Find Latest Group of Size M

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Arrays

Given an array arr that represents a permutation of numbers from 1 to n.

You have a binary string of size n that initially has all its bits set to zero. At each step i (assuming both the binary string and arr are 1-indexed) from 1 to n, the bit at position arr[i] is set to 1.

You are also given an integer m. Find the latest step at which there exists a group of ones of length m. A group of ones is a contiguous substring of 1's such that it cannot be extended in either direction.

Return the latest step at which there exists a group of ones of length exactly m. If no such group exists, return -1.

Example 1:

Input: arr = [3,5,1,2,4], m = 1
Output: 4
Explanation: 
Step 1: "00100", groups: ["1"]
Step 2: "00101", groups: ["1", "1"]
Step 3: "10101", groups: ["1", "1", "1"]
Step 4: "11101", groups: ["111", "1"]
Step 5: "11111", groups: ["11111"]
The latest step at which there exists a group of size 1 is step 4.

Example 2:

Input: arr = [3,1,5,4,2], m = 2
Output: -1
Explanation: 
Step 1: "00100", groups: ["1"]
Step 2: "10100", groups: ["1", "1"]
Step 3: "10101", groups: ["1", "1", "1"]
Step 4: "10111", groups: ["1", "111"]
Step 5: "11111", groups: ["11111"]
No group of size 2 exists during any step.

Constraints:

n == arr.length
1 <= m <= n <= 10⁵
1 <= arr[i] <= n
All integers in arr are distinct.

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 maximum possible value for `m` relative to the length of the input array?
If there are no groups of size `m` at any point, what value should I return?
Are the values in the input array `arr` guaranteed to be unique and within the range [1, n]?
If multiple groups of size `m` become the latest at the same step, should I return the index of the last such step?
What data type should I use for the input array `arr`, and can I assume it is non-null and properly formatted?

Brute Force Solution

Approach

We want to find the last moment when a group of consecutive numbers has a specific size. The brute force approach involves going through each moment and checking every possible group of consecutive numbers to see if any has the desired size.

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

Look at the first moment in time.
Check every possible sequence of consecutive numbers to see if there's a sequence of the required size.
If we find a sequence of the required size, we note down this moment.
Move on to the next moment in time.
Again, check every possible sequence of consecutive numbers for a group of the required size.
If we find a sequence of the required size, we note down this moment.
Repeat this process for every moment in time.
After checking all moments, find the latest moment that we noted down. This will be the answer.

Code Implementation

def find_latest_group_of_size_m_brute_force(array_of_ones, group_size):
    latest_moment = -1
    array_length = len(array_of_ones)

    for moment in range(array_length):
        current_arrangement = [0] * array_length
        # Simulate placing 1s up to the current moment
        for i in range(moment + 1):
            current_arrangement[array_of_ones[i] - 1] = 1

        # Check all possible sequences to find a group of the required size
        for start_index in range(array_length):
            for end_index in range(start_index, array_length):
                sub_array = current_arrangement[start_index:end_index+1]
                if len(sub_array) >= group_size:
                    is_consecutive_group = True
                    group_length = 0
                    for element in sub_array:
                        if element == 1:
                            group_length += 1
                        else:
                            group_length = 0

                        if group_length == group_size:
                            # We found a group, so we update latest_moment
                            latest_moment = moment + 1
                            is_consecutive_group = False
                            break

                    if not is_consecutive_group:
                        break

    return latest_moment

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm iterates through each moment in time, which is determined by the length of the input array arr, so there are n iterations in the outer loop. Inside each iteration, it checks every possible consecutive sequence. In the worst case, this check requires iterating through the updated array to identify the consecutive sequences which will take O(n) time. Therefore, the nested structure results in a time complexity of O(n * n). Simplifying, the overall time complexity is O(n^2).

Space Complexity

O(1) – The provided brute force approach checks every possible consecutive sequence at each moment in time. It only notes down the latest moment when a group of the required size M is found. This implies using a single variable to store the latest moment. No auxiliary data structures that scale with the input size N (where N is the length of the input array representing moments in time) are created. Therefore, the space complexity is constant.

