LeetCode #893 — MEDIUM

Groups of Special-Equivalent Strings

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 strings of the same length words.

In one move, you can swap any two even indexed characters or any two odd indexed characters of a string words[i].

Two strings words[i] and words[j] are special-equivalent if after any number of moves, words[i] == words[j].

For example, words[i] = "zzxy" and words[j] = "xyzz" are special-equivalent because we may make the moves "zzxy" -> "xzzy" -> "xyzz".

A group of special-equivalent strings from words is a non-empty subset of words such that:

Every pair of strings in the group are special equivalent, and
The group is the largest size possible (i.e., there is not a string words[i] not in the group such that words[i] is special-equivalent to every string in the group).

Return the number of groups of special-equivalent strings from words.

Example 1:

Input: words = ["abcd","cdab","cbad","xyzz","zzxy","zzyx"]
Output: 3
Explanation: 
One group is ["abcd", "cdab", "cbad"], since they are all pairwise special equivalent, and none of the other strings is all pairwise special equivalent to these.
The other two groups are ["xyzz", "zzxy"] and ["zzyx"].
Note that in particular, "zzxy" is not special equivalent to "zzyx".

Example 2:

Input: words = ["abc","acb","bac","bca","cab","cba"]
Output: 3

Constraints:

1 <= words.length <= 1000
1 <= words[i].length <= 20
words[i] consist of lowercase English letters.
All the strings are of the same length.

Step 01

Brute Force Baseline

Problem summary: You are given an array of strings of the same length words. In one move, you can swap any two even indexed characters or any two odd indexed characters of a string words[i]. Two strings words[i] and words[j] are special-equivalent if after any number of moves, words[i] == words[j]. For example, words[i] = "zzxy" and words[j] = "xyzz" are special-equivalent because we may make the moves "zzxy" -> "xzzy" -> "xyzz". A group of special-equivalent strings from words is a non-empty subset of words such that: Every pair of strings in the group are special equivalent, and The group is the largest size possible (i.e., there is not a string words[i] not in the group such that words[i] is special-equivalent to every string in the group). Return the number of groups of special-equivalent strings from words.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

["abcd","cdab","cbad","xyzz","zzxy","zzyx"]

Example 2

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

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #893: Groups of Special-Equivalent Strings
class Solution {
    public int numSpecialEquivGroups(String[] words) {
        Set<String> s = new HashSet<>();
        for (String word : words) {
            s.add(convert(word));
        }
        return s.size();
    }

    private String convert(String word) {
        List<Character> a = new ArrayList<>();
        List<Character> b = new ArrayList<>();
        for (int i = 0; i < word.length(); ++i) {
            char ch = word.charAt(i);
            if (i % 2 == 0) {
                a.add(ch);
            } else {
                b.add(ch);
            }
        }
        Collections.sort(a);
        Collections.sort(b);
        StringBuilder sb = new StringBuilder();
        for (char c : a) {
            sb.append(c);
        }
        for (char c : b) {
            sb.append(c);
        }
        return sb.toString();
    }
}

// Accepted solution for LeetCode #893: Groups of Special-Equivalent Strings
func numSpecialEquivGroups(words []string) int {
	s := map[string]bool{}
	for _, word := range words {
		a, b := []rune{}, []rune{}
		for i, c := range word {
			if i&1 == 1 {
				a = append(a, c)
			} else {
				b = append(b, c)
			}
		}
		sort.Slice(a, func(i, j int) bool {
			return a[i] < a[j]
		})
		sort.Slice(b, func(i, j int) bool {
			return b[i] < b[j]
		})
		s[string(a)+string(b)] = true
	}
	return len(s)
}

# Accepted solution for LeetCode #893: Groups of Special-Equivalent Strings
class Solution:
    def numSpecialEquivGroups(self, words: List[str]) -> int:
        s = {''.join(sorted(word[::2]) + sorted(word[1::2])) for word in words}
        return len(s)

// Accepted solution for LeetCode #893: Groups of Special-Equivalent Strings
struct Solution;

use std::collections::BTreeMap;
use std::collections::HashSet;

#[derive(Debug, PartialEq, Eq, Clone, Hash)]
struct Count {
    even: BTreeMap<char, usize>,
    odd: BTreeMap<char, usize>,
}

impl Count {
    fn new(s: String) -> Self {
        let mut even: BTreeMap<char, usize> = BTreeMap::new();
        let mut odd: BTreeMap<char, usize> = BTreeMap::new();
        for (i, c) in s.chars().enumerate() {
            if i % 2 == 0 {
                *even.entry(c).or_default() += 1;
            } else {
                *odd.entry(c).or_default() += 1;
            }
        }
        Count { even, odd }
    }
}

impl Solution {
    fn num_special_equiv_groups(a: Vec<String>) -> i32 {
        let mut hs: HashSet<Count> = HashSet::new();
        for s in a {
            hs.insert(Count::new(s));
        }
        hs.len() as i32
    }
}

#[test]
fn test() {
    let a: Vec<String> = vec_string!["a", "b", "c", "a", "c", "c"];
    assert_eq!(Solution::num_special_equiv_groups(a), 3);
    let a: Vec<String> = vec_string!["aa", "bb", "ab", "ba"];
    assert_eq!(Solution::num_special_equiv_groups(a), 4);
    let a: Vec<String> = vec_string!["abc", "acb", "bac", "bca", "cab", "cba"];
    assert_eq!(Solution::num_special_equiv_groups(a), 3);
    let a: Vec<String> = vec_string!["abcd", "cdab", "adcb", "cbad"];
    assert_eq!(Solution::num_special_equiv_groups(a), 1);
}

// Accepted solution for LeetCode #893: Groups of Special-Equivalent Strings
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #893: Groups of Special-Equivalent Strings
// class Solution {
//     public int numSpecialEquivGroups(String[] words) {
//         Set<String> s = new HashSet<>();
//         for (String word : words) {
//             s.add(convert(word));
//         }
//         return s.size();
//     }
// 
//     private String convert(String word) {
//         List<Character> a = new ArrayList<>();
//         List<Character> b = new ArrayList<>();
//         for (int i = 0; i < word.length(); ++i) {
//             char ch = word.charAt(i);
//             if (i % 2 == 0) {
//                 a.add(ch);
//             } else {
//                 b.add(ch);
//             }
//         }
//         Collections.sort(a);
//         Collections.sort(b);
//         StringBuilder sb = new StringBuilder();
//         for (char c : a) {
//             sb.append(c);
//         }
//         for (char c : b) {
//             sb.append(c);
//         }
//         return sb.toString();
//     }
// }

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 or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.