Find Maximum Non-decreasing Array Length

Hard

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ArraysGreedy Algorithms

You are given a 0-indexed integer array nums.

You can perform any number of operations, where each operation involves selecting a subarray of the array and replacing it with the sum of its elements. For example, if the given array is [1,3,5,6] and you select subarray [3,5] the array will convert to [1,8,6].

Return the maximum length of a non-decreasing array that can be made after applying operations.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [5,2,2]
Output: 1
Explanation: This array with length 3 is not non-decreasing.
We have two ways to make the array length two.
First, choosing subarray [2,2] converts the array to [5,4].
Second, choosing subarray [5,2] converts the array to [7,2].
In these two ways the array is not non-decreasing.
And if we choose subarray [5,2,2] and replace it with [9] it becomes non-decreasing. 
So the answer is 1.

Example 2:

Input: nums = [1,2,3,4]
Output: 4
Explanation: The array is non-decreasing. So the answer is 4.

Example 3:

Input: nums = [4,3,2,6]
Output: 3
Explanation: Replacing [3,2] with [5] converts the given array to [4,5,6] that is non-decreasing.
Because the given array is not non-decreasing, the maximum possible answer is 3.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 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 are the constraints on the size of the input array `nums` and the range of values within it?
Can the input array `nums` be empty or null?
If no operations are performed and the array is already non-decreasing, should I return the original length of the array?
If an element cannot be split to form a non-decreasing array, what should the algorithm do?
Are there any constraints on the number of times an element can be split?

Brute Force Solution

Approach

The brute force approach to finding the longest non-decreasing array involves checking absolutely every possible subarray. We try every starting point and every ending point to consider every possible subarray combination. By checking every possible combination, we're guaranteed to find the longest non-decreasing one.

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

Start by considering only the first number in the entire set.
Then, check the first two numbers together, and see if the second number is at least as big as the first. If it is, remember this pair.
Next, check the first three numbers together, again seeing if each number is at least as big as the one before it. If so, remember this set of three.
Keep doing this, adding one more number each time from the beginning of the set, until you've included all the numbers.
Now, move on to the second number in the original set, and repeat the same process: check the second number alone, then the second and third, then the second, third, and fourth, and so on.
Continue this pattern, each time starting with the next number in the original set and checking all the sets that follow it.
As you check each set of numbers, keep track of how many numbers are in each non-decreasing set you find.
Once you've checked every possible set, choose the one with the most numbers. That will be the longest non-decreasing set.

Code Implementation

def find_maximum_non_decreasing_length(numbers):
    maximum_length = 0

    for start_index in range(len(numbers)):
        for end_index in range(start_index, len(numbers)):
            subarray = numbers[start_index:end_index + 1]
            
            # Handle edge case where subarray is empty
            if not subarray:
                continue

            is_non_decreasing = True

            # Check if the current subarray is non-decreasing
            for index in range(1, len(subarray)):
                if subarray[index] < subarray[index - 1]:
                    is_non_decreasing = False
                    break
            
            # If the subarray is non-decreasing, update maximum length
            if is_non_decreasing:
                current_length = len(subarray)

                # Update maximum_length if needed.
                if current_length > maximum_length:
                    maximum_length = current_length

    return maximum_length

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible subarrays of the input array of size n. The outer loop determines the starting index, which ranges from 0 to n-1. The inner loop determines the ending index of the subarray, starting from the current starting index and going up to n-1. Therefore, for each starting index, the inner loop can iterate up to n times in the worst case. Thus, the total number of operations approximates n * n/2, which simplifies to O(n²).

Space Complexity

O(1) – The brute force approach, as described, doesn't use any auxiliary data structures that scale with the input size N. The algorithm only needs to maintain a few variables, such as the starting and ending indices of the subarrays being checked, and the length of the longest non-decreasing subarray found so far. These variables consume a constant amount of space regardless of the input array's size. Therefore, the auxiliary space complexity is constant, or O(1).

Optimal Solution

Approach

The key is to build the longest possible non-decreasing sequence by making the smartest choice at each step. We focus on only keeping track of the potential end values of these sequences, and updating them in a clever way to achieve the longest length.

