Longest String Chain - Chain of words in an Array
Last Updated :
23 Jul, 2025
Given an array of words where each word consists of lowercase English letters, we need to find the longest word chain possible.
A word 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.
The task is to return the length of the longest possible word chain with words chosen from the given list of words in any order.
Examples:
Input: words = ["ba", "b", "a", "bca", "bda", "bdca"]
Output: 4
Explanation: One of the longest word chains is ["a", "ba", "bda", "bdca"].
Input: words = ["abcd", "dbqca"]
Output: 1
Explanation: The trivial word chain ["abcd"] is one of the longest word chains.
[Naive Approach] Using Recursion - Exponential Time
If we take a closer look, we can notice that this problem is a variation of Longest Increasing Subsequence Problem. For every word, we find the longest chain ending with it. At the end, we return the maximum of all longest chains.
1. Consider every word w one by one
2. Find the length of the longest chain where w is the last word (or longest word). We compute this using a recursive function longestEndCurr().
3. Return the maximum of all values obtained in step 2.
C++
// C++ code for Longest String Chain
// using Recursion
#include <bits/stdc++.h>
using namespace std;
// Function to check if the second string can be obtained
// by removing one character from the first string
bool isPred(const string &a, const string &b) {
if (a.size() != b.size() + 1) return false;
int i = 0, j = 0;
while (i < a.size() && j < b.size()) {
if (a[i] == b[j]) { i++; j++; }
else { i++; }
}
return (j == b.size());
}
// Recursive function to find the length of the longest
// chain that ends with curr
int longestEndCurr(vector<string> &words, string &curr) {
int maxEnd = 1;
for (string &w : words) {
if (isPred(curr, w)) {
maxEnd = max(maxEnd, 1 + longestEndCurr(words, w));
}
}
return maxEnd;
}
// Naive function to find the longest string chain
int longestStringChain(vector<string> &words) {
// Find the longest chain ending with every
// word and return max of all
int res = 1;
for (string &w : words) {
res = max(res, longestEndCurr(words, w));
}
return res;
}
// Driver code
int main() {
vector<string> s = { "a", "b", "ba", "bca", "bda", "bdca" };
cout << longestStringChain(s);
return 0;
}
// C code for Longest String Chain
// using Recursion
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdlib.h>
// Function to check if the second string can be obtained
// by removing one character from the first string
bool isPred(const char *a, const char *b) {
int len_a = strlen(a);
int len_b = strlen(b);
if (len_a != len_b + 1) return false;
int i = 0, j = 0;
while (i < len_a && j < len_b) {
if (a[i] == b[j]) { i++; j++; }
else { i++; }
}
return (j == len_b);
}
// Recursive function to find the length of the longest
// chain that ends with curr
int longestEndCurr(char **words, int n, const char *curr) {
int maxEnd = 1;
for (int i = 0; i < n; i++) {
if (isPred(curr, words[i])) {
int length = longestEndCurr(words, n, words[i]);
if (length + 1 > maxEnd) {
maxEnd = length + 1;
}
}
}
return maxEnd;
}
// Naive function to find the longest string chain
int longestStringChain(char **words, int n) {
// Find the longest chain ending with every
// word and return max of all
int res = 1;
for (int i = 0; i < n; i++) {
int length = longestEndCurr(words, n, words[i]);
if (length > res) {
res = length;
}
}
return res;
}
// Driver code
int main() {
char *s[] = { "a", "b", "ba", "bca", "bda", "bdca" };
int n = sizeof(s) / sizeof(s[0]);
printf("%d\n", longestStringChain(s, n));
return 0;
}
// Java code for Longest String Chain
// using Recursion
import java.util.Arrays;
public class GfG {
// Function to check if b can be obtained by removing
// one character from a
static boolean isPred(String a, String b) {
if (a.length() != b.length() + 1) return false;
int i = 0, j = 0;
// Compare characters of both strings
while (i < a.length() && j < b.length()) {
if (a.charAt(i) == b.charAt(j)) {
i++;
j++;
} else {
i++; // Skip one character in a
}
}
return (j == b.length());
}
// Recursive function to find the longest chain that
// ends with the given word 'curr'
