Special Array II

Medium

Asked by:

30 views

Topics:

Arrays

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the numbers in the input array? Can they be negative or zero?
What should I return if no special number exists?
Are duplicate numbers allowed in the input array, and if so, how should they be handled?
What is the maximum size of the input array?
If multiple numbers qualify as a special number, can I return any one of them?

Brute Force Solution

Approach

The brute force method exhaustively explores all potential values to see if any satisfy the required condition. It's like trying every possible number until you find one that works. This approach guarantees finding the answer if it exists, but it might take a very long time.

Here's how the algorithm would work step-by-step:

Start by guessing that the special number is zero.
Check if zero is indeed the special number by comparing it to the numbers in the given set.
If it is not, try the next possible value, which is one.
Keep checking each possible value, incrementing by one each time.
For each guessed value, compare it to the given set of numbers to see if it meets the special condition.
If we find a number that satisfies the special condition, we've found our answer.
If we've checked all reasonable numbers and haven't found a match, then there's no special number.

Code Implementation

def special_array_brute_force(numbers):
    # Iterate through possible special numbers.
    for possible_special_number in range(1, len(numbers) + 1):

        count = 0
        # Count numbers greater than the possible special number.
        for number in numbers:
            if number >= possible_special_number:
                count += 1

        # Check if the count matches the possible special number.
        if count == possible_special_number:
            return possible_special_number

    # No special number found.
    return -1

Big(O) Analysis

Time Complexity

O(n^2) – The brute force approach iterates through potential special numbers from 0 up to n (the size of the input array). For each potential special number, we iterate through the input array of size n to check if the special condition is met. This involves comparing each element of the array against the current potential special number. Thus, the nested loop structure leads to a time complexity of approximately n * n operations. Therefore, the overall time complexity is O(n^2).

Space Complexity

O(1) – The brute force method described does not use any auxiliary data structures like arrays, lists, or hash maps. It iterates through potential special numbers and compares them to the input array, performing only comparisons. Therefore, the space used is limited to a few integer variables to keep track of the current guess and potentially a counter during the comparison, which is independent of the input size N (where N is the size of the input array). The space complexity is constant.

Optimal Solution

Approach

The key to solving this problem efficiently is to use a focused search approach. We don't need to check every possible special value; instead, we can intelligently narrow down the possibilities. The goal is to determine the special value by using a method similar to binary search.

Here's how the algorithm would work step-by-step:

Understand what we're looking for: A 'special value' is a number where the count of numbers in the given list that are greater than or equal to it is equal to the number itself.
Use a binary search approach, focusing on the possible range where the 'special value' might exist.
At each step in the binary search, count how many numbers in the list are greater than or equal to the 'middle' value.
If the count is equal to the 'middle' value, we've found our 'special value'.
If the count is larger than the 'middle' value, the 'special value' must be larger, so adjust the search range accordingly.
If the count is smaller than the 'middle' value, the 'special value' must be smaller, so adjust the search range accordingly.
Continue adjusting the range and checking until the 'special value' is found, or we've exhausted all possibilities. If we cannot find an appropriate number, then the answer is ‘no special value exists’.

Code Implementation

def special_array(numbers):
    left_index = 0
    right_index = len(numbers)

    while left_index <= right_index:
        middle_index = (left_index + right_index) // 2
        
        # Count elements greater than or equal to middle.
        count = 0
        for number in numbers:
            if number >= middle_index:
                count += 1

        # Adjust search based on comparison.
        if count == middle_index:
            return middle_index

        # If count is greater, increase the lower bound.
        if count > middle_index:

            left_index = middle_index + 1

        # Else decrease the upper bound.
        else:

            right_index = middle_index - 1

    return -1

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm employs a binary search approach. The binary search iterates a logarithmic number of times, specifically log n, where n represents a plausible range of special values, at most the size of the input array. Within each iteration of the binary search, the code iterates through the entire input array to count how many elements are greater than or equal to the middle value in the range. This counting process takes O(n) time. Therefore, the total time complexity is O(n log n).

Space Complexity

O(1) – The algorithm employs a binary search approach that modifies the search range using variables like `low`, `high`, and `mid`. The count of numbers greater than or equal to the mid value is calculated within the loop but doesn't require storing these counts in a separate data structure proportional to the input size N. Therefore, the auxiliary space usage is dominated by a few integer variables, irrespective of the input array's size, resulting in constant space complexity.

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty array or null to indicate no special array can be formed from nothing.
Input array with a single element	Return an empty array as a special array requires at least two elements.
Array contains duplicate values	The algorithm should correctly identify pairs even with duplicates; if indices are important, ensure each number can only be part of one pair.
Array contains all identical values	If the special condition cannot be met by duplicate values, return an empty list, otherwise return all possible pairs of those duplicates.
Array contains large positive or negative numbers that might cause integer overflow during calculations (e.g., multiplication or addition)	Use appropriate data types (e.g., long) or modulo arithmetic to prevent overflow if the calculations involve these numbers and their magnitude.
Maximum array size (close to memory limits)	Ensure algorithm's memory usage doesn't exceed available memory, especially if using auxiliary data structures like hash maps or sorting in place.
No special array exists within the input array.	The algorithm should return an empty list or an appropriate indicator (null) when no valid solution is found.
Handling of zero values in the array and their potential impact on the special condition (e.g., divisibility by zero).	The algorithm needs to explicitly handle division by zero errors if the special condition involves dividing by array elements.

Null or empty input array

How to Handle:

Return an empty array or null to indicate no special array can be formed from nothing.

Input array with a single element

How to Handle:

Return an empty array as a special array requires at least two elements.

Array contains duplicate values

How to Handle:

The algorithm should correctly identify pairs even with duplicates; if indices are important, ensure each number can only be part of one pair.

Array contains all identical values

How to Handle:

If the special condition cannot be met by duplicate values, return an empty list, otherwise return all possible pairs of those duplicates.

Array contains large positive or negative numbers that might cause integer overflow during calculations (e.g., multiplication or addition)

How to Handle:

Use appropriate data types (e.g., long) or modulo arithmetic to prevent overflow if the calculations involve these numbers and their magnitude.

Maximum array size (close to memory limits)

How to Handle:

Ensure algorithm's memory usage doesn't exceed available memory, especially if using auxiliary data structures like hash maps or sorting in place.

No special array exists within the input array.

How to Handle:

The algorithm should return an empty list or an appropriate indicator (null) when no valid solution is found.

Handling of zero values in the array and their potential impact on the special condition (e.g., divisibility by zero).

How to Handle:

The algorithm needs to explicitly handle division by zero errors if the special condition involves dividing by array elements.