Introduction

In this blog post, we will explore how to solve the LeetCode problem "1481" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, 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

Given an array of integers arr and an integer k. Find the least number of unique integers after removing exactly k elements. Example 1: Input: arr = [5,5,4], k = 1 Output: 1 Explanation: Remove the