LeetCode #3138 — MEDIUM

Minimum Length of Anagram Concatenation

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

The Problem

Problem Statement

You are given a string s, which is known to be a concatenation of anagrams of some string t.

Return the minimum possible length of the string t.

An anagram is formed by rearranging the letters of a string. For example, "aab", "aba", and, "baa" are anagrams of "aab".

Example 1:

Input: s = "abba"

Output: 2

Explanation:

One possible string t could be "ba".

Example 2:

Input: s = "cdef"

Output: 4

Explanation:

One possible string t could be "cdef", notice that t can be equal to s.

Example 2:

Input: s = "abcbcacabbaccba"

Output: 3

Constraints:

1 <= s.length <= 10⁵
s consist only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s, which is known to be a concatenation of anagrams of some string t. Return the minimum possible length of the string t. An anagram is formed by rearranging the letters of a string. For example, "aab", "aba", and, "baa" are anagrams of "aab".

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"abba"

Example 2

"cdef"

Example 3

"abcbcacabbaccba"

Step 02

Core Insight

What unlocks the optimal approach

The answer should be a divisor of <code>s.length</code>.
Check each candidate naively.

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 #3138: Minimum Length of Anagram Concatenation
class Solution {
    private int n;
    private char[] s;
    private int[] cnt = new int[26];

    public int minAnagramLength(String s) {
        n = s.length();
        this.s = s.toCharArray();
        for (int i = 0; i < n; ++i) {
            ++cnt[this.s[i] - 'a'];
        }
        for (int i = 1;; ++i) {
            if (n % i == 0 && check(i)) {
                return i;
            }
        }
    }

    private boolean check(int k) {
        for (int i = 0; i < n; i += k) {
            int[] cnt1 = new int[26];
            for (int j = i; j < i + k; ++j) {
                ++cnt1[s[j] - 'a'];
            }
            for (int j = 0; j < 26; ++j) {
                if (cnt1[j] * (n / k) != cnt[j]) {
                    return false;
                }
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #3138: Minimum Length of Anagram Concatenation
func minAnagramLength(s string) int {
	n := len(s)
	cnt := [26]int{}
	for _, c := range s {
		cnt[c-'a']++
	}
	check := func(k int) bool {
		for i := 0; i < n; i += k {
			cnt1 := [26]int{}
			for j := i; j < i+k; j++ {
				cnt1[s[j]-'a']++
			}
			for j, v := range cnt {
				if cnt1[j]*(n/k) != v {
					return false
				}
			}
		}
		return true
	}
	for i := 1; ; i++ {
		if n%i == 0 && check(i) {
			return i
		}
	}
}

# Accepted solution for LeetCode #3138: Minimum Length of Anagram Concatenation
class Solution:
    def minAnagramLength(self, s: str) -> int:
        def check(k: int) -> bool:
            for i in range(0, n, k):
                cnt1 = Counter(s[i : i + k])
                for c, v in cnt.items():
                    if cnt1[c] * (n // k) != v:
                        return False
            return True

        cnt = Counter(s)
        n = len(s)
        for i in range(1, n + 1):
            if n % i == 0 and check(i):
                return i

// Accepted solution for LeetCode #3138: Minimum Length of Anagram Concatenation
/**
 * [3138] Minimum Length of Anagram Concatenation
 */
pub struct Solution {}

// submission codes start here
use std::collections::HashMap;

impl Solution {
    pub fn min_anagram_length(s: String) -> i32 {
        let s: Vec<char> = s.chars().collect();
        let length = s.len();

        for i in 1..length {
            if length % i != 0 {
                continue;
            }

            let count = length / i;
            let mut standard_map = HashMap::new();

            for k in 0..i {
                let entry = standard_map.entry(s[k]).or_insert(0);
                *entry += 1;
            }

            let mut flag = true;
            for j in 1..count {
                let mut map = HashMap::new();
                for k in (j * i)..(i * (j + 1)) {
                    let entry = map.entry(s[k]).or_insert(0);
                    *entry += 1;
                }

                flag = map == standard_map;
                if !flag {
                    break;
                }
            }

            if flag {
                return i as i32;
            }
        }

        s.len() as i32
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3138() {
        assert_eq!(2, Solution::min_anagram_length("abba".to_owned()));
        assert_eq!(4, Solution::min_anagram_length("cdef".to_owned()));
    }
}

// Accepted solution for LeetCode #3138: Minimum Length of Anagram Concatenation
function minAnagramLength(s: string): number {
    const n = s.length;
    const cnt: Record<string, number> = {};
    for (let i = 0; i < n; i++) {
        cnt[s[i]] = (cnt[s[i]] || 0) + 1;
    }
    const check = (k: number): boolean => {
        for (let i = 0; i < n; i += k) {
            const cnt1: Record<string, number> = {};
            for (let j = i; j < i + k; j++) {
                cnt1[s[j]] = (cnt1[s[j]] || 0) + 1;
            }
            for (const [c, v] of Object.entries(cnt)) {
                if (cnt1[c] * ((n / k) | 0) !== v) {
                    return false;
                }
            }
        }
        return true;
    };
    for (let i = 1; ; ++i) {
        if (n % i === 0 && check(i)) {
            return i;
        }
    }
}

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(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.