LeetCode #3152 — MEDIUM

Special Array II

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

The Problem

Problem Statement

An array is considered special if every pair of its adjacent elements contains two numbers with different parity.

You are given an array of integer nums and a 2D integer matrix queries, where for queries[i] = [from_i, to_i] your task is to check that subarray nums[from_i..to_i] is special or not.

Return an array of booleans answer such that answer[i] is true if nums[from_i..to_i] is special.

Example 1:

Input: nums = [3,4,1,2,6], queries = [[0,4]]

Output: [false]

Explanation:

The subarray is [3,4,1,2,6]. 2 and 6 are both even.

Example 2:

Input: nums = [4,3,1,6], queries = [[0,2],[2,3]]

Output: [false,true]

Explanation:

The subarray is [4,3,1]. 3 and 1 are both odd. So the answer to this query is false.
The subarray is [1,6]. There is only one pair: (1,6) and it contains numbers with different parity. So the answer to this query is true.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵
1 <= queries.length <= 10⁵
queries[i].length == 2
0 <= queries[i][0] <= queries[i][1] <= nums.length - 1

Step 01

Brute Force Baseline

Problem summary: An array is considered special if every pair of its adjacent elements contains two numbers with different parity. You are given an array of integer nums and a 2D integer matrix queries, where for queries[i] = [fromi, toi] your task is to check that subarray nums[fromi..toi] is special or not. Return an array of booleans answer such that answer[i] is true if nums[fromi..toi] is special.

Baseline thinking

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

Pattern signal: Array · Binary Search

Example 1

[3,4,1,2,6]
[[0,4]]

Example 2

[4,3,1,6]
[[0,2],[2,3]]

Step 02

Core Insight

What unlocks the optimal approach

Try to split the array into some non-intersected continuous special subarrays.
For each query check that the first and the last elements of that query are in the same subarray or not.

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 #3152: Special Array II
class Solution {
    public boolean[] isArraySpecial(int[] nums, int[][] queries) {
        int n = nums.length;
        int[] d = new int[n];
        for (int i = 1; i < n; ++i) {
            if (nums[i] % 2 != nums[i - 1] % 2) {
                d[i] = d[i - 1];
            } else {
                d[i] = i;
            }
        }
        int m = queries.length;
        boolean[] ans = new boolean[m];
        for (int i = 0; i < m; ++i) {
            ans[i] = d[queries[i][1]] <= queries[i][0];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3152: Special Array II
func isArraySpecial(nums []int, queries [][]int) (ans []bool) {
	n := len(nums)
	d := make([]int, n)
	for i := range d {
		d[i] = i
	}
	for i := 1; i < len(nums); i++ {
		if nums[i]%2 != nums[i-1]%2 {
			d[i] = d[i-1]
		}
	}
	for _, q := range queries {
		ans = append(ans, d[q[1]] <= q[0])
	}
	return
}

# Accepted solution for LeetCode #3152: Special Array II
class Solution:
    def isArraySpecial(self, nums: List[int], queries: List[List[int]]) -> List[bool]:
        n = len(nums)
        d = list(range(n))
        for i in range(1, n):
            if nums[i] % 2 != nums[i - 1] % 2:
                d[i] = d[i - 1]
        return [d[t] <= f for f, t in queries]

// Accepted solution for LeetCode #3152: Special Array II
// 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 #3152: Special Array II
// class Solution {
//     public boolean[] isArraySpecial(int[] nums, int[][] queries) {
//         int n = nums.length;
//         int[] d = new int[n];
//         for (int i = 1; i < n; ++i) {
//             if (nums[i] % 2 != nums[i - 1] % 2) {
//                 d[i] = d[i - 1];
//             } else {
//                 d[i] = i;
//             }
//         }
//         int m = queries.length;
//         boolean[] ans = new boolean[m];
//         for (int i = 0; i < m; ++i) {
//             ans[i] = d[queries[i][1]] <= queries[i][0];
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3152: Special Array II
function isArraySpecial(nums: number[], queries: number[][]): boolean[] {
    const n = nums.length;
    const d: number[] = Array.from({ length: n }, (_, i) => i);
    for (let i = 1; i < n; ++i) {
        if (nums[i] % 2 !== nums[i - 1] % 2) {
            d[i] = d[i - 1];
        }
    }
    return queries.map(([from, to]) => d[to] <= from);
}

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

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.