# Solving Leetcode Interviews in Seconds with AI: Maximum Number of Pairs in Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2341" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given a 0-indexed integer array nums. In one operation, you may do the following:  Choose two integers in nums that are equal. Remove both integers from nums, forming a pair.  The operation is done on nums as many times as possible. Return a 0-indexed integer array answer of size 2 where answer[0] is the number of pairs that are formed and answer[1] is the number of leftover integers in nums after doing the operation as many times as possible.   Example 1:  Input: nums = [1,3,2,1,3,2,2] Output: [3,1] Explanation: Form a pair with nums[0] and nums[3] and remove them from nums. Now, nums = [3,2,3,2,2]. Form a pair with nums[0] and nums[2] and remove them from nums. Now, nums = [2,2,2]. Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [2]. No more pairs can be formed. A total of 3 pairs have been formed, and there is 1 number leftover in nums.  Example 2:  Input: nums = [1,1] Output: [1,0] Explanation: Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = []. No more pairs can be formed. A total of 1 pair has been formed, and there are 0 numbers leftover in nums.  Example 3:  Input: nums = [0] Output: [0,1] Explanation: No pairs can be formed, and there is 1 number leftover in nums.    Constraints:  1 <= nums.length <= 100 0 <= nums[i] <= 100  

	# Explanation
	Here's a solution approach and the Python code:

*   **Count Frequencies:** Use a dictionary (or a simple array, given the constraint on the number range) to count the frequency of each number in `nums`.
*   **Calculate Pairs and Leftovers:** Iterate through the frequencies. For each number, the number of pairs is the frequency divided by 2 (integer division). The number of leftovers is the frequency modulo 2.
*   **Return the Result:** Return an array containing the total number of pairs and the total number of leftovers.

*   **Runtime Complexity:** O(n), where n is the length of `nums`.
*   **Storage Complexity:** O(1), because the maximum size of the frequency counter is bounded by the constraint on the range of numbers (0-100). We can use an array of size 101 as a frequency counter

	
	# Code
	```python
	def numberOfPairs(nums):
    """
    Finds the number of pairs and leftover integers in an array.

    Args:
        nums: A list of integers.

    Returns:
        A list containing the number of pairs and leftovers.
    """

    counts = {}
    for num in nums:
        counts[num] = counts.get(num, 0) + 1

    pairs = 0
    leftovers = 0
    for count in counts.values():
        pairs += count // 2
        leftovers += count % 2

    return [pairs, leftovers]
	```