Maximum of Minimum Values in All Subarrays

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You are given an integer array nums of size n. You need to find the maximum possible minimum value among all subarrays of nums of different lengths.

More formally, for each possible length k from 1 to n, consider all subarrays of nums of length k. Find the minimum value within each of these subarrays, and then take the maximum of all these minimums. Your goal is to return an array result of size n, where result[i] is the maximum of minimum values among all subarrays of length i+1.

Example 1:

Input: nums = [1,3,2,4]
Output: [4,3,2,1]
Explanation:
For length 1, the subarrays are [1], [3], [2], [4]. The minimums are 1, 3, 2, 4. The maximum of these minimums is 4.
For length 2, the subarrays are [1,3], [3,2], [2,4]. The minimums are 1, 2, 2. The maximum of these minimums is 2.
For length 3, the subarrays are [1,3,2], [3,2,4]. The minimums are 1, 2. The maximum of these minimums is 2.
For length 4, the subarray is [1,3,2,4]. The minimum is 1. The maximum of the minimum is 1.
So the result is [4, 2, 2, 1].

Example 2:

Input: nums = [3,5,4,7,6,2]
Output: [7,6,4,4,3,2]

Constraints:

1 <= n <= 10⁵
1 <= nums[i] <= 10⁹

Try coding it in LeetCode

Maximum of Minimum Values in All Subarrays

Medium

Asked by:

4 views

Topics:

ArraysStacks

You are given an integer array nums of size n. You need to find the maximum possible minimum value among all subarrays of nums of different lengths.

More formally, for each possible length k from 1 to n, consider all subarrays of nums of length k. Find the minimum value within each of these subarrays, and then take the maximum of all these minimums. Your goal is to return an array result of size n, where result[i] is the maximum of minimum values among all subarrays of length i+1.

Example 1:

Input: nums = [1,3,2,4]
Output: [4,3,2,1]
Explanation:
For length 1, the subarrays are [1], [3], [2], [4]. The minimums are 1, 3, 2, 4. The maximum of these minimums is 4.
For length 2, the subarrays are [1,3], [3,2], [2,4]. The minimums are 1, 2, 2. The maximum of these minimums is 2.
For length 3, the subarrays are [1,3,2], [3,2,4]. The minimums are 1, 2. The maximum of these minimums is 2.
For length 4, the subarray is [1,3,2,4]. The minimum is 1. The maximum of the minimum is 1.
So the result is [4, 2, 2, 1].

Example 2:

Input: nums = [3,5,4,7,6,2]
Output: [7,6,4,4,3,2]

Constraints:

1 <= n <= 10⁵
1 <= nums[i] <= 10⁹

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 expected data type of the input array elements, and what is the range of possible values?
Can the input array be empty or contain null values?
If there are multiple subarrays with the same maximum of minimum values, should I return any one of them, or is there a specific criterion for choosing one?
Are there any constraints on the size of the subarrays?
Could you provide an example input and the corresponding expected output for better understanding?

Brute Force Solution

Approach

The brute force method tackles this by looking at every possible group of numbers within the main list. For each of these groups, it finds the smallest number, and then compares all these 'smallest numbers' to find the largest among them.

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

First, consider all groups of size one, then all groups of size two, then size three, and so on, until you look at a group containing all the numbers.
For each group you pick, identify the smallest number within that group.
Keep track of all the 'smallest numbers' that you found from all the different groups.
Once you've considered all possible groups and found their smallest numbers, find the largest number among those 'smallest numbers'. This largest number is your answer.