static int longestEndCurr(String[] words, String curr) {
int maxEnd = 1;
// Check for all words in the array
for (String w : words) {
if (isPred(curr, w)) {
// Recursively find the longest chain
maxEnd = Math.max(maxEnd, 1 + longestEndCurr(words, w));
}
}
return maxEnd;
}
// Function to find the length of the longest string chain
static int longestStringChain(String[] words) {
int res = 1;
// Find the longest chain ending at each word
for (String w : words) {
res = Math.max(res, longestEndCurr(words, w));
}
return res;
}
// Driver code
public static void main(String[] args) {
String[] s = {"a", "b", "ba", "bca", "bda", "bdca"};
// Print the length of the longest string chain
System.out.println(longestStringChain(s));
}
}
# Python code for Longest String Chain
# using Recursion
# Function to check if b can be obtained by removing
# one character from a
def isPred(a, b):
if len(a) != len(b) + 1:
return False
i = j = 0
# Compare characters of both strings
while i < len(a) and j < len(b):
if a[i] == b[j]:
i += 1
j += 1
else:
i += 1 # Skip one character in a
return j == len(b)
# Recursive function to find the longest chain that
# ends with the given word 'curr'
def longestEndCurr(words, curr):
maxEnd = 1
# Check for all words in the array
for w in words:
if isPred(curr, w):
# Recursively find the longest chain
maxEnd = max(maxEnd, 1 + longestEndCurr(words, w))
return maxEnd
# Function to find the length of the longest string chain
def longestStringChain(words):
res = 1
# Find the longest chain ending at each word
for w in words:
res = max(res, longestEndCurr(words, w))
return res
# Driver code
if __name__ == "__main__":
s = ["a", "b", "ba", "bca", "bda", "bdca"]
# Print the length of the longest string chain
print(longestStringChain(s))
// C# code for Longest String Chain
// using Recursion
using System;
class GfG {
// Function to check if b can be obtained by removing
// one character from a
static bool IsPred(string a, string b) {
if (a.Length != b.Length + 1) return false;
int i = 0, j = 0;
// Compare characters of both strings
while (i < a.Length && j < b.Length) {
if (a[i] == b[j]) {
i++;
j++;
} else {
i++; // Skip one character in a
}
}
return j == b.Length;
}
// Recursive function to find the longest chain that
// ends with the given word 'curr'
static int LongestEndCurr(string[] words, string curr) {
int maxEnd = 1;
// Check for all words in the array
foreach (string w in words) {
if (IsPred(curr, w)) {
// Recursively find the longest chain
maxEnd = Math.Max(maxEnd, 1 + LongestEndCurr(words, w));
}
}
return maxEnd;
}
// Function to find the length of the longest string chain
static int LongestStringChain(string[] words) {
int res = 1;
// Find the longest chain ending at each word
foreach (string w in words) {
res = Math.Max(res, LongestEndCurr(words, w));
}
return res;
}
// Driver code
static void Main() {
string[] s = { "a", "b", "ba", "bca", "bda", "bdca" };
// Print the length of the longest string chain
Console.WriteLine(LongestStringChain(s));
}
}
// Javascript code for Longest String Chain
// using Recursion
// Function to check if the second string can be obtained
// by removing one character from the first string
function isPred(a, b) {
if (a.length !== b.length + 1) return false;
let i = 0, j = 0;
while (i < a.length && j < b.length) {
if (a[i] === b[j]) { i++; j++; }
else { i++; }
}
return j === b.length;
}
// Recursive function to find the length of the longest
// chain that ends with curr
function longestEndCurr(words, curr) {
let maxEnd = 1;
for (let w of words) {
if (isPred(curr, w)) {
maxEnd = Math.max(maxEnd, 1 + longestEndCurr(words, w));
}
}
return maxEnd;
}
// Naive function to find the longest string chain
function longestStringChain(words) {
// Find the longest chain ending with every
// word and return max of all
let res = 1;
for (let w of words) {
res = Math.max(res, longestEndCurr(words, w));
}
return res;
}
// Driver code
const s = ["a", "b", "ba", "bca", "bda", "bdca"];
console.log(longestStringChain(s));
[Better Approach] Using DP(Tabulation) - O(nlogn + n²m) Time and O(n) Space
The idea is to build the longest possible word chain by checking if a word can be formed by adding one letter to a shorter word.
