LeetCode #3719 — MEDIUM

Longest Balanced Subarray I

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

The Problem

Problem Statement

You are given an integer array nums.

A subarray is called balanced if the number of distinct even numbers in the subarray is equal to the number of distinct odd numbers.

Return the length of the longest balanced subarray.

Example 1:

Input: nums = [2,5,4,3]

Output: 4

Explanation:

The longest balanced subarray is [2, 5, 4, 3].
It has 2 distinct even numbers [2, 4] and 2 distinct odd numbers [5, 3]. Thus, the answer is 4.

Example 2:

Input: nums = [3,2,2,5,4]

Output: 5

Explanation:

The longest balanced subarray is [3, 2, 2, 5, 4].
It has 2 distinct even numbers [2, 4] and 2 distinct odd numbers [3, 5]. Thus, the answer is 5.

Example 3:

Input: nums = [1,2,3,2]

Output: 3

Explanation:

The longest balanced subarray is [2, 3, 2].
It has 1 distinct even number [2] and 1 distinct odd number [3]. Thus, the answer is 3.

Constraints:

1 <= nums.length <= 1500
1 <= nums[i] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. A subarray is called balanced if the number of distinct even numbers in the subarray is equal to the number of distinct odd numbers. Return the length of the longest balanced subarray.

Baseline thinking

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

Pattern signal: Array · Hash Map · Segment Tree

Example 1

[2,5,4,3]

Example 2

[3,2,2,5,4]

Example 3

[1,2,3,2]

Step 02

Core Insight

What unlocks the optimal approach

Use brute force
Try every subarray and use a map/set data structure to track the distinct even and odd numbers

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 #3719: Longest Balanced Subarray I
class Solution {
    public int longestBalanced(int[] nums) {
        int n = nums.length;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            Set<Integer> vis = new HashSet<>();
            int[] cnt = new int[2];
            for (int j = i; j < n; ++j) {
                if (vis.add(nums[j])) {
                    ++cnt[nums[j] & 1];
                }
                if (cnt[0] == cnt[1]) {
                    ans = Math.max(ans, j - i + 1);
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3719: Longest Balanced Subarray I
func longestBalanced(nums []int) (ans int) {
	n := len(nums)
	for i := 0; i < n; i++ {
		vis := map[int]bool{}
		cnt := [2]int{}
		for j := i; j < n; j++ {
			if !vis[nums[j]] {
				vis[nums[j]] = true
				cnt[nums[j]&1]++
			}
			if cnt[0] == cnt[1] {
				ans = max(ans, j-i+1)
			}
		}
	}
	return
}

# Accepted solution for LeetCode #3719: Longest Balanced Subarray I
class Solution:
    def longestBalanced(self, nums: List[int]) -> int:
        n = len(nums)
        ans = 0
        for i in range(n):
            cnt = [0, 0]
            vis = set()
            for j in range(i, n):
                if nums[j] not in vis:
                    cnt[nums[j] & 1] += 1
                    vis.add(nums[j])
                if cnt[0] == cnt[1]:
                    ans = max(ans, j - i + 1)
        return ans

// Accepted solution for LeetCode #3719: Longest Balanced Subarray I
use std::collections::HashSet;

impl Solution {
    pub fn longest_balanced(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut ans: i32 = 0;

        for i in 0..n {
            let mut vis: HashSet<i32> = HashSet::new();
            let mut cnt = [0i32; 2];

            for j in i..n {
                if !vis.contains(&nums[j]) {
                    vis.insert(nums[j]);
                    let idx = (nums[j] & 1) as usize;
                    cnt[idx] += 1;
                }
                if cnt[0] == cnt[1] {
                    ans = ans.max((j - i + 1) as i32);
                }
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #3719: Longest Balanced Subarray I
function longestBalanced(nums: number[]): number {
    const n = nums.length;
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        const vis = new Set<number>();
        const cnt: number[] = Array(2).fill(0);
        for (let j = i; j < n; ++j) {
            if (!vis.has(nums[j])) {
                vis.add(nums[j]);
                ++cnt[nums[j] & 1];
            }
            if (cnt[0] === cnt[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^2)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n × q) time

O(1) space

For each of q queries, scan the entire range to compute the aggregate (sum, min, max). Each range scan takes up to O(n) for a full-array query. With q queries: O(n × q) total. Point updates are O(1) but queries dominate.

SEGMENT TREE

O(n + q log n) time

O(n) space

Building the tree is O(n). Each query or update traverses O(log n) nodes (tree height). For q queries: O(n + q log n) total. Space is O(4n) ≈ O(n) for the tree array. Lazy propagation adds O(1) per node for deferred updates.

Shortcut: Build O(n), query/update O(log n) each. When you need both range queries AND point updates.

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.