Number of Equal Numbers Blocks

Medium

Asked by:

34 views

Topics:

Arrays

You are given an integer n representing the total number of blocks.

You need to divide these blocks into one or more non-overlapping groups such that each group forms a perfect square. In other words, the number of blocks in each group must be a perfect square (e.g., 1, 4, 9, 16, ...).

Two groups are considered equal if they have the same number of blocks. For example, two groups with 4 blocks are considered equal.

A numbers block is a set of groups where all groups in the set are equal. For example, a numbers block could consist of three groups of 4 blocks each.

Your task is to determine the maximum number of non-overlapping numbers blocks that can be formed from the given n blocks.

Example 1:

Input: n = 12
Output: 3
Explanation:
You can create three numbers blocks:
- Numbers block 1: One group of 9 blocks (9 is a perfect square).
- Numbers block 2: One group of 1 block (1 is a perfect square).
- Numbers block 3: One group of 1 block (1 is a perfect square).

Example 2:

Input: n = 16
Output: 1
Explanation:
You can create one numbers block consisting of one group of 16 blocks.

Example 3:

Input: n = 17
Output: 2
Explanation:
You can create two numbers blocks:
- Numbers block 1: One group of 16 blocks.
- Numbers block 2: One group of 1 block.

Constraints:

1 <= n <= 10⁵

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, and can it contain negative numbers, zeros, or floating-point numbers?
What should I return if there are no equal number blocks found in the input array?
Could you define more explicitly what constitutes a 'block'? Specifically, does a block need to be contiguous?
If multiple maximal equal number blocks exist, should I return the leftmost one, the longest one, or any valid one?
What are the constraints on the size of the input array, and do I need to consider memory usage or potential overflow issues?

Brute Force Solution

Approach

The basic idea is to look at every possible way the input data can be divided into blocks. Then, for each of those possibilities, we'll check if it meets the requirements of the problem.

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

First, consider just the very first portion of the data as one block.
Then, consider the first two portions together as a block, then the first three, and so on, all the way up to considering the entire data set as one big block.
For each of these possible first blocks, we then consider all the ways to form a second block using the remaining data.
We repeat this process for each subsequent block until we've divided the entire input into blocks.
For every single possible grouping of the data into blocks that we come up with, we check if all the numbers inside that specific block are equal. If they are NOT, we discard that grouping.
Finally, after checking all the possibilities and discarding the invalid ones, we count how many groupings are valid, and that gives us the answer.

Code Implementation

def number_of_equal_numbers_blocks_brute_force(numbers):
    number_of_blocks = 0
    def is_valid_block(block):
        if not block:
            return True
        first_number = block[0]
        for number in block:
            if number != first_number:
                return False
        return True

    def generate_blocks(index, current_blocks):
        nonlocal number_of_blocks
        # Base case: if we've reached the end of the list
        if index == len(numbers):
            number_of_blocks += 1
            return

        # Iterate through possible block sizes starting from index
        for block_end in range(index + 1, len(numbers) + 1):

            current_block = numbers[index:block_end]
            # Check if the current block is valid
            if is_valid_block(current_block):

                generate_blocks(block_end, current_blocks + [current_block])

    generate_blocks(0, [])
    return number_of_blocks

Big(O) Analysis

Time Complexity

O(3^n) – The described algorithm explores all possible ways to divide the input of size n into blocks. For each element, we have three choices: start a new block, extend the current block, or continue searching. Since, in the worst-case scenario, each element gives a branching factor of 3 for forming the blocks, and the problem is solved via brute force checking. In total, checking all possibilities will be proportional to 3 to the power of n. Each block's equality check contributes a factor that's linear with the size of the block but doesn't change the overall growth rate of 3^n.

Space Complexity

O(N) – The provided explanation suggests an exhaustive exploration of all possible block groupings using a recursive approach. Each level of recursion corresponds to a potential starting point for a new block. In the worst-case scenario, the recursion depth could reach N (the size of the input data) as we recursively divide the data into smaller blocks. This leads to a call stack whose space complexity is O(N). Other local variables used in each recursive call have constant size and do not affect the overall space complexity.

Optimal Solution

Approach

The most efficient way to count the number of equal blocks focuses on identifying consecutive groups of identical numbers. Instead of comparing every possible block, we look for where these groups start and end, and how many there are.

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

Go through the numbers one by one, keeping track of how long each streak of the same number is.
Every time the number changes, record the length of the previous streak.
Once you've gone through all the numbers, you'll have a list of the lengths of all the streaks.
Finally, just count how many of those streak lengths are the same. That's your answer.

Code Implementation

def equal_numbers_blocks(numbers):
    streak_lengths = []
    current_streak_length = 0
    previous_number = None

    for number in numbers:
        if number == previous_number:
            current_streak_length += 1
        else:
            # Record the previous streak length when the number changes
            if previous_number is not None:
                streak_lengths.append(current_streak_length)
            current_streak_length = 1
            previous_number = number

    # Add the length of the last streak
    if numbers:
        streak_lengths.append(current_streak_length)

    equal_streaks_count = 0

    # Count the number of streak lengths that appear more than once
    for length in set(streak_lengths):
        if streak_lengths.count(length) > 1:
            equal_streaks_count += 1

    return equal_streaks_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of numbers once to identify and record the lengths of consecutive streaks of equal numbers. The length of this iteration is directly proportional to n, where n is the number of elements in the input array. After obtaining the streak lengths, it iterates through this list (which has at most n elements) once to count the occurrences of each length. Thus, the overall time complexity is dominated by the initial linear scan and the final counting, resulting in O(n).

Space Complexity

O(N) – The space complexity is determined by the list of streak lengths that the algorithm creates. In the worst-case scenario, where the input array alternates between different numbers (e.g., [1, 2, 1, 2, ...]), the length of each streak will be 1. Consequently, we will store N streak lengths. Therefore, the auxiliary space used grows linearly with the input size N, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Empty or null input list	Return 0 or throw an IllegalArgumentException since no blocks can be formed.
List with only one element	Return 0 since a block requires at least two elements.
List with all elements being the same number	Count all adjacent pairs as blocks, or N-1 blocks for N same elements.
List with alternating equal and different numbers, e.g. [1,1,2,2,1,1]	The solution must correctly identify and count all separated blocks.
List with negative numbers	The solution should handle negative numbers without issues, treating them the same as positive ones.
Large input list to check for time complexity and memory usage	The chosen algorithm should have a time complexity of O(n) or O(n log n) to scale efficiently.
Integer overflow if counting blocks (if the count is very large)	Use a data type with sufficient range (e.g., long) for the block count if necessary.
List contains zero values	Zeros should be treated the same as any other number and compared for equality.

Empty or null input list

How to Handle:

Return 0 or throw an IllegalArgumentException since no blocks can be formed.

List with only one element

How to Handle:

Return 0 since a block requires at least two elements.

List with all elements being the same number

How to Handle:

Count all adjacent pairs as blocks, or N-1 blocks for N same elements.

List with alternating equal and different numbers, e.g. [1,1,2,2,1,1]

How to Handle:

The solution must correctly identify and count all separated blocks.

List with negative numbers

How to Handle:

The solution should handle negative numbers without issues, treating them the same as positive ones.

Large input list to check for time complexity and memory usage

How to Handle:

The chosen algorithm should have a time complexity of O(n) or O(n log n) to scale efficiently.

Integer overflow if counting blocks (if the count is very large)

How to Handle:

Use a data type with sufficient range (e.g., long) for the block count if necessary.

List contains zero values

How to Handle:

Zeros should be treated the same as any other number and compared for equality.