Code Implementation

def find_maximum_of_minimums(numbers):
    maximum_of_minimum_values = float('-inf')

    # Iterate through all possible subarray lengths
    for subarray_length in range(1, len(numbers) + 1):

        # Iterate through all possible starting positions for the subarray
        for start_index in range(len(numbers) - subarray_length + 1):
            end_index = start_index + subarray_length
            current_subarray = numbers[start_index:end_index]

            # Find the minimum value within the current subarray
            minimum_value_in_subarray = min(current_subarray)

            # Update the maximum of minimums if necessary
            # This keeps track of the largest min we've seen
            maximum_of_minimum_values = max(maximum_of_minimum_values, minimum_value_in_subarray)

    return maximum_of_minimum_values

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm iterates through all possible subarray lengths, from 1 to n. For each subarray length 'k', it iterates through all possible starting positions, resulting in approximately n-k+1 subarrays of length k. Finding the minimum element in each subarray of length k takes O(k) time. Therefore, the overall time complexity can be approximated as the sum of (n-k+1) * k for k from 1 to n, which expands to approximately n^3/6, so the time complexity is O(n^3).

Space Complexity

O(1) – The provided brute force algorithm, as described, does not use any significant auxiliary space. It iterates through all possible subarrays and calculates the minimum within each subarray. While it 'keeps track' of the smallest numbers found, it only needs to store the maximum of these minimums at any given time. Therefore, the auxiliary space used is constant and independent of the input size N (where N is the length of the input list).

Optimal Solution

Approach

The goal is to find the biggest value that's also the smallest in some smaller section of the data. Instead of checking every section separately, we cleverly use information about how sections relate to each other to speed things up.

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

Think about each value in the data one at a time, and consider it as a potential minimum.
For each value, figure out the biggest section around it where that value is actually the smallest value in that entire section.
The length of this biggest section tells you how many smaller sections that value is the minimum for.
Multiply the value (the potential minimum) by the length of that biggest section.
Keep track of the largest product you find across all these values.
The largest product you tracked is the answer; it's the biggest possible result of multiplying a minimum value by how many sections it's the minimum for.

Code Implementation

def maximum_of_minimum_values(data):
    maximum_product = 0
    data_length = len(data)

    for i in range(data_length):
        # Consider each element as a potential minimum
        current_minimum = data[i]
        left_boundary = i
        right_boundary = i

        # Expand left until a smaller element is found
        while left_boundary > 0 and data[left_boundary - 1] >= current_minimum:
            left_boundary -= 1

        # Expand right until a smaller element is found
        while right_boundary < data_length - 1 and data[right_boundary + 1] >= current_minimum:
            right_boundary += 1

        subarray_length = right_boundary - left_boundary + 1

        # Calculate product: min value * subarray length
        current_product = current_minimum * subarray_length

        #Update maximum product
        maximum_product = max(maximum_product, current_product)

    return maximum_product

Big(O) Analysis

Time Complexity

O(n) – The solution iterates through each of the n elements in the input array. For each element, it determines the largest subarray where that element is the minimum. This is done efficiently, likely using a stack-based approach to find the next smaller elements on the left and right. Therefore, finding the largest subarray for each element takes O(1) on average. Because the solution iterates through each element of the array once and performs a constant time operation for each, the overall time complexity is O(n).

Space Complexity

O(N) – The algorithm calculates, for each element, the length of the largest subarray where that element is the minimum. To determine these lengths, the algorithm implicitly needs to store information about the left and right boundaries for each element, often achievable using auxiliary data structures like stacks or arrays of size N. Thus, the extra space used is proportional to the input size N. This results in a linear space complexity.

Edge Cases

Case	How to Handle
Empty array input	Return an empty list immediately as there are no subarrays.
Array with a single element	Return a list containing just that single element as it is the only possible minimum.
Array with all identical values	The maximum of minimums will simply be that identical value, and the code should handle it correctly.
Array with extreme boundary values (very large or very small)	The code needs to handle large numbers without overflow; consider using long integers if necessary.
Array with all negative numbers	The code must correctly identify the largest negative number as the maximum of minimums.
Array sorted in ascending order	The minimum of each subarray will always be the first element of that subarray.
Array sorted in descending order	The minimum of each subarray will always be the last element of that subarray; important for testing boundary conditions within the subarray loop.
Very large array size potentially leading to time limit exceed	Ensure the algorithm has optimal time complexity, ideally O(n) or O(n log n), where n is the array size.