Problem Description

You are given an integer array nums that is sorted in non-decreasing order and an integer k. Your task is to determine if you can divide this array into one or more disjoint increasing subsequences where each subsequence has a length of at least k.

A few key points to understand:

• The array is already sorted in non-decreasing order (elements can be equal or increasing)
• You need to partition all elements into groups
• Each group must form a strictly increasing subsequence (no equal elements allowed within a group)
• Each group must contain at least k elements
• The groups must be disjoint (each element belongs to exactly one group)

For example, if nums = [1, 1, 2, 2, 3, 3] and k = 3, you could potentially divide it into two subsequences like [1, 2, 3] and [1, 2, 3], making it valid.

The solution leverages a key insight: if a number appears cnt times in the array, those cnt occurrences must go into different subsequences (since subsequences must be strictly increasing). Therefore, you need at least cnt subsequences. Since each subsequence needs at least k elements, and you have n total elements, the condition cnt × k ≤ n must be satisfied. The algorithm finds the maximum frequency of any element and checks if this condition holds.

Return true if such a division is possible, false otherwise.

Intuition

Let's think about what constraints we face when trying to divide the array into strictly increasing subsequences.

Since the array is sorted in non-decreasing order, we might have duplicate values. The critical observation is that duplicate values cannot be placed in the same subsequence because subsequences must be strictly increasing. If a value appears cnt times, we need at least cnt different subsequences to accommodate all occurrences of that value.

Consider the value that appears most frequently in the array - let's say it appears max_count times. This becomes our bottleneck. We absolutely need at least max_count subsequences because these duplicate values must be separated into different groups.

Now, given that we need at least max_count subsequences, and each subsequence must have at least k elements, we need a total of at least max_count × k elements in our array. If our array has n elements, then for a valid division to exist, we must have max_count × k ≤ n.

Why is this condition sufficient? Because the array is sorted, we can greedily distribute elements across our max_count subsequences. We can place the first occurrence of each distinct value in the first subsequence, the second occurrence in the second subsequence, and so on. Since the array is sorted, this naturally creates strictly increasing subsequences.

Therefore, the problem reduces to finding the maximum frequency of any element in the array and checking if max_frequency × k ≤ array_length. This elegant reduction transforms a seemingly complex partitioning problem into a simple counting problem.

Solution Approach

The implementation follows directly from our intuition that we need to find the maximum frequency of any element and check if max_frequency × k ≤ n.

The solution uses Python's groupby function from the itertools module to efficiently count consecutive occurrences of each value in the sorted array. Here's how the implementation works:

1. Group consecutive equal elements: Since the array is already sorted, all occurrences of the same value are consecutive. The groupby(nums) function creates groups where each group contains all consecutive occurrences of the same value.

2. Find maximum group size: For each group created by groupby, we convert it to a list and find its length. The expression max(len(list(x)) for _, x in groupby(nums)) iterates through all groups and finds the maximum size among them. This gives us the count of the most frequent element.

3. Check the feasibility condition: We multiply the maximum frequency mx by k and check if it's less than or equal to the total length of the array. If mx * k <= len(nums), we can successfully divide the array into subsequences, so we return true. Otherwise, we return false.

The algorithm runs in O(n) time complexity where n is the length of the array, as we need to traverse the array once to group elements and find the maximum frequency. The space complexity is O(1) if we don't count the space used by the iterator, or O(m) where m is the maximum frequency of any element (for temporarily storing the group as a list).

This approach elegantly transforms the complex problem of partitioning into subsequences into a simple counting problem, leveraging the sorted nature of the input array.

Example Walkthrough

Let's walk through an example with nums = [1, 2, 2, 3, 3, 3, 4, 4, 5] and k = 3.