First, we sort the words by length to ensure that we process shorter words first. We use a dp array, where dp[i] stores the longest chain ending at words[i].
For each word, we check all previous words to see if they are its predecessor (by removing one character). If a valid predecessor is found, we update dp[i] accordingly. The maximum value in dp gives the longest string chain.
Below is the implementation of the above approach:
C++
// C++ code for Longest String Chain
// using Tabulation (Bottom-Up DP)
#include <bits/stdc++.h>
using namespace std;
// Function to check if b is a predecessor of a
bool isPred(const string &a, const string &b) {
if (a.size() != b.size() + 1) {
return false;
}
int i = 0, j = 0;
while (i < a.size() && j < b.size()) {
if (a[i] == b[j]) {
i++;
j++;
} else {
i++;
}
}
return (j == b.size());
}
// Function to find the longest string chain using DP
int longestStringChain(vector<string> &words) {
// Sort words based on length
sort(words.begin(), words.end(), [](const string &a,
const string &b) {
return a.size() < b.size();
});
int n = words.size();
vector<int> dp(n, 1);
int res = 1;
// Iterate over each word
for (int i = 0; i < n; i++) {
for (int j = 0; j < i; j++) {
if (isPred(words[i], words[j])) {
dp[i] = max(dp[i], dp[j] + 1);
}
}
res = max(res, dp[i]);
}
return res;
}
// Driver code
int main() {
vector<string> s = { "a", "b", "ba", "bca", "bda", "bdca" };
cout << longestStringChain(s);
return 0;
}
// Java code for Longest String Chain
// using Tabulation (Bottom-Up DP)
import java.util.*;
class GfG {
// Function to check if b is a predecessor of a
private static boolean isPred(String a, String b) {
if (a.length() != b.length() + 1) {
return false;
}
int i = 0, j = 0;
while (i < a.length() && j < b.length()) {
if (a.charAt(i) == b.charAt(j)) {
i++;
j++;
} else {
i++;
}
}
return j == b.length();
}
// Function to find the longest string chain using DP
public static int longestStringChain(String words[]) {
// Sort words based on length
Arrays.sort(words, Comparator.comparingInt(String::length));
int n = words.length;
int dp[] = new int[n];
Arrays.fill(dp, 1);
int res = 1;
// Iterate over each word
for (int i = 0; i < n; i++) {
for (int j = 0; j < i; j++) {
if (isPred(words[i], words[j])) {
dp[i] = Math.max(dp[i], dp[j] + 1);
}
}
res = Math.max(res, dp[i]);
}
return res;
}
// Driver code
public static void main(String[] args) {
String words[] = { "a", "b", "ba", "bca", "bda", "bdca" };
System.out.println(longestStringChain(words));
}
}
# Python code for Longest String Chain
# using Tabulation (Bottom-Up DP)
# Function to check if b is a predecessor of a
def is_pred(a, b):
if len(a) != len(b) + 1:
return False
i, j = 0, 0
while i < len(a) and j < len(b):
if a[i] == b[j]:
i += 1
j += 1
else:
i += 1
return j == len(b)
# Function to find the longest string chain using DP
def longestStringChain(words):
# Sort words based on length
words.sort(key=len)
n = len(words)
dp = [1] * n
res = 1
# Iterate over each word
for i in range(n):
for j in range(i):
if is_pred(words[i], words[j]):
dp[i] = max(dp[i], dp[j] + 1)
res = max(res, dp[i])
