LeetCode #3713 — MEDIUM

Longest Balanced Substring I

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

The Problem

Problem Statement

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

A substring of s is called balanced if all distinct characters in the substring appear the same number of times.

Return the length of the longest balanced substring of s.

Example 1:

Input: s = "abbac"

Output: 4

Explanation:

The longest balanced substring is "abba" because both distinct characters 'a' and 'b' each appear exactly 2 times.

Example 2:

Input: s = "zzabccy"

Output: 4

Explanation:

The longest balanced substring is "zabc" because the distinct characters 'z', 'a', 'b', and 'c' each appear exactly 1 time.

Example 3:

Input: s = "aba"

Output: 2

Explanation:

One of the longest balanced substrings is "ab" because both distinct characters 'a' and 'b' each appear exactly 1 time. Another longest balanced substring is "ba".

Constraints:

1 <= s.length <= 1000
s consists of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting of lowercase English letters. A substring of s is called balanced if all distinct characters in the substring appear the same number of times. Return the length of the longest balanced substring of s.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"abbac"

Example 2

"zzabccy"

Example 3

"aba"

Step 02

Core Insight

What unlocks the optimal approach

Use bruteforce over all substrings

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 #3713: Longest Balanced Substring I
class Solution {
    public int longestBalanced(String s) {
        int n = s.length();
        int[] cnt = new int[26];
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            Arrays.fill(cnt, 0);
            int mx = 0, v = 0;
            for (int j = i; j < n; ++j) {
                int c = s.charAt(j) - 'a';
                if (++cnt[c] == 1) {
                    ++v;
                }
                mx = Math.max(mx, cnt[c]);
                if (mx * v == j - i + 1) {
                    ans = Math.max(ans, j - i + 1);
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3713: Longest Balanced Substring I
func longestBalanced(s string) (ans int) {
	n := len(s)
	for i := 0; i < n; i++ {
		cnt := [26]int{}
		mx, v := 0, 0
		for j := i; j < n; j++ {
			c := s[j] - 'a'
			cnt[c]++
			if cnt[c] == 1 {
				v++
			}
			mx = max(mx, cnt[c])
			if mx*v == j-i+1 {
				ans = max(ans, j-i+1)
			}
		}
	}

	return ans
}

# Accepted solution for LeetCode #3713: Longest Balanced Substring I
class Solution:
    def longestBalanced(self, s: str) -> int:
        n = len(s)
        ans = 0
        for i in range(n):
            cnt = Counter()
            mx = v = 0
            for j in range(i, n):
                cnt[s[j]] += 1
                mx = max(mx, cnt[s[j]])
                if cnt[s[j]] == 1:
                    v += 1
                if mx * v == j - i + 1:
                    ans = max(ans, j - i + 1)
        return ans

// Accepted solution for LeetCode #3713: Longest Balanced Substring I
impl Solution {
    pub fn longest_balanced(s: String) -> i32 {
        let n: i32 = s.len() as i32;
        let bytes = s.as_bytes();
        let mut ans: i32 = 0;

        for i in 0..n {
            let mut cnt: [i32; 26] = [0; 26];
            let mut mx: i32 = 0;
            let mut v: i32 = 0;

            for j in i..n {
                let c: usize = (bytes[j as usize] - b'a') as usize;
                cnt[c] += 1;

                if cnt[c] == 1 {
                    v += 1;
                }

                mx = mx.max(cnt[c]);

                if mx * v == j - i + 1 {
                    ans = ans.max(j - i + 1);
                }
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #3713: Longest Balanced Substring I
function longestBalanced(s: string): number {
    const n = s.length;
    let ans: number = 0;
    for (let i = 0; i < n; ++i) {
        const cnt: number[] = Array(26).fill(0);
        let [mx, v] = [0, 0];
        for (let j = i; j < n; ++j) {
            const c = s[j].charCodeAt(0) - 97;
            if (++cnt[c] === 1) {
                ++v;
            }
            mx = Math.max(mx, cnt[c]);
            if (mx * v === j - i + 1) {
                ans = Math.max(ans, j - i + 1);
            }
        }
    }
    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.