Problem Description

You have an integer array nums that contains exactly 2 * n integers.

Your task is to determine if you can divide all elements in nums into n pairs where:

• Every element in the array must be assigned to exactly one pair (no element can be left unpaired or used in multiple pairs)
• Both elements in each pair must have the same value (equal elements)

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

For example, if nums = [3,2,3,2,2,2], you can form 3 pairs: (3,3), (2,2), and (2,2), so the answer would be true.

The key insight is that for pairing to be possible, each unique value in the array must appear an even number of times. If any value appears an odd number of times, it's impossible to pair all its occurrences, making the division impossible.

Intuition

Think about what it means to form pairs with equal elements. If we have a value that appears in the array, we need to pair each occurrence of that value with another occurrence of the same value.

Consider a simple example: if the number 5 appears 3 times in the array, can we pair all three occurrences? We can pair two of them together to form one pair (5,5), but we'll have one 5 left over with no partner. This tells us that any value appearing an odd number of times makes pairing impossible.

On the other hand, if a value appears an even number of times (like 2, 4, 6, etc.), we can always pair them up completely. For instance, if 7 appears 4 times, we can form 2 pairs: (7,7) and (7,7).

This observation leads to a simple rule: for the array to be divisible into n pairs of equal elements, every unique value must appear an even number of times. If even one value appears an odd number of times, we cannot form the required pairs.

Therefore, the solution becomes straightforward - count the frequency of each element in the array and check if all frequencies are even. We can use a Counter or hash table to track frequencies, then verify that each count is divisible by 2 (i.e., count % 2 == 0).

Solution Approach

The implementation uses a counting approach with a hash table to track element frequencies.

Step 1: Count Element Frequencies

We use Python's Counter from the collections module to create a frequency map of all elements in the array. The Counter(nums) creates a dictionary-like object where:

• Keys are the unique elements from nums
• Values are the count of how many times each element appears

For example, if nums = [1,2,3,3,2,1], then cnt = Counter(nums) would give us {1: 2, 2: 2, 3: 2}.

Step 2: Check All Frequencies are Even

We iterate through all the frequency values using cnt.values() and check if each frequency is even using the modulo operator v % 2 == 0. This expression returns True if the value is even (divisible by 2) and False if it's odd.

The all() function returns True only if all elements in the iterable are True. So all(v % 2 == 0 for v in cnt.values()) will:

• Return True if every unique element appears an even number of times
• Return False if at least one element appears an odd number of times

Time Complexity: O(n) where n is the length of the array, as we traverse the array once to count frequencies and then check each unique value once.

Space Complexity: O(k) where k is the number of unique elements in the array, needed to store the frequency map.

This solution elegantly determines if pairing is possible by verifying the necessary and sufficient condition: all elements must have even frequencies.

Example Walkthrough

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

Step 1: Count Element Frequencies

First, we create a frequency map using Counter:

• Element 1 appears 2 times
• Element 2 appears 4 times
• Element 3 appears 2 times

So our frequency map is: {1: 2, 2: 4, 3: 2}

Step 2: Check if All Frequencies are Even

Now we check each frequency value:

• Frequency of 1 is 2 → 2 % 2 = 0 ✓ (even)
• Frequency of 2 is 4 → 4 % 2 = 0 ✓ (even)
• Frequency of 3 is 2 → 2 % 2 = 0 ✓ (even)

Since all frequencies are even, we return true.

We can verify this is correct by forming the pairs:

• Two 1s form pair (1,1)
• Four 2s form pairs (2,2) and (2,2)
• Two 3s form pair (3,3)

This gives us 4 pairs from 8 elements, satisfying the requirement.

Counter-example: If we had nums = [1,2,3,4], the frequency map would be {1: 1, 2: 1, 3: 1, 4: 1}. Since all frequencies are odd (1), we'd return false - it's impossible to pair elements when each appears only once.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm iterates through the input array nums once to build the frequency counter using Counter(nums), which takes O(n) time where n is the length of the array. Then, it iterates through all values in the counter dictionary to check if each frequency is even. In the worst case, every element is unique, resulting in n values to check, but this is still O(n). The modulo operation for each value takes O(1) time. Therefore, the overall time complexity is O(n) + O(n) = O(n).

Space Complexity: O(n)

The Counter object stores the frequency of each unique element in the array. In the worst case, when all elements are unique, the counter will store n key-value pairs, requiring O(n) space. Even in the best case where all elements are the same, it would only store one key-value pair (O(1)), but we consider the worst-case scenario for space complexity analysis. Therefore, the space complexity is O(n).

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

Common Pitfalls

1. Forgetting Edge Cases with Empty or Odd-Length Arrays

A common mistake is not considering that the input array must have an even length for pairing to be possible. While the problem states the array contains exactly 2 * n integers, defensive programming suggests validating this assumption.

Pitfall Example:

def divideArray(self, nums: List[int]) -> bool:
    # Directly counting without checking array length
    frequency_count = Counter(nums)
    return all(count % 2 == 0 for count in frequency_count.values())

Solution:

def divideArray(self, nums: List[int]) -> bool:
    # First check if array length is even
    if len(nums) % 2 != 0:
        return False
  
    frequency_count = Counter(nums)
    return all(count % 2 == 0 for count in frequency_count.values())

2. Manual Frequency Counting with Incorrect Logic

When implementing without Counter, developers often make mistakes in the counting logic or checking phase.

Pitfall Example:

def divideArray(self, nums: List[int]) -> bool:
    freq = {}
    for num in nums:
        freq[num] = 1  # Wrong: This resets count to 1 each time
  
    for count in freq.values():
        if count % 2 != 0:
            return False
    return True

Solution:

def divideArray(self, nums: List[int]) -> bool:
    freq = {}
    for num in nums:
        freq[num] = freq.get(num, 0) + 1  # Correct: Increment the count
  
    for count in freq.values():
        if count % 2 != 0:
            return False
    return True

3. Misunderstanding the Pairing Requirement

Some might think they need to actually create the pairs or track which elements are paired together, leading to unnecessary complexity.

Pitfall Example:

def divideArray(self, nums: List[int]) -> bool:
    nums.sort()
    pairs = []
    i = 0
    while i < len(nums) - 1:
        if nums[i] == nums[i + 1]:
            pairs.append((nums[i], nums[i + 1]))
            i += 2
        else:
            return False
    return len(pairs) == len(nums) // 2

Solution: The frequency-based approach is simpler and more efficient. We only need to verify that pairing is possible, not actually construct the pairs:

def divideArray(self, nums: List[int]) -> bool:
    frequency_count = Counter(nums)
    return all(count % 2 == 0 for count in frequency_count.values())

4. Using Sets Instead of Frequency Counting

Attempting to use sets to check for duplicates without considering the actual count of each element.

Pitfall Example:

def divideArray(self, nums: List[int]) -> bool:
    unique_nums = set(nums)
    return len(nums) == 2 * len(unique_nums)  # Wrong logic

This incorrectly assumes each unique number appears exactly twice, which isn't always the case.

Solution: Always use frequency counting to track the exact number of occurrences for each element, as the same number might need to form multiple pairs.