algoadvance

Given a list of integers nums, return the total number of pairs in the list and the leftover elements after all the pairs have been removed. A pair is considered as two identical elements from the list.

Example:

Input: nums = [1,3,2,1,3,2,2]
Output: [3,1]

Input: nums = [1,1]
Output: [1,0]

Input: nums = [0]
Output: [0,1]

Clarifying Questions

1. Q: Can the elements in the list be negative? A: Yes, integers in the list can be negative, zero, or positive.

2. Q: What is the range of the length of the input list? A: The length of the input list can range from 1 to 10^5.

Strategy

To solve this problem, we need to count how many times each element appears in the list. We can use a dictionary to keep track of these counts.

1. Count Frequencies: Traverse the list and count the occurrences of each element using a dictionary.
2. Calculate Pairs and Leftovers: For each unique element, determine how many pairs can be formed (using integer division by 2) and how many elements are leftover (using modulus 2).
3. Sum Up Pairs and Leftovers: Aggregate the counts of pairs and leftovers to get the final result.

Code

def numberOfPairs(nums):
    from collections import Counter
    
    # Count the frequency of each number
    freq = Counter(nums)
    
    # Initialize pairs and leftovers
    pairs = 0
    leftovers = 0
    
    # Calculate pairs and leftovers from the frequency map
    for count in freq.values():
        pairs += count // 2
        leftovers += count % 2
    
    return [pairs, leftovers]

# Test cases
print(numberOfPairs([1, 3, 2, 1, 3, 2, 2]))  # Output: [3, 1]
print(numberOfPairs([1, 1]))                # Output: [1, 0]
print(numberOfPairs([0]))                   # Output: [0, 1]

Time Complexity

• Counting Frequencies: O(n), where n is the number of elements in the list.
• Calculating Pairs and Leftovers: O(k), where k is the number of unique elements in the list. In the worst case, k = n.

Hence, the overall time complexity is O(n), making this solution efficient even for large inputs.

Space Complexity

• Using a dictionary to store the frequency count involves O(k) space.
• Therefore, the space complexity is O(n) in the worst case when all elements are unique.

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