LeetCode #3714 — MEDIUM

Longest Balanced Substring II

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

The Problem

Problem Statement

You are given a string s consisting only of the characters 'a', 'b', and 'c'.

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 = "aabcc"

Output: 3

Explanation:

The longest balanced substring is "abc" because all distinct characters '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 <= 10⁵
s contains only the characters 'a', 'b', and 'c'.

Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting only of the characters 'a', 'b', and 'c'. 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

"aabcc"

Example 3

"aba"

Step 02

Core Insight

What unlocks the optimal approach

Solve for three cases: all-equal characters, exactly two distinct characters, and all three characters present. Treat each case separately and take the maximum length.
Case 1: single character: the longest balanced substring is the longest run of the same character; report its length.
Case 2: two distinct characters: reduce to that pair (ignore the third character) and use prefix differences of their counts; equal counts between two indices mean the substring between them is balanced for those two chars.
Case 3: all three characters: use prefix counts and hash the pair <code>(count_b - count_a, count_c - count_a)</code> for each prefix; if the same pair appears at two indices the substring between them has equal counts for a, b, and c. Store earliest index per pair to get maximal length.

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 #3714: Longest Balanced Substring II
class Solution {
    public int longestBalanced(String s) {
        char[] cs = s.toCharArray();
        int x = calc1(cs);
        int y = Math.max(calc2(cs, 'a', 'b'), Math.max(calc2(cs, 'b', 'c'), calc2(cs, 'a', 'c')));
        int z = calc3(cs);
        return Math.max(x, Math.max(y, z));
    }

    private int calc1(char[] s) {
        int res = 0;
        int i = 0, n = s.length;
        while (i < n) {
            int j = i + 1;
            while (j < n && s[j] == s[i]) {
                j++;
            }
            res = Math.max(res, j - i);
            i = j;
        }
        return res;
    }

    private int calc2(char[] s, char a, char b) {
        int res = 0;
        int i = 0, n = s.length;
        while (i < n) {
            while (i < n && s[i] != a && s[i] != b) {
                i++;
            }
            Map<Integer, Integer> pos = new HashMap<>();
            pos.put(0, i - 1);
            int d = 0;
            while (i < n && (s[i] == a || s[i] == b)) {
                d += (s[i] == a) ? 1 : -1;
                Integer prev = pos.get(d);
                if (prev != null) {
                    res = Math.max(res, i - prev);
                } else {
                    pos.put(d, i);
                }
                i++;
            }
        }
        return res;
    }

    private int calc3(char[] s) {
        Map<Long, Integer> pos = new HashMap<>();
        pos.put(f(0, 0), -1);

        int[] cnt = new int[3];
        int res = 0;

        for (int i = 0; i < s.length; i++) {
            char c = s[i];
            ++cnt[c - 'a'];
            int x = cnt[0] - cnt[1];
            int y = cnt[1] - cnt[2];
            long k = f(x, y);

            Integer prev = pos.get(k);
            if (prev != null) {
                res = Math.max(res, i - prev);
            } else {
                pos.put(k, i);
            }
        }
        return res;
    }

    private long f(int x, int y) {
        return (x + 100000) << 20 | (y + 100000);
    }
}

// Accepted solution for LeetCode #3714: Longest Balanced Substring II
func longestBalanced(s string) int {
	x := calc1(s)
	y := max(calc2(s, 'a', 'b'), calc2(s, 'b', 'c'), calc2(s, 'a', 'c'))
	z := calc3(s)
	return max(x, max(y, z))
}

func calc1(s string) int {
	res := 0
	n := len(s)
	i := 0
	for i < n {
		j := i + 1
		for j < n && s[j] == s[i] {
			j++
		}
		if j-i > res {
			res = j - i
		}
		i = j
	}
	return res
}

func calc2(s string, a, b byte) int {
	res := 0
	n := len(s)
	i := 0
	for i < n {
		for i < n && s[i] != a && s[i] != b {
			i++
		}
		pos := map[int]int{0: i - 1}
		d := 0
		for i < n && (s[i] == a || s[i] == b) {
			if s[i] == a {
				d++
			} else {
				d--
			}
			if prev, ok := pos[d]; ok {
				if i-prev > res {
					res = i - prev
				}
			} else {
				pos[d] = i
			}
			i++
		}
	}
	return res
}

type key struct {
	x, y int
}

func calc3(s string) int {
	pos := make(map[key]int)
	pos[key{0, 0}] = -1

	cnt := [3]int{}
	res := 0

	for i := 0; i < len(s); i++ {
		c := s[i]
		cnt[c-'a']++
		x := cnt[0] - cnt[1]
		y := cnt[1] - cnt[2]
		k := key{x, y}

		if j, ok := pos[k]; ok {
			if i-j > res {
				res = i - j
			}
		} else {
			pos[k] = i
		}
	}
	return res
}

