Problem Description

You are given an integer array nums and an integer k.

Your task is to determine whether you can partition all elements of nums into groups where:

• Each group contains exactly k distinct elements
• Every element in nums must be assigned to exactly one group

Return true if such a partition is possible, otherwise return false.

For example, if nums = [1, 2, 3, 3, 4, 4] and k = 3, you could form two groups: [1, 2, 3] and [3, 4, 4]. Each group has exactly 3 distinct elements, and all elements from the original array are used exactly once.

The key insight is that if we have n total elements and want groups of size k, we need n to be divisible by k to form m = n/k groups. Additionally, since each element can appear in multiple groups but each group needs distinct elements, no element can appear more than m times in the array. If any element appears more than m times, we won't be able to distribute all its occurrences across the m groups while maintaining the distinct element requirement within each group.

Intuition

Let's think about what constraints we have when trying to partition the array into groups of k distinct elements.

First, if we have n total elements and each group must have exactly k elements, then we need n to be perfectly divisible by k. If not, we'll have leftover elements that can't form a complete group. This gives us our first condition: n % k == 0.

Now, assuming n is divisible by k, we'll have exactly m = n / k groups to form.

Here's the crucial observation: since each group must contain distinct elements, an element that appears multiple times in the array must be distributed across different groups. For instance, if element 5 appears 3 times in the array, these three occurrences must go into 3 different groups.

This leads to our key constraint: if any element appears more than m times, it's impossible to satisfy the requirements. Why? Because we only have m groups total, and each occurrence of that element must go into a different group (to maintain the "distinct elements per group" rule). If an element appears more than m times, we simply don't have enough groups to place all its occurrences.

Conversely, if every element appears at most m times, we can always find a valid partition. We can think of it as a matching problem where we're assigning element occurrences to groups, and as long as no element exceeds the number of available groups, a valid assignment exists.

Therefore, the solution reduces to two simple checks:

1. Is n divisible by k?
2. Does any element appear more than m = n/k times?

If both conditions are satisfied, we can partition the array as required.

Solution Approach

The implementation follows directly from our intuition and consists of three main steps:

1. Check divisibility: First, we calculate the quotient m and remainder mod when dividing the length of the array n by k:

   m, mod = divmod(len(nums), k)

   If mod is not zero (meaning n is not divisible by k), we immediately return False since we cannot form complete groups.

2. Count element frequencies: We use Python's Counter from the collections module to count the occurrence of each element in the array:

   Counter(nums)

   This gives us a dictionary where keys are the unique elements and values are their frequencies.

3. Check maximum frequency: We find the maximum frequency among all elements using:

   max(Counter(nums).values())

   If this maximum frequency exceeds m (the number of groups we can form), we return False. Otherwise, we return True.

The complete solution elegantly combines these checks:

def partitionArray(self, nums: List[int], k: int) -> bool:
    m, mod = divmod(len(nums), k)
    if mod:
        return False
    return max(Counter(nums).values()) <= m

Time Complexity: O(n) where n is the length of the array, as we need to count the frequency of each element.

Space Complexity: O(d) where d is the number of distinct elements in the array, needed to store the frequency count.

The beauty of this solution lies in its simplicity - instead of trying to construct the actual groups, we identify the necessary and sufficient conditions for a valid partition to exist and check only those conditions.

Example Walkthrough

Let's walk through the solution with nums = [1, 2, 3, 3, 4, 4, 4] and k = 3.

Step 1: Check if we can form complete groups

• Array length n = 7
• Calculate m = n // k = 7 // 3 = 2 (number of groups)
• Calculate mod = n % k = 7 % 3 = 1 (remainder)
• Since mod = 1 ≠ 0, we cannot divide 7 elements into groups of exactly 3

At this point, we return False immediately. We need 1 more element to form complete groups.

Let's modify the example to nums = [1, 2, 3, 3, 4, 4] and k = 3 for a complete walkthrough:

Step 1: Check divisibility

• Array length n = 6
• Calculate m = n // k = 6 // 3 = 2 (we can form 2 groups)
• Calculate mod = n % k = 6 % 3 = 0 (no remainder)
• Since mod = 0, we can potentially form complete groups ✓

Step 2: Count element frequencies

• Element frequencies: {1: 1, 2: 1, 3: 2, 4: 2}
• Maximum frequency = 2

Step 3: Check if max frequency ≤ number of groups

• Maximum frequency (2) ≤ number of groups (2)? Yes ✓
• Return True

This works because:

• We can form Group 1: [1, 2, 3] (using one occurrence of 3)
• We can form Group 2: [3, 4, 4] (using the second occurrence of 3 and both 4s)

