Maximum Number of Distinct Elements After Operations

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

You are allowed to perform the following operation on each element of the array at most once:

• Add an integer in the range [-k, k] to the element.

Return the maximum possible number of distinct elements in nums after performing the operations.

Example 1:

Input: nums = [1,2,2,3,3,4], k = 2 Output: 6 Explanation: nums changes to [-1, 0, 1, 2, 3, 4] after performing operations on the first four elements.

Example 2:

Input: nums = [4,4,4,4], k = 1 Output: 3 Explanation: By adding -1 to nums[0] and 1 to nums[1], nums changes to [3, 5, 4, 4].

from collections import Counter

def max_distinct_elements(nums, k):
    """
    Calculates the maximum possible number of distinct elements in nums after performing operations.

    Args:
        nums (List[int]): The input integer array.
        k (int): The maximum allowed change to each element.

    Returns:
        int: The maximum number of distinct elements after operations.
    """
    counts = Counter(nums)
    distinct_count = len(counts)
    duplicates = []
    for num, count in counts.items():
        if count > 1:
            duplicates.append(count - 1)

    duplicates.sort()
    
    for i in range(len(duplicates)):
        if k >= duplicates[i]:
            k -= duplicates[i]
            distinct_count += 1
        else:
            distinct_count += k
            k = 0
            break
            
    distinct_count += min(k, len(nums) - distinct_count)

    return distinct_count


# Example usage:
nums1 = [1, 2, 2, 3, 3, 4]
k1 = 2
print(f"Example 1: nums = {nums1}, k = {k1}, Output = {max_distinct_elements(nums1, k1)}")  # Output: 6

nums2 = [4, 4, 4, 4]
k2 = 1
print(f"Example 2: nums = {nums2}, k = {k2}, Output = {max_distinct_elements(nums2, k2)}")  # Output: 3

nums3 = [1,1,3,3,3]
k3 = 2
print(f"Example 3: nums = {nums3}, k = {k3}, Output = {max_distinct_elements(nums3, k3)}") # Output: 4

nums4 = [1,10,3,3,3,4,4]
k4 = 3
print(f"Example 4: nums = {nums4}, k = {k4}, Output = {max_distinct_elements(nums4, k4)}") # Output: 7

nums5 = [4,1,5,4,1,5]
k5 = 2
print(f"Example 5: nums = {nums5}, k = {k5}, Output = {max_distinct_elements(nums5, k5)}") # Output: 6

nums6 = [2,2,5,7,7,7,9,9,10,10]
k6 = 2
print(f"Example 6: nums = {nums6}, k = {k6}, Output = {max_distinct_elements(nums6, k6)}") # Output: 9

Explanation:

1. Count Frequencies: Use a Counter to count the frequency of each number in the input array nums. This helps identify how many duplicates exist for each number.
2. Identify Duplicates: Iterate through the counts and store the counts of duplicates (count - 1) in a list called duplicates. Sort this list to process elements with fewer duplicates first.
3. Greedy Approach: Iterate through the sorted duplicates list. For each duplicate count, check if k is sufficient to eliminate the duplicates. If yes, decrement k by the number of duplicates and increment the distinct_count. Otherwise, add k to the distinct_count and set k to 0, breaking the loop.
4. Handle Remaining k: After processing duplicates, if there's any remaining k, it means we can potentially make more elements distinct. We add the minimum of k and the number of non-distinct elements to the distinct_count. The number of non-distinct elements is calculated as len(nums) - distinct_count.
5. Return Result: Return the final distinct_count.

Time and Space Complexity:

Time Complexity: O(N log N), where N is the length of the input array nums. This is due to the sorting of the duplicates list. The rest of the operations (counting frequencies, iterating through counts and duplicates) take O(N) time.

Space Complexity: O(N), where N is the length of the input array nums. This is because the Counter can store up to N distinct elements, and the duplicates list can also contain up to N elements in the worst case.

Edge Cases:

1. Empty Input Array: If the input array nums is empty, the function should return 0. This case is implicitly handled as the loop will not execute.
2. k is Zero: If k is 0, the function should return the number of distinct elements in the original array. This is handled correctly as no duplicates can be removed, and no additional distinct elements can be created.
3. All Elements are Same: If all elements in the array are the same, the function should return 1 + min(k, len(nums) - 1). The min part is important when k is large. E.g. nums = [1,1,1,1], k = 5, Output should be 4, not 6.
4. k is Very Large: If k is very large, it should not result in more than len(nums) distinct elements. This condition is handled correctly in the final step where distinct_count += min(k, len(nums) - distinct_count).
5. Negative Numbers: The algorithm should work correctly with negative numbers in the input array because the operations involve adding or subtracting integers in the range [-k, k].
6. Large Numbers: The algorithm should also handle large numbers without overflowing since we are only working with counts and indices.