Step 1: Identify consecutive groups Using groupby, we identify groups of consecutive equal elements:

• Group 1: [1] - one occurrence of 1
• Group 2: [2, 2] - two occurrences of 2
• Group 3: [3, 3, 3] - three occurrences of 3
• Group 4: [4, 4] - two occurrences of 4
• Group 5: [5] - one occurrence of 5

Step 2: Find maximum frequency We find the maximum group size:

• Group sizes: 1, 2, 3, 2, 1
• Maximum frequency mx = 3 (from the value 3 appearing three times)

Step 3: Check feasibility condition

• We need at least mx = 3 subsequences (since the three 3's must go into different subsequences)
• Each subsequence needs at least k = 3 elements
• Total elements needed: mx × k = 3 × 3 = 9
• Total elements available: len(nums) = 9
• Since 9 ≤ 9, the condition is satisfied ✓

Step 4: Visualize the division (for understanding) We can divide the array into 3 subsequences:

• Subsequence 1: [1, 2, 3]
• Subsequence 2: [2, 3, 4]
• Subsequence 3: [3, 4, 5]

Each subsequence is strictly increasing and has exactly 3 elements, meeting our requirements.

Counter-example: If k = 4 instead:

• We'd need mx × k = 3 × 4 = 12 elements
• But we only have 9 elements
• Since 12 > 9, we'd return false

The algorithm correctly identifies that when an element appears mx times, we need exactly mx subsequences, and with each needing k elements, we can quickly determine feasibility by checking if mx × k ≤ n.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. This is because:

• The groupby function iterates through the array once to create groups of consecutive equal elements, which takes O(n) time
• For each group created by groupby, we convert it to a list using list(x) and calculate its length
• Finding the maximum among all group lengths requires iterating through all groups once
• The total number of elements across all groups is n, so the overall time complexity remains O(n)

The space complexity is O(n) in the worst case, not O(1) as stated in the reference answer. This is because:

• The list(x) operation inside the generator expression materializes each group into a list
• In the worst case (when all elements are the same), a single group would contain all n elements, requiring O(n) space
• Even though we process groups one at a time in the generator expression, each group still needs to be converted to a list to calculate its length

Note: If the code were modified to count group sizes without materializing lists (e.g., using sum(1 for _ in x)), the space complexity could be reduced to O(1).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Empty Array Handling

The current implementation will raise a ValueError when the input array is empty because max() is called on an empty sequence.

Why it happens: When nums is empty, groupby(nums) produces no groups, making the generator expression empty. Calling max() on an empty sequence raises an exception.

Solution:

def canDivideIntoSubsequences(self, nums: List[int], k: int) -> bool:
    if not nums:
        return k == 0  # or return True, depending on problem requirements

    max_frequency = max(len(list(group)) for _, group in groupby(nums))
    return max_frequency * k <= len(nums)

Pitfall 2: Misunderstanding the Strictly Increasing Requirement

Developers might incorrectly assume that the sorted array itself forms a valid subsequence, forgetting that strictly increasing means no duplicate values are allowed within the same subsequence.

Wrong assumption: If nums = [1, 2, 2, 3] and k = 4, one might think this can form a single subsequence, but [1, 2, 2, 3] is not strictly increasing due to the duplicate 2s.

Correct understanding: The two 2s must go into different subsequences, requiring at least 2 subsequences. With k = 4, we'd need at least 8 elements total, but we only have 4, so it's impossible.

Pitfall 3: Manual Frequency Counting Instead of Using groupby

Some might attempt to manually count frequencies using a dictionary or Counter, which works but is less efficient for sorted arrays:

# Less efficient approach for sorted arrays:
from collections import Counter
max_frequency = max(Counter(nums).values())  # Works but unnecessary

Why groupby is better: Since the array is already sorted, consecutive equal elements are grouped together. Using groupby avoids the overhead of hashing and provides a more elegant solution that takes advantage of the sorted property.

Pitfall 4: Consuming the groupby Iterator Multiple Times

The group iterator from groupby can only be consumed once. If you try to use it multiple times, you'll get unexpected results:

# Wrong approach:
for value, group in groupby(nums):
    print(len(list(group)))  # Consumes the iterator
    print(len(list(group)))  # Will always print 0!

Solution: Always convert to a list immediately if you need to use the group multiple times or access its length.