Now let's see a failing case with nums = [1, 1, 1, 2, 3, 4] and k = 2:

Step 1: Check divisibility

• Array length n = 6
• Calculate m = n // k = 6 // 2 = 3 (we can form 3 groups)
• Calculate mod = n % k = 6 % 2 = 0 ✓

Step 2: Count element frequencies

• Element frequencies: {1: 3, 2: 1, 3: 1, 4: 1}
• Maximum frequency = 3

Step 3: Check if max frequency ≤ number of groups

• Maximum frequency (3) ≤ number of groups (3)? Yes ✓
• Return True

This works: Group 1: [1, 2], Group 2: [1, 3], Group 3: [1, 4]

Solution Implementation

Time and Space Complexity

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

• Computing len(nums) takes O(1) time
• divmod(len(nums), k) takes O(1) time
• Counter(nums) iterates through all n elements once to count their frequencies, which takes O(n) time
• Counter(nums).values() returns a view of the values in O(1) time
• max(Counter(nums).values()) iterates through all unique values in the counter to find the maximum, which in the worst case is O(n) when all elements are unique

The space complexity is O(n) or O(M), where n is the length of the array and M is the number of unique elements in the array. This is because:

• Counter(nums) creates a dictionary that stores each unique element as a key with its frequency as the value
• In the worst case when all elements are unique, the counter stores n key-value pairs, resulting in O(n) space
• In the best case when all elements are the same, the counter stores only 1 key-value pair, resulting in O(1) space
• Generally, the space complexity is O(M) where M is the number of distinct elements in nums, and M ≤ n

Common Pitfalls

Pitfall 1: Misunderstanding the Problem Requirements

The Issue: The most critical pitfall is misinterpreting what "k distinct elements" means. Many developers initially think the problem asks for groups where each group contains k identical elements (like [1,1,1] when k=3), when it actually requires k DIFFERENT elements in each group (like [1,2,3] when k=3).

Why It Happens: The example [3, 4, 4] in the problem description can be misleading - it shows a group with duplicate values, but the key is that it has 3 distinct elements (3, 4, and 4 are three positions with two distinct values).

The Fix: Re-read the problem carefully. The constraint is that each group must have exactly k elements where all elements within a single group are distinct from each other. This means:

• If k=3, valid groups look like [1,2,3], [4,5,6], not [1,1,1]
• No element can repeat within the same group

Pitfall 2: Incorrect Frequency Check Logic

The Issue: The current solution checks if max(element_frequency.values()) <= num_groups, but this is actually checking the wrong constraint. This would be correct if we needed groups of identical elements, but for groups of distinct elements, the logic is different.

The Correct Approach: For groups of k distinct elements:

def partitionArray(self, nums: List[int], k: int) -> bool:
    n = len(nums)
  
    # Array length must be divisible by k
    if n % k != 0:
        return False
  
    num_groups = n // k
    element_frequency = Counter(nums)
  
    # We need exactly k distinct elements per group
    # So we need at least k distinct elements total
    if len(element_frequency) < k:
        return False
  
    # Each element can appear at most num_groups times
    # (once per group maximum since elements must be distinct within groups)
    for freq in element_frequency.values():
        if freq > num_groups:
            return False
  
    return True

Pitfall 3: Edge Case Handling

Missing Edge Cases:

1. Empty array: nums = [] with any k
2. k larger than array length: nums = [1,2] with k = 3
3. k equals 1: Every element forms its own group
4. All elements are identical: nums = [1,1,1,1] with k = 2

Enhanced Solution with Edge Cases:

def partitionArray(self, nums: List[int], k: int) -> bool:
    # Handle empty array
    if not nums:
        return True
  
    n = len(nums)
  
    # Handle k = 1 (each element is its own group)
    if k == 1:
        return True
  
    # Basic divisibility check
    if n % k != 0:
        return False
  
    num_groups = n // k
    element_frequency = Counter(nums)
  
    # Need at least k distinct elements to form even one valid group
    if len(element_frequency) < k:
        return False
  
    # Check frequency constraint
    return all(freq <= num_groups for freq in element_frequency.values())

Pitfall 4: Performance Optimization Oversight

The Issue: Using max(Counter(nums).values()) creates the Counter twice if you also need to check the number of distinct elements, which is inefficient.

Better Approach:

def partitionArray(self, nums: List[int], k: int) -> bool:
    if not nums or k == 1:
        return True
      
    n = len(nums)
    if n % k != 0:
        return False
  
    num_groups = n // k
    element_frequency = Counter(nums)  # Create once, use multiple times
  
    # Check both constraints using the same Counter
    if len(element_frequency) < k:
        return False
  
    return max(element_frequency.values()) <= num_groups