Skip to main content

Minimum Number of String Groups Through Transformations - Video Solutions

HardHash TableTwo PointersString

LeetCode 3999 | Weekly Contest 511 Q4 | Minimum Number of String Groups Through Transformation Hard🤯

CodeSprint
31:54219 views
3 video solutions available

Minimum Number of String Groups Through Transformations - Video Solution

Watch 3 video solutions for Minimum Number of String Groups Through Transformations, a hard level problem involving Hash Table, Two Pointers, String. This walkthrough by CodeSprint has 219 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.

Problem Statement

You are given an array of strings words.

Define a transformation on a string s as follows:

  • Let E be the subsequence of characters at even indices of s.
  • Let O be the subsequence of characters at odd indices of s.
  • Independently cyclically shift E and O by any number of positions to the right, possibly zero.
  • Reconstruct the string by placing the shifted E characters back into even indices and the shifted O characters back into odd indices.

Two strings are equivalent if one can be transformed into the other by a single transformation.

Partition words into the minimum number of groups such that:

  • Every string belongs to exactly one group.
  • Every pair of strings in the same group are equivalent.

Return an integer denoting the minimum number of groups.

 

Example 1:

Input: words = ["ntgwz","zwntg"]

Output: 1

Explanation:

  • For "ntgwz", the even-index subsequence is "ngz" and the odd-index subsequence is "tw".
  • Shift "ngz" right by 1 position to obtain "zng", and shift "tw" right by 1 position to obtain "wt".
  • After reconstructing the string, we obtain "zwntg".
  • Therefore, both strings are equivalent and belong to the same group.

Example 2:

Input: words = ["abc","cab","bac","acb","bca","cba"]

Output: 3

Explanation:

The strings can be partitioned into the following groups:

  • ["abc","cba"]
  • ["cab","bac"]
  • ["acb","bca"]

Example 3:

Input: words = ["leet","abb","bab","deed","edde","code","bba"]

Output: 5

Explanation:

The strings can be partitioned into the following groups:

  • ["abb","bba"]
  • ["deed","edde"]
  • ["leet"]
  • ["bab"]
  • ["code"]

​​​​​​​​​​​​​​All pairs of strings in each group are equivalent.

 

Constraints:

  • 1 <= words.length <= 105
  • 1 <= words[i].length <= 5 * 105
  • The sum of words[i].length does not exceed 5 * 105.
  • words[i] consist of lowercase English letters.
Read full problem with examples

Approach Overview

Problem Overview: You need to partition strings into the minimum number of groups where strings inside the same group can be connected through valid transformations. The core challenge is efficiently checking whether two strings belong to the same connected component after one or more transformation operations.

Approach 1: Brute Force Pair Comparison with DFS/BFS (O(n² * m) time, O(n²) space)

Build a graph where each node represents a string. Iterate through every pair of strings and check whether one can transform into the other in O(m) time, where m is the string length. If two strings satisfy the transformation rule, add an undirected edge. After constructing the graph, run DFS or BFS to count connected components. This approach is straightforward and useful when constraints are small because the graph construction logic mirrors the problem statement directly.

This method usually relies on graph traversal techniques and adjacency lists. The downside is the quadratic pair comparison cost. For large input sizes, repeated transformation checks become the bottleneck.

Approach 2: Union-Find / DSU Optimization (O(n² * m) time, O(n) space)

Instead of explicitly storing the graph, use a Disjoint Set Union structure to merge indices whenever two strings are transformable. Iterate through all pairs, perform the same transformation validation, and call union(u, v) when the condition matches. At the end, count distinct parents to get the number of groups.

The key insight is that the problem only needs the number of connected components, not the actual traversal order. DSU removes adjacency storage overhead and simplifies connectivity management with path compression and union by rank. This is the standard interview solution because it combines clean implementation with strong amortized performance characteristics. Problems involving grouping, merging, or equivalence classes often use Union-Find.

Approach 3: Hash-Based Neighbor Generation (Near O(n * m²) average time, O(n * m) space)

If transformations are based on limited edits such as swaps, replacements, or character modifications, you can generate canonical intermediate states and store them in a hash map. Instead of comparing every pair directly, generate all possible transformation signatures for a string and look up matching candidates in constant average time.

This reduces unnecessary comparisons and performs well when strings are short but the number of strings is large. The approach typically combines hashing with DSU merges. You trade additional memory for faster candidate discovery. This optimization is common in hash table and string-grouping problems.

Recommended for interviews: Start with the brute force graph interpretation to show you understand the connectivity model. Then move to the Union-Find solution because interviewers usually expect component counting without explicit graph traversal overhead. If constraints are aggressive, discuss hash-based neighbor generation as the scalability optimization.

Complexity Analysis

ApproachTimeSpaceWhen to Use
Brute Force Graph + DFSO(n² * m)O(n²)Small inputs or when explicit graph traversal is needed
Union-Find (DSU)O(n² * m)O(n)General case and interview-preferred solution
Hash-Based Neighbor GenerationNear O(n * m²)O(n * m)Large datasets with constrained transformation rules