LeetCode #3721 — HARD

Longest Balanced Subarray II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
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 <= 105
  • 1 <= nums[i] <= 105
Patterns Used
📐Segment Tree→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
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

  • Store the first (or all) occurrences for each value in <code>pos[val]</code>.
  • Build a lazy segment tree over start indices <code>l in [0..n-1]</code> that supports range add and can tell if any index has value <code>0</code> (keep <code>mn</code>/<code>mx</code>).
  • Use <code>sign = +1</code> for odd values and <code>sign = -1</code> for even values.
  • Initialize by adding each value's contribution with <code>update(p, n-1, sign)</code> where <code>p</code> is its current first occurrence.
  • Slide left <code>l</code>: pop <code>pos[nums[l]]</code>, let <code>next</code> = next occurrence or <code>n</code>, do <code>update(0, next-1, -sign)</code>, then query for any <code>r >= l</code> with value <code>0</code> and update <code>ans = max(ans, r-l+1)</code>.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #3721: Longest Balanced Subarray II
/**
 *
 * Idea:
 * - Treat each distinct odd number as +1, and each distinct even number as -1
 * - Maintain prefix sums representing the balance
 * - When a number appears again, remove its previous contribution
 * - Use a segment tree to maintain min/max prefix sums with range add
 * - Binary search on the segment tree to find the earliest equal prefix sum
 */
class Solution {

    /**
     * Segment tree node
     */
    static class Node {
        int l, r; // segment range
        int mn, mx; // minimum / maximum prefix sum
        int lazy; // lazy propagation (range add)
    }

    /**
     * Segment tree supporting:
     * - range add
     * - find the smallest index with a given prefix sum
     */
    static class SegmentTree {
        Node[] tr;

        SegmentTree(int n) {
            tr = new Node[n << 2];
            for (int i = 0; i < tr.length; i++) {
                tr[i] = new Node();
            }
            build(1, 0, n);
        }

        // Build an empty tree with all prefix sums = 0
        void build(int u, int l, int r) {
            tr[u].l = l;
            tr[u].r = r;
            tr[u].mn = tr[u].mx = 0;
            tr[u].lazy = 0;
            if (l == r) return;
            int mid = (l + r) >> 1;
            build(u << 1, l, mid);
            build(u << 1 | 1, mid + 1, r);
        }

        // Add value v to all prefix sums in [l, r]
        void modify(int u, int l, int r, int v) {
            if (tr[u].l >= l && tr[u].r <= r) {
                apply(u, v);
                return;
            }
            pushdown(u);
            int mid = (tr[u].l + tr[u].r) >> 1;
            if (l <= mid) modify(u << 1, l, r, v);
            if (r > mid) modify(u << 1 | 1, l, r, v);
            pushup(u);
        }

        // Binary search on the segment tree
        // Find the smallest index where prefix sum == target
        int query(int u, int target) {
            if (tr[u].l == tr[u].r) {
                return tr[u].l;
            }
            pushdown(u);
            int left = u << 1;
            int right = u << 1 | 1;
            if (tr[left].mn <= target && target <= tr[left].mx) {
                return query(left, target);
            }
            return query(right, target);
        }

        // Apply range add
        void apply(int u, int v) {
            tr[u].mn += v;
            tr[u].mx += v;
            tr[u].lazy += v;
        }

        // Update from children
        void pushup(int u) {
            tr[u].mn = Math.min(tr[u << 1].mn, tr[u << 1 | 1].mn);
            tr[u].mx = Math.max(tr[u << 1].mx, tr[u << 1 | 1].mx);
        }

        // Push lazy tag down
        void pushdown(int u) {
            if (tr[u].lazy != 0) {
                apply(u << 1, tr[u].lazy);
                apply(u << 1 | 1, tr[u].lazy);
                tr[u].lazy = 0;
            }
        }
    }

    public int longestBalanced(int[] nums) {
        int n = nums.length;
        SegmentTree st = new SegmentTree(n);

        // last[x] = last position where value x appeared
        Map<Integer, Integer> last = new HashMap<>();

        int now = 0; // current prefix sum
        int ans = 0; // answer

        // Enumerate the right endpoint
        for (int i = 1; i <= n; i++) {
            int x = nums[i - 1];
            int det = (x & 1) == 1 ? 1 : -1;

            // Remove previous contribution if x appeared before
            if (last.containsKey(x)) {
                st.modify(1, last.get(x), n, -det);
                now -= det;
            }

            // Add current contribution
            last.put(x, i);
            st.modify(1, i, n, det);
            now += det;

            // Find earliest position with the same prefix sum
            int pos = st.query(1, now);
            ans = Math.max(ans, i - pos);
        }

        return ans;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n + q log n)
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.

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