LeetCode #3692 — EASY

Majority Frequency Characters

Build confidence with an intuition-first walkthrough focused on hash map fundamentals.

The Problem

Problem Statement

You are given a string s consisting of lowercase English letters.

The frequency group for a value k is the set of characters that appear exactly k times in s.

The majority frequency group is the frequency group that contains the largest number of distinct characters.

Return a string containing all characters in the majority frequency group, in any order. If two or more frequency groups tie for that largest size, pick the group whose frequency k is larger.

Example 1:

Input: s = "aaabbbccdddde"

Output: "ab"

Explanation:

Frequency (k)	Distinct characters in group	Group size	Majority?
4	{d}	1	No
3	{a, b}	2	Yes
2	{c}	1	No
1	{e}	1	No

Both characters 'a' and 'b' share the same frequency 3, they are in the majority frequency group. "ba" is also a valid answer.

Example 2:

Input: s = "abcd"

Output: "abcd"

Explanation:

Frequency (k)	Distinct characters in group	Group size	Majority?
1	{a, b, c, d}	4	Yes

All characters share the same frequency 1, they are all in the majority frequency group.

Example 3:

Input: s = "pfpfgi"

Output: "fp"

Explanation:

Frequency (k)	Distinct characters in group	Group size	Majority?
2	{p, f}	2	Yes
1	{g, i}	2	No (tied size, lower frequency)

Both characters 'p' and 'f' share the same frequency 2, they are in the majority frequency group. There is a tie in group size with frequency 1, but we pick the higher frequency: 2.

Constraints:

1 <= s.length <= 100
s consists only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting of lowercase English letters. The frequency group for a value k is the set of characters that appear exactly k times in s. The majority frequency group is the frequency group that contains the largest number of distinct characters. Return a string containing all characters in the majority frequency group, in any order. If two or more frequency groups tie for that largest size, pick the group whose frequency k is larger.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"aaabbbccdddde"

Example 2

"abcd"

Example 3

"pfpfgi"

Step 02

Core Insight

What unlocks the optimal approach

Start by counting the frequency of each character in the string.
Group characters based on their frequencies.
Identify the group with the largest number of distinct characters.
If multiple groups tie, select the one with the higher frequency value.
Construct the answer string using all characters from that chosen group.

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 #3692: Majority Frequency Characters
class Solution {
    public String majorityFrequencyGroup(String s) {
        int[] cnt = new int[26];
        for (char c : s.toCharArray()) {
            ++cnt[c - 'a'];
        }
        Map<Integer, StringBuilder> f = new HashMap<>();
        for (int i = 0; i < cnt.length; ++i) {
            if (cnt[i] > 0) {
                f.computeIfAbsent(cnt[i], k -> new StringBuilder()).append((char) ('a' + i));
            }
        }
        int mx = 0;
        int mv = 0;
        String ans = "";
        for (var e : f.entrySet()) {
            int v = e.getKey();
            var cs = e.getValue();
            if (mx < cs.length() || (mx == cs.length() && mv < v)) {
                mx = cs.length();
                mv = v;
                ans = cs.toString();
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3692: Majority Frequency Characters
func majorityFrequencyGroup(s string) string {
	cnt := make([]int, 26)
	for _, c := range s {
		cnt[c-'a']++
	}

	f := make(map[int][]byte)
	for i, v := range cnt {
		if v > 0 {
			f[v] = append(f[v], byte('a'+i))
		}
	}

	mx, mv := 0, 0
	var ans []byte
	for v, cs := range f {
		if len(cs) > mx || (len(cs) == mx && v > mv) {
			mx = len(cs)
			mv = v
			ans = cs
		}
	}
	return string(ans)
}

# Accepted solution for LeetCode #3692: Majority Frequency Characters
class Solution:
    def majorityFrequencyGroup(self, s: str) -> str:
        cnt = Counter(s)
        f = defaultdict(list)
        for c, v in cnt.items():
            f[v].append(c)
        mx = mv = 0
        ans = []
        for v, cs in f.items():
            if mx < len(cs) or (mx == len(cs) and mv < v):
                mx = len(cs)
                mv = v
                ans = cs
        return "".join(ans)

// Accepted solution for LeetCode #3692: Majority Frequency Characters
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3692: Majority Frequency Characters
// class Solution {
//     public String majorityFrequencyGroup(String s) {
//         int[] cnt = new int[26];
//         for (char c : s.toCharArray()) {
//             ++cnt[c - 'a'];
//         }
//         Map<Integer, StringBuilder> f = new HashMap<>();
//         for (int i = 0; i < cnt.length; ++i) {
//             if (cnt[i] > 0) {
//                 f.computeIfAbsent(cnt[i], k -> new StringBuilder()).append((char) ('a' + i));
//             }
//         }
//         int mx = 0;
//         int mv = 0;
//         String ans = "";
//         for (var e : f.entrySet()) {
//             int v = e.getKey();
//             var cs = e.getValue();
//             if (mx < cs.length() || (mx == cs.length() && mv < v)) {
//                 mx = cs.length();
//                 mv = v;
//                 ans = cs.toString();
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3692: Majority Frequency Characters
function majorityFrequencyGroup(s: string): string {
    const cnt: Record<string, number> = {};
    for (const c of s) {
        cnt[c] = (cnt[c] || 0) + 1;
    }
    const f = new Map<number, string[]>();
    for (const [c, v] of Object.entries(cnt)) {
        if (!f.has(v)) {
            f.set(v, []);
        }
        f.get(v)!.push(c);
    }
    let [mx, mv] = [0, 0];
    let ans = '';
    f.forEach((cs, v) => {
        if (mx < cs.length || (mx == cs.length && mv < v)) {
            mx = cs.length;
            mv = v;
            ans = cs.join('');
        }
    });
    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(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.