#3152

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Special Array II

Determine whether each subarray meets the special condition of alternating parity for all adjacent elements efficiently using prefix sums.

array binary search prefix sum

Problem Statement

You are given an integer array nums and a 2D integer array queries where each query queries[i] = [fromi, toi] asks whether the subarray nums[fromi..toi] is special. An array is considered special if every pair of adjacent elements has different parity, meaning one is even and the other is odd.

Return a boolean array answer such that answer[i] is true if the subarray nums[fromi..toi] satisfies the special condition and false otherwise. Ensure your solution handles large arrays efficiently using a pattern of prefix sums and binary search over potential violation points.

Examples

Example 1

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

Output: [false]

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]

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

Solution Approach

Precompute Parity Violations

Scan the nums array once to mark positions where adjacent elements have the same parity. Store these in a prefix sum array so that any subarray query can check in O(1) if a violation exists.

Query Evaluation

For each query [fromi, toi], use the prefix sum of violations to determine if the subarray contains any parity conflicts. If the difference in prefix sums for the range is zero, the subarray is special; otherwise, it is not.

Leverage Binary Search Insights

Although prefix sums answer each query in constant time, the underlying pattern involves thinking in terms of splitting arrays into continuous special subarrays. Binary search over potential split points ensures you recognize non-overlapping segments that preserve the special property.

Complexity Analysis

Metric	Value
Time	O(M + N)
Space	O(M)

Time complexity is O(M + N), where N is the length of nums and M is the number of queries, because we precompute violations in one pass and answer each query in O(1). Space complexity is O(N) for storing the prefix sum array of violations.

What Interviewers Usually Probe

Expect discussion of edge cases with consecutive even or odd numbers.
Look for recognition of prefix sums as a method to optimize multiple queries over large arrays.
Watch for attempts to split arrays and reasoning about continuous special subarrays using binary search patterns.

Common Pitfalls or Variants

Common pitfalls

Failing to check that the prefix sum correctly reflects violations at the last element of the subarray.
Assuming the subarray is special if the first and last elements are different parity, ignoring internal violations.
Overcomplicating with full nested loops instead of precomputing violations efficiently.

Follow-up variants

Check for special subarrays where the parity difference must alternate every two elements instead of each adjacent pair.
Determine the maximum length of a special subarray within a single query range.
Modify nums dynamically and answer queries in real-time, requiring a segment tree or binary indexed tree.

FAQ

What defines a special array in Special Array II?

A special array requires that every pair of adjacent elements has different parity, meaning no two consecutive numbers are both even or both odd.

Can prefix sums handle all queries efficiently?

Yes, by precomputing where parity violations occur, each query can be answered in O(1) using prefix sums.

Why use binary search over the valid answer space here?

Binary search guides thinking about splitting the array into non-overlapping continuous special subarrays, ensuring correct evaluations of large ranges.

What happens if the subarray has only one element?

Single-element subarrays are trivially special since there are no adjacent pairs to violate the parity condition.

How does GhostInterview validate subarray correctness instantly?

It uses precomputed parity violations and prefix sums to check any query range instantly, showing true if no violations exist and false otherwise.

terminal

Solution

Solution 1: Record the Leftmost Special Array Position for Each Position

We can define an array $d$ to record the leftmost special array position for each position, initially $d[i] = i$. Then we traverse the array $nums$ from left to right. If $nums[i]$ and $nums[i - 1]$ have different parities, then $d[i] = d[i - 1]$.

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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]

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