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

Imagine you are building a series of increasing steps where each step represents the end of a valid non-decreasing sequence.
Maintain a list that stores the smallest possible ending value for each possible sequence length. For example, the first number in the list represents the smallest ending value of any sequence of length 1, the second number represents the smallest ending value of any sequence of length 2, and so on.
As you go through the input, you check where each number would fit within these existing sequences. If a number is larger than the smallest ending value of an existing sequence, it means you can extend that sequence.
If the number can extend a sequence, see if using the number makes for a better (smaller) ending value for the new, longer sequence. If so, update the ending value.
By always aiming for the smallest ending value for each sequence length, we guarantee that we are maximizing our ability to extend those sequences later on.
The length of the longest possible sequence is then the length of the list of smallest ending values you are tracking.

Code Implementation

def find_maximum_non_decreasing_array_length(input_array):
    smallest_end_values = []

    for number in input_array:
        if not smallest_end_values or number >= smallest_end_values[-1]:
            # Extends the longest sequence
            smallest_end_values.append(number)

        else:
            # Find the first element greater than the current number
            left_index = 0
            right_index = len(smallest_end_values) - 1

            while left_index < right_index:
                middle_index = (left_index + right_index) // 2

                if smallest_end_values[middle_index] <= number:
                    left_index = middle_index + 1
                else:
                    right_index = middle_index

            # This ensures the smallest ending value for a sequence of same length
            smallest_end_values[left_index] = number

    return len(smallest_end_values)

Big(O) Analysis

Time Complexity

O(n log n) – We iterate through each of the n elements in the input array. For each element, we perform a binary search on the list of smallest ending values to find the appropriate place to insert or update. Binary search takes O(log k) time, where k is the length of the list of smallest ending values. Since k can be at most n, each iteration takes O(log n) time. Therefore, the overall time complexity is O(n log n).

Space Complexity

O(N) – The provided algorithm maintains a list that stores the smallest possible ending value for each possible sequence length. In the worst-case scenario, the input array is strictly increasing, leading to a new sequence of length i for each element in the input. Therefore, this list can grow up to the size of the input array, N, where N is the number of elements in the input array. This list represents the auxiliary space used by the algorithm. Thus the space complexity is O(N).

Edge Cases

Case	How to Handle
Empty or null input array	Return 0, as no non-decreasing array can be formed.
Array with a single element	Return 1, as a single element is always non-decreasing.
Array with all elements equal	Each element can be split into many 1s, resulting in array length equal to sum of elements; the solution should handle this efficiently.
Array with elements already in non-decreasing order	Return the original array length directly as no splits are needed.
Very large input numbers potentially causing integer overflow when split repeatedly.	Ensure that the splitting logic and calculations avoid exceeding the maximum integer value using appropriate data types or modulo operations if necessary.
Array with a mix of small and extremely large numbers	The splitting strategy should handle both cases efficiently, potentially needing to split large numbers more often to accommodate small numbers.
Maximum sized input array (constrained by problem definition)	Ensure the solution scales efficiently (ideally O(n log n) or better) to avoid exceeding the time limit.
Input array containing only the value 1	The array's initial state is already optimal, return the initial length.

Empty or null input array

How to Handle:

Return 0, as no non-decreasing array can be formed.

Array with a single element

How to Handle:

Return 1, as a single element is always non-decreasing.

Array with all elements equal

How to Handle:

Each element can be split into many 1s, resulting in array length equal to sum of elements; the solution should handle this efficiently.

Array with elements already in non-decreasing order

How to Handle:

Return the original array length directly as no splits are needed.

Very large input numbers potentially causing integer overflow when split repeatedly.

How to Handle:

Ensure that the splitting logic and calculations avoid exceeding the maximum integer value using appropriate data types or modulo operations if necessary.

Array with a mix of small and extremely large numbers

How to Handle:

The splitting strategy should handle both cases efficiently, potentially needing to split large numbers more often to accommodate small numbers.

Maximum sized input array (constrained by problem definition)

How to Handle:

Ensure the solution scales efficiently (ideally O(n log n) or better) to avoid exceeding the time limit.

Input array containing only the value 1

How to Handle:

The array's initial state is already optimal, return the initial length.