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Special Array I

#626 Most AskedEasy

Special Array I #626 Most Asked

Easy

Microsoft

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Topics:

Arrays

An array is considered special if the parity of every pair of adjacent elements is different. In other words, one element in each pair must be even, and the other must be odd.

You are given an array of integers nums. Return true if nums is a special array, otherwise, return false.

Example 1:

Input: nums = [1]

Output: true

Explanation:

There is only one element. So the answer is true.

Example 2:

Input: nums = [2,1,4]

Output: true

Explanation:

There is only two pairs: (2,1) and (1,4), and both of them contain numbers with different parity. So the answer is true.

Example 3:

Input: nums = [4,3,1,6]

Output: false

Explanation:

nums[1] and nums[2] are both odd. So the answer is false.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100

Try coding it in LeetCode

Special Array I

#626 Most AskedEasy

Special Array I #626 Most Asked

Easy

Microsoft

7 views

Topics:

Arrays

An array is considered special if the parity of every pair of adjacent elements is different. In other words, one element in each pair must be even, and the other must be odd.

You are given an array of integers nums. Return true if nums is a special array, otherwise, return false.

Example 1:

Input: nums = [1]

Output: true

Explanation:

There is only one element. So the answer is true.

Example 2:

Input: nums = [2,1,4]

Output: true

Explanation:

There is only two pairs: (2,1) and (1,4), and both of them contain numbers with different parity. So the answer is true.

Example 3:

Input: nums = [4,3,1,6]

Output: false

Explanation:

nums[1] and nums[2] are both odd. So the answer is false.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100

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 within the input array?
Can the input array be empty or null?
If multiple values of 'x' satisfy the condition, should I return any one of them?
Are all the numbers in the input array integers?
What should I return if no such 'x' exists?

Brute Force Solution

Approach

The brute force approach involves guessing a special number and checking if it meets the problem's condition. We will try every possible number in a range until we find the special number or exhaust all options.

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

Start with the number zero.
Check if the number of values in the input that are greater than or equal to zero matches zero itself.
If the match occurs, we have found the special number.
If there's no match, increase the number we're checking by one.
Repeat the checking process with this new number.
Continue incrementing and checking until we find a match or we've tried all possible numbers within the size of the input.

Code Implementation

def special_array_i_brute_force(numbers):
    array_length = len(numbers)

    for potential_special_value in range(array_length + 1):
        # Count elements greater or equal to the potential value
        count = 0
        for number in numbers:
            if number >= potential_special_value:
                count += 1

        # Check if potential value is the special number
        if count == potential_special_value:
            return potential_special_value

    # No special value found, after checking
    # all potential values.
    return -1

Big(O) Analysis

Time Complexity

O(n²) – The brute force solution iterates from 0 to n, where n is the size of the input array. In each iteration, it counts the number of elements in the array that are greater than or equal to the current number. Counting the elements in the array takes O(n) time. Since this counting is done for each number from 0 to n, the overall time complexity becomes O(n * n), which simplifies to O(n²).

Space Complexity

O(1) – The provided approach iterates through potential special numbers, checking a condition each time. It doesn't create any auxiliary data structures like arrays, lists, or hash maps to store intermediate results or visited elements. Only a single variable is used to represent the number being checked, which occupies constant space regardless of the input array's size (N). Therefore, the space complexity is constant.

Optimal Solution

Approach

The problem asks us to find a special number within a list. A number is 'special' if exactly that many numbers in the list are greater than or equal to it. The optimal approach efficiently searches for this special number without needing to check every possibility by utilizing a clever strategy of elimination.

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

First, sort the list of numbers in increasing order.
Then, start guessing potential special numbers. A good starting point is the largest number in the list.
For each guess, count how many numbers in the list are greater than or equal to that guess.
If the count matches the guess, you've found the special number, and you're done.
If the count is too low, your guess was too high, so try a lower number. Since the list is sorted, we can use this fact to efficiently skip over numbers.
If the count is too high, your guess was too low, so try a higher number. Again, the sorted list helps us do this efficiently.
Continue adjusting your guess until you find the special number, or until you determine that no such number exists in the list.

Code Implementation

def find_special_integer(numbers):
    numbers.sort()
    list_length = len(numbers)
    left_index = 0
    right_index = list_length - 1

    while left_index <= right_index:
        potential_special_number = (left_index + right_index) // 2
        count = 0

        # Count elements greater or equal
        for number in numbers:
            if number >= numbers[potential_special_number]:
                count += 1

        # Check if it's the special integer
        if count == numbers[potential_special_number]:
            return numbers[potential_special_number]

        # Adjust the search range if it is too low
        if count < numbers[potential_special_number]:
            right_index = potential_special_number - 1

        # Adjust the search range if it is too high
        else:
            left_index = potential_special_number + 1

    return -1

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation is sorting the input list nums, which takes O(n log n) time. While there is a search process to find the special number, it involves iterating and comparing, but the sorting step determines the overall time complexity. Since the search for the special number after sorting is O(n) at most, the overall complexity is determined by sorting. Therefore, the algorithm has a time complexity of O(n log n).