return res
# Driver code
if __name__ == "__main__":
words = ["a", "b", "ba", "bca", "bda", "bdca"]
print(longestStringChain(words))
// C# code for Longest String Chain
// using Tabulation (Bottom-Up DP)
using System;
class GfG {
// Function to check if b is a predecessor of a
static bool IsPred(string a, string b) {
if (a.Length != b.Length + 1) {
return false;
}
int i = 0, j = 0;
while (i < a.Length && j < b.Length) {
if (a[i] == b[j]) {
i++;
j++;
} else {
i++;
}
}
return j == b.Length;
}
// Function to find the longest string chain using DP
public static int LongestStringChain(string[] words) {
// Sort words based on length
Array.Sort(words, (a, b) => a.Length - b.Length);
int n = words.Length;
int[] dp = new int[n];
Array.Fill(dp, 1);
int res = 1;
// Iterate over each word
for (int i = 0; i < n; i++) {
for (int j = 0; j < i; j++) {
if (IsPred(words[i], words[j])) {
dp[i] = Math.Max(dp[i], dp[j] + 1);
}
}
res = Math.Max(res, dp[i]);
}
return res;
}
// Driver code
public static void Main() {
string[] words = { "a", "b", "ba",
"bca", "bda", "bdca" };
Console.WriteLine(LongestStringChain(words));
}
}
// JavaScript code for Longest String Chain
// using Tabulation (Bottom-Up DP) without Map
// Function to check if b is a predecessor of a
function isPred(a, b) {
if (a.length !== b.length + 1) {
return false;
}
let i = 0, j = 0;
while (i < a.length && j < b.length) {
if (a[i] === b[j]) {
i++;
j++;
} else {
i++;
}
}
return j === b.length;
}
// Function to find the longest string chain using DP
function longestStringChain(words) {
// Sort words based on length
words.sort((a, b) => a.length - b.length);
let n = words.length;
let dp = new Array(n).fill(1);
let res = 1;
// Iterate over each word
for (let i = 0; i < n; i++) {
for (let j = 0; j < i; j++) {
if (isPred(words[i], words[j])) {
dp[i] = Math.max(dp[i], dp[j] + 1);
}
}
res = Math.max(res, dp[i]);
}
return res;
}
// Driver code
const words = ["a", "b", "ba", "bca", "bda", "bdca"];
console.log(longestStringChain(words));
Time Complexity: O(nlogn + n²m), as sorting takes O(n log n), and checking predecessors takes O(n^2 * m), where n is the number of words and m is the maximum word length.
Auxiliary Space: O(n), as the dp array stores one value per word, requiring O(n) space.
[Expected Approach] DP + Hashing- O(nlogn + n*m²) Time and O(n*m) Space
Using dynamic programming, we store the longest chain length for each word in a hash table. For each word, we generate possible predecessors by removing one character and check if they exist in the table. If a valid predecessor is found, we update the chain length, ensuring we track the longest possible chain efficiently.
Steps to implement the above idea:
- Sort the words by length to process shorter words first.
- Use a DP table (array or map) to store the longest chain length for each word.
- Iterate through each word and initialize its chain length to 1.
- Generate predecessors by removing one character at a time.
- Check if the predecessor exists in the DP table and update the chain length.
- Track the maximum chain length found and return the result.