Optimal Solution

Approach

The core idea is to keep track of how many consecutive ones are around each new one we place. We focus on how placing a 'one' affects the lengths of its neighboring groups, allowing us to find the latest time a group of the required size appears.

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

Imagine a line of empty spaces initially represented as zeros.
Process each number from the input list one by one, treating each number as a location to fill with a 'one'.
When filling a location with 'one', check the lengths of the group of 'ones' to its left and to its right.
Update the total length of the new combined group of 'ones' created by filling this location.
As groups are formed, keep track of how many groups of each length exist at each point.
If a group of the desired length appears or disappears, record the current time (the number we are processing) because this is the point when a valid group was either formed or broken.
After processing all the numbers, return the latest time when a group of the desired length existed. If the desired group never existed, return negative one.

Code Implementation

def find_latest_group(array, group_size):
    array_length = len(array)
    length_of_groups = [0] * (array_length + 2)
    count_of_groups = [0] * (array_length + 1)
    latest_time = -1

    for current_time, current_position in enumerate(array):
        left_length = length_of_groups[current_position - 1]
        right_length = length_of_groups[current_position + 1]

        # Total length of the new combined group.
        total_length = left_length + right_length + 1

        # Decrement count of groups that are now part of the larger group.
        count_of_groups[left_length] -= 1
        count_of_groups[right_length] -= 1

        # Increment count of the new group.
        count_of_groups[total_length] += 1

        # Update the length for the entire new group.
        length_of_groups[current_position - left_length] = total_length
        length_of_groups[current_position + right_length] = total_length
        length_of_groups[current_position] = total_length

        # Check if we have a group of desired size.
        if count_of_groups[group_size] > 0:
            latest_time = current_time + 1

    return latest_time

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of length n once. Inside the loop, the primary operations involve checking the lengths of neighboring groups and updating counts, which take constant time O(1). These O(1) operations are performed for each of the n elements in the input array. Therefore, the overall time complexity is proportional to n * O(1), which simplifies to O(n).

Space Complexity

O(N) – The algorithm uses auxiliary space primarily for two data structures: an array representing the line of empty spaces which grows linearly with the input size N, and a data structure for keeping track of how many groups of each length exist, which in the worst case, can also grow linearly with N (e.g., if each 'one' forms a group of size 1). Therefore, the overall auxiliary space required is proportional to the input size N, resulting in a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty input array arr	Return -1 immediately as no groups can be formed.
m is zero	If m is zero, return n (length of arr) because every index constitutes a group of size zero after all numbers are written.
m is greater than n	Return -1 because no group of size m can be formed if m is larger than the total number of elements.
arr contains duplicate numbers	The problem specifies distinct integers, so the provided 'arr' should not contain duplicates.
No group of size m exists	Return -1 after iterating through all elements if no group of size m was ever formed at any point.
Integer overflow in intermediate calculations	The problem constraints specify the integer range, and intermediate calculations should be kept within those bounds by using int data type.
Large n (array size) impacting performance	Optimize the solution to use a disjoint set or similar data structure for efficient component size tracking to maintain acceptable time complexity.
m is equal to n	If m equals n, the result will be n if all elements are present at any point, otherwise, it should still return -1.

Empty input array arr

How to Handle:

Return -1 immediately as no groups can be formed.

m is zero

How to Handle:

If m is zero, return n (length of arr) because every index constitutes a group of size zero after all numbers are written.

m is greater than n

How to Handle:

Return -1 because no group of size m can be formed if m is larger than the total number of elements.

arr contains duplicate numbers

How to Handle:

The problem specifies distinct integers, so the provided 'arr' should not contain duplicates.

No group of size m exists

How to Handle:

Return -1 after iterating through all elements if no group of size m was ever formed at any point.

Integer overflow in intermediate calculations

How to Handle:

The problem constraints specify the integer range, and intermediate calculations should be kept within those bounds by using int data type.

Large n (array size) impacting performance

How to Handle:

Optimize the solution to use a disjoint set or similar data structure for efficient component size tracking to maintain acceptable time complexity.

m is equal to n

How to Handle:

If m equals n, the result will be n if all elements are present at any point, otherwise, it should still return -1.