# Accepted solution for LeetCode #3714: Longest Balanced Substring II
class Solution:
    def longestBalanced(self, s: str) -> int:
        def calc1(s: str) -> int:
            res = 0
            i, n = 0, len(s)
            while i < n:
                j = i + 1
                while j < n and s[j] == s[i]:
                    j += 1
                res = max(res, j - i)
                i = j
            return res

        def calc2(s: str, a: str, b: str) -> int:
            res = 0
            i, n = 0, len(s)
            while i < n:
                while i < n and s[i] not in (a, b):
                    i += 1
                pos = {0: i - 1}
                d = 0
                while i < n and s[i] in (a, b):
                    d += 1 if s[i] == a else -1
                    if d in pos:
                        res = max(res, i - pos[d])
                    else:
                        pos[d] = i
                    i += 1
            return res

        def calc3(s: str) -> int:
            pos = {(0, 0): -1}
            cnt = Counter()
            res = 0
            for i, c in enumerate(s):
                cnt[c] += 1
                k = (cnt["a"] - cnt["b"], cnt["b"] - cnt["c"])
                if k in pos:
                    res = max(res, i - pos[k])
                else:
                    pos[k] = i
            return res

        x = calc1(s)
        y = max(calc2(s, "a", "b"), calc2(s, "b", "c"), calc2(s, "a", "c"))
        z = calc3(s)
        return max(x, y, z)

// Accepted solution for LeetCode #3714: Longest Balanced Substring II
use std::collections::HashMap;

impl Solution {
    pub fn longest_balanced(s: String) -> i32 {
        let x = Self::calc1(&s);
        let y = Self::calc2(&s, 'a', 'b')
            .max(Self::calc2(&s, 'b', 'c'))
            .max(Self::calc2(&s, 'a', 'c'));
        let z = Self::calc3(&s);
        x.max(y).max(z)
    }

    fn calc1(s: &str) -> i32 {
        let bytes = s.as_bytes();
        let mut res = 0i32;
        let mut i = 0usize;
        let n = bytes.len();

        while i < n {
            let mut j = i + 1;
            while j < n && bytes[j] == bytes[i] {
                j += 1;
            }
            res = res.max((j - i) as i32);
            i = j;
        }
        res
    }

    fn calc2(s: &str, a: char, b: char) -> i32 {
        let bytes = s.as_bytes();
        let a = a as u8;
        let b = b as u8;

        let mut res = 0i32;
        let mut i = 0usize;
        let n = bytes.len();

        while i < n {
            while i < n && bytes[i] != a && bytes[i] != b {
                i += 1;
            }

            let mut pos: HashMap<i32, i32> = HashMap::new();
            pos.insert(0, i as i32 - 1);

            let mut d = 0i32;
            while i < n && (bytes[i] == a || bytes[i] == b) {
                if bytes[i] == a {
                    d += 1;
                } else {
                    d -= 1;
                }

                if let Some(&pre) = pos.get(&d) {
                    res = res.max(i as i32 - pre);
                } else {
                    pos.insert(d, i as i32);
                }
                i += 1;
            }
        }

        res
    }

    fn f(x: i32, y: i32) -> i64 {
        (((x + 100000) as i64) << 20) | ((y + 100000) as i64)
    }

    fn calc3(s: &str) -> i32 {
        let mut pos: HashMap<i64, i32> = HashMap::new();
        pos.insert(Self::f(0, 0), -1);

        let mut cnt = [0i32; 3];
        let mut res = 0i32;

        for (i, c) in s.bytes().enumerate() {
            cnt[(c - b'a') as usize] += 1;

            let x = cnt[0] - cnt[1];
            let y = cnt[1] - cnt[2];
            let k = Self::f(x, y);

            if let Some(&pre) = pos.get(&k) {
                res = res.max(i as i32 - pre);
            } else {
                pos.insert(k, i as i32);
            }
        }

        res
    }
}

// Accepted solution for LeetCode #3714: Longest Balanced Substring II
function longestBalanced(s: string): number {
    const x = calc1(s);
    const y = Math.max(calc2(s, 'a', 'b'), calc2(s, 'b', 'c'), calc2(s, 'a', 'c'));
    const z = calc3(s);
    return Math.max(x, y, z);
}

function calc1(s: string): number {
    let res = 0;
    const n = s.length;
    let i = 0;
    while (i < n) {
        let j = i + 1;
        while (j < n && s[j] === s[i]) j++;
        res = Math.max(res, j - i);
        i = j;
    }
    return res;
}

function calc2(s: string, a: string, b: string): number {
    let res = 0;
    const n = s.length;
    let i = 0;

    while (i < n) {
        while (i < n && s[i] !== a && s[i] !== b) i++;

        const pos = new Map<number, number>();
        pos.set(0, i - 1);

        let d = 0;
        while (i < n && (s[i] === a || s[i] === b)) {
            d += s[i] === a ? 1 : -1;

            const prev = pos.get(d);
            if (prev !== undefined) {
                res = Math.max(res, i - prev);
            } else {
                pos.set(d, i);
            }
            i++;
        }
    }
    return res;
}

function calc3(s: string): number {
    const pos = new Map<string, number>();
    pos.set(key(0, 0), -1);

    const cnt = [0, 0, 0];
    let res = 0;

    for (let i = 0; i < s.length; i++) {
        const c = s.charCodeAt(i) - 97;
        cnt[c]++;

        const x = cnt[0] - cnt[1];
        const y = cnt[1] - cnt[2];
        const k = key(x, y);

        const prev = pos.get(k);
        if (prev !== undefined) {
            res = Math.max(res, i - prev);
        } else {
            pos.set(k, i);
        }
    }
    return res;
}

function key(x: number, y: number): string {
    return x + '#' + y;
}

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.