algoadvance

You are given an integer array nums consisting of 2 * n integers. You need to divide nums into n pairs such that:

• Each pair consists of exactly two elements.
• Both elements in a pair are equal.

Return true if you can divide the array into n pairs, otherwise, return false.

Clarifying Questions

1. Input Constraints:
   • Will the length of the array always be even? Given the problem states it is 2 * n, we can assume it is.
   • What range of numbers can appear in the array?
2. Output:
   • We need to return a boolean value, true or false.
3. Edge Cases:
   • What if the array contains negative numbers?
   • What should we return if the array is empty? Though given the constraints, it’s implied it will always contain elements.

Strategy

To determine if it is possible to divide the array into pairs where each pair contains two equal elements, we can use the following approach:

1. Count Frequencies: Count the frequency of each element using a dictionary.
2. Check Pairs: Iterate through the frequency dictionary and check if each frequency is even. If all the frequencies are even, it is possible to form pairs of equal elements.

Code

def divideArray(nums):
    from collections import Counter
    
    # Utilize Counter to get the frequency of each element
    freq_dict = Counter(nums)
    
    # Check if all frequencies are even
    for frequency in freq_dict.values():
        if frequency % 2 != 0:
            return False
        
    return True

Explanation

1. Counting Frequencies:
   • Counter(nums) creates a dictionary where keys are the elements from the array and values are their corresponding counts.
2. Checking Each Frequency:
   • The loop for frequency in freq_dict.values() iterates over the count of each element.
   • The condition if frequency % 2 != 0: checks if the frequency is odd. If any frequency is odd, it returns False since it will not be possible to form pairs with that element.
3. Returning the Result:
   • If the loop completes without finding any odd frequency, the function returns True, indicating it is possible to divide the array into pairs of equal elements.

Time Complexity

• Time Complexity: O(n), where n is the number of elements in the array. This includes the time to count the elements and the time to iterate through the frequency dictionary.
• Space Complexity: O(n) for storing the frequency count in a dictionary.

This solution is efficient and should work well within the problem constraints.

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