Below is the implementation of the above approach:
C++
// C++ code for Longest String Chain
// using DP + Hashing
#include <bits/stdc++.h>
using namespace std;
// Custom comparison function to sort strings by their lengths
bool static cmp(const string& a, const string& b) {
return a.size() < b.size();
}
// Function to find the length of the longest string chain
int longestStringChain(vector<string>& words) {
// Sort the words by length
sort(words.begin(), words.end(), cmp);
// Create an unordered map to store the
// maximum chain length for each word
unordered_map<string, int> dp;
int res = 1;
// Iterate through each word in the sorted list of words
for (const string& w : words) {
dp[w] = 1; // Initialize length for the current word
// Try removing one character at a time to
// check if it forms a valid predecessor
for (int i = 0; i < w.size(); ++i) {
string pred = w.substr(0, i) + w.substr(i + 1);
if (dp.find(pred) != dp.end()) {
dp[w] = max(dp[w], dp[pred] + 1);
}
}
res = max(res, dp[w]);
}
return res;
}
// Driver code
int main() {
vector<string> words = { "a", "b", "ba", "bca", "bda", "bdca" };
cout << longestStringChain(words) << endl;
return 0;
}
// Java code for Longest String Chain
// using DP + Hashing
import java.util.Arrays;
import java.util.HashMap;
class GfG {
// Custom comparator to sort strings by length
public static int longestStringChain(String[] words) {
// Sort words by length
Arrays.sort(words, (a, b) -> a.length() - b.length());
// HashMap to store the maximum chain length for each word
HashMap<String, Integer> dp = new HashMap<>();
int res = 1;
// Iterate through each word in the sorted array
for (String w : words) {
dp.put(w, 1); // Initialize length for current word
// Try removing one character at a time
for (int i = 0; i < w.length(); i++) {
String pred = w.substring(0, i) + w.substring(i + 1);
if (dp.containsKey(pred)) {
dp.put(w, Math.max(dp.get(w), dp.get(pred) + 1));
}
}
res = Math.max(res, dp.get(w));
}
return res;
}
// Driver code
public static void main(String[] args) {
String[] words = {"a", "b", "ba", "bca", "bda", "bdca"};
System.out.println(longestStringChain(words));
}
}
# Python code for Longest String Chain
# using DP + Hashing
def longestStringChain(words):
# Sort words by length
words.sort(key=len)
# Dictionary to store the maximum chain length for each word
dp = {}
res = 1
# Iterate through each word in the sorted list
for w in words:
dp[w] = 1 # Initialize length for current word
# Try removing one character at a time
for i in range(len(w)):
pred = w[:i] + w[i+1:]
if pred in dp:
dp[w] = max(dp[w], dp[pred] + 1)
res = max(res, dp[w])
return res
# Driver code
if __name__ == "__main__":
words = ["a", "b", "ba", "bca", "bda", "bdca"]
print(longestStringChain(words))
// C# code for Longest String Chain
// using DP + Hashing
using System;
using System.Collections.Generic;
class GfG {
// Function to find the length of the longest string chain
public static int LongestStringChain(string[] words) {
// Sort words by length
Array.Sort(words, (a, b) => a.Length - b.Length);
// Dictionary to store the maximum chain length for each word
Dictionary<string, int> dp = new Dictionary<string, int>();
int res = 1;
// Iterate through each word in the sorted array
foreach (string w in words) {
dp[w] = 1; // Initialize length for current word
// Try removing one character at a time
for (int i = 0; i < w.Length; i++) {
string pred = w.Substring(0, i) + w.Substring(i + 1);
if (dp.ContainsKey(pred)) {
dp[w] = Math.Max(dp[w], dp[pred] + 1);
}
}
res = Math.Max(res, dp[w]);
}
return res;
}
// Driver code
public static void Main(string[] args) {
string[] words = {"a", "b", "ba", "bca", "bda", "bdca"};
Console.WriteLine(LongestStringChain(words));
}
}
// Javascript code for Longest String Chain
// using DP + Hashing
function longestStringChain(words) {
// Sort words by length
words.sort((a, b) => a.length - b.length);
// Map to store the maximum chain length for each word
let dp = new Map();
let res = 1;
// Iterate through each word in the sorted list
for (let w of words) {
dp.set(w, 1); // Initialize length for current word
// Try removing one character at a time
for (let i = 0; i < w.length; i++) {
let pred = w.slice(0, i) + w.slice(i + 1);
if (dp.has(pred)) {
dp.set(w, Math.max(dp.get(w), dp.get(pred) + 1));
}
}
res = Math.max(res, dp.get(w));
}
return res;
}
// Driver code
let words = ["a", "b", "ba", "bca", "bda", "bdca"];
console.log(longestStringChain(words));
Time Complexity: O(nlogn + n*m²), where n is the number of words and m is the maximum word length (sorting takes O(n * log n), and checking predecessors takes O(n*m²).
Auxiliary Space: O(n*m), as we store n words in the DP table, each of length up to m.
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14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
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