x);

            ret = std::cmp::max(ret, x);
        }

        ret
    }
}

// Accepted solution for LeetCode #1048: Longest String Chain
function longestStrChain(words: string[]): number {
    words.sort((a, b) => a.length - b.length);
    let ans = 0;
    let hashTable = new Map();
    for (let word of words) {
        let c = 1;
        for (let i = 0; i < word.length; i++) {
            let pre = word.substring(0, i) + word.substring(i + 1);
            c = Math.max(c, (hashTable.get(pre) || 0) + 1);
        }
        hashTable.set(word, c);
        ans = Math.max(ans, c);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.