Space Complexity

O(1) – The provided solution primarily uses in-place sorting (or a sorting algorithm with minimal auxiliary space). After sorting, the algorithm iterates and performs comparisons using a few integer variables. The number of integer variables used is constant and independent of the input list's size, denoted as N. Thus, the auxiliary space required does not scale with the input and is considered constant.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as there's no array to process.
Input array with a single element	Return 0 immediately as no 'special' number can exist.
Array with all elements being 0	The algorithm should iterate and count elements correctly, correctly returning the 'special' number or 0.
Array sorted in descending order	The algorithm should handle this without issues by iterating and comparing elements with their index.
Array with very large integers	The algorithm compares the integer values with their index so these should not affect complexity or cause issues if within acceptable bounds for the programming language.
Array containing duplicates and a 'special' number exists	The 'special' number should still be found correctly despite duplicates, if it exists.
No 'special' number exists in the input	The algorithm should correctly return 0 after checking all possibilities.
Array of maximum size allowed by memory constraints	The algorithm must iterate through the array in O(n) time and not allocate unnecessary memory to avoid memory constraints.

Null or empty input array

How to Handle:

Return 0 immediately as there's no array to process.

Input array with a single element

How to Handle:

Return 0 immediately as no 'special' number can exist.

Array with all elements being 0

How to Handle:

The algorithm should iterate and count elements correctly, correctly returning the 'special' number or 0.

Array sorted in descending order

How to Handle:

The algorithm should handle this without issues by iterating and comparing elements with their index.

Array with very large integers

How to Handle:

The algorithm compares the integer values with their index so these should not affect complexity or cause issues if within acceptable bounds for the programming language.

Array containing duplicates and a 'special' number exists

How to Handle:

The 'special' number should still be found correctly despite duplicates, if it exists.

No 'special' number exists in the input

How to Handle:

The algorithm should correctly return 0 after checking all possibilities.

Array of maximum size allowed by memory constraints

How to Handle:

The algorithm must iterate through the array in O(n) time and not allocate unnecessary memory to avoid memory constraints.

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 within the input array?
Can the input array be empty or null?
If multiple values of 'x' satisfy the condition, should I return any one of them?
Are all the numbers in the input array integers?
What should I return if no such 'x' exists?

Brute Force Solution

Approach

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

Start with the number zero.
Check if the number of values in the input that are greater than or equal to zero matches zero itself.
If the match occurs, we have found the special number.
If there's no match, increase the number we're checking by one.
Repeat the checking process with this new number.
Continue incrementing and checking until we find a match or we've tried all possible numbers within the size of the input.

Code Implementation

def special_array_i_brute_force(numbers):
    array_length = len(numbers)

    for potential_special_value in range(array_length + 1):
        # Count elements greater or equal to the potential value
        count = 0
        for number in numbers:
            if number >= potential_special_value:
                count += 1

        # Check if potential value is the special number
        if count == potential_special_value:
            return potential_special_value

    # No special value found, after checking
    # all potential values.
    return -1

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, sort the list of numbers in increasing order.
Then, start guessing potential special numbers. A good starting point is the largest number in the list.
For each guess, count how many numbers in the list are greater than or equal to that guess.
If the count matches the guess, you've found the special number, and you're done.
If the count is too low, your guess was too high, so try a lower number. Since the list is sorted, we can use this fact to efficiently skip over numbers.
If the count is too high, your guess was too low, so try a higher number. Again, the sorted list helps us do this efficiently.
Continue adjusting your guess until you find the special number, or until you determine that no such number exists in the list.

Code Implementation

def find_special_integer(numbers):
    numbers.sort()
    list_length = len(numbers)
    left_index = 0
    right_index = list_length - 1

    while left_index <= right_index:
        potential_special_number = (left_index + right_index) // 2
        count = 0

        # Count elements greater or equal
        for number in numbers:
            if number >= numbers[potential_special_number]:
                count += 1

        # Check if it's the special integer
        if count == numbers[potential_special_number]:
            return numbers[potential_special_number]

        # Adjust the search range if it is too low
        if count < numbers[potential_special_number]:
            right_index = potential_special_number - 1

        # Adjust the search range if it is too high
        else:
            left_index = potential_special_number + 1

    return -1

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as there's no array to process.
Input array with a single element	Return 0 immediately as no 'special' number can exist.
Array with all elements being 0	The algorithm should iterate and count elements correctly, correctly returning the 'special' number or 0.
Array sorted in descending order	The algorithm should handle this without issues by iterating and comparing elements with their index.
Array with very large integers	The algorithm compares the integer values with their index so these should not affect complexity or cause issues if within acceptable bounds for the programming language.
Array containing duplicates and a 'special' number exists	The 'special' number should still be found correctly despite duplicates, if it exists.
No 'special' number exists in the input	The algorithm should correctly return 0 after checking all possibilities.
Array of maximum size allowed by memory constraints	The algorithm must iterate through the array in O(n) time and not allocate unnecessary memory to avoid memory constraints.

Null or empty input array

How to Handle:

Return 0 immediately as there's no array to process.

Input array with a single element

How to Handle:

Return 0 immediately as no 'special' number can exist.

Array with all elements being 0

How to Handle:

The algorithm should iterate and count elements correctly, correctly returning the 'special' number or 0.

Array sorted in descending order

How to Handle:

The algorithm should handle this without issues by iterating and comparing elements with their index.

Array with very large integers

How to Handle:

The algorithm compares the integer values with their index so these should not affect complexity or cause issues if within acceptable bounds for the programming language.

Array containing duplicates and a 'special' number exists

How to Handle:

The 'special' number should still be found correctly despite duplicates, if it exists.

No 'special' number exists in the input

How to Handle:

The algorithm should correctly return 0 after checking all possibilities.

Array of maximum size allowed by memory constraints

How to Handle:

The algorithm must iterate through the array in O(n) time and not allocate unnecessary memory to avoid memory constraints.

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