3152. Special Array II

3152. Special Array II

Difficulty: Medium

Topics: Array, Binary Search, Prefix Sum

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 subarray¹ 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.

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:

1. The subarray is [4,3,1]. 3 and 1 are both odd. So the answer to this query is false.
2. 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

Hint:

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

Solution:

We need to determine whether a subarray of nums is "special," i.e., every pair of adjacent elements in the subarray must have different parity (one must be odd, and the other must be even).

Approach:

1. Identify Parity Transitions: We can preprocess the array to mark positions where the parity changes. For example:
   • 0 represents an even number.
   • 1 represents an odd number.

The idea is to identify all positions where adjacent elements have different parity. This will help us efficiently determine if a subarray is special by checking if the positions in the query are part of the same "special" block.

1. Preprocessing:
   Create a binary array parity_change where each element is 1 if the adjacent elements have different parity, otherwise 0. For example:

   • If nums[i] and nums[i+1] have different parity, set parity_change[i] = 1, otherwise 0.

2. Prefix Sum Array:
   Construct a prefix sum array prefix_sum where each entry at index i represents the cumulative number of parity transitions up to that index. This helps quickly check if all pairs within a subarray have different parity.

3. Query Processing:
   For each query [from, to], check if there is any position in the range [from, to-1] where the parity doesn't change. This can be done by checking the difference in the prefix sum values: prefix_sum[to] - prefix_sum[from].

Let's implement this solution in PHP: 3152. Special Array II

<?php
/**
 * @param Integer[] $nums
 * @param Integer[][] $queries
 * @return Boolean[]
 */
function specialArray($nums, $queries) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Example usage
$nums1 = [3,4,1,2,6];
$queries1 = [[0, 4]];
print_r(specialArray($nums1, $queries1)); // [false]

$nums2 = [4,3,1,6];
$queries2 = [[0, 2], [2, 3]];
print_r(specialArray($nums2, $queries2)); // [false, true]
?>

Explanation:

1. Preprocessing Parity Transitions:
   We calculate parity_change[i] = 1 if the elements nums[i] and nums[i+1] have different parity. Otherwise, we set it to 0.

2. Prefix Sum Array:
   The prefix_sum[i] stores the cumulative count of parity transitions from the start of the array up to index i. This allows us to compute how many transitions occurred in any subarray [from, to] in constant time using the formula:
   

   $transition_count = $prefix_sum[$to] - $prefix_sum[$from];

1. Query Evaluation: For each query, if the number of transitions is equal to the length of the subarray minus 1, the subarray is special, and we return true. Otherwise, we return false.

Time Complexity:

• Preprocessing the parity transitions takes O(n).
• Constructing the prefix sum array takes O(n).
• Each query can be answered in O(1) using the prefix sum array.
• Hence, the total time complexity is O(n + q), where n is the length of the array and q is the number of queries.

This solution efficiently handles the problem constraints with an optimized approach.

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1. Subarray A subarray is a contiguous sequence of elements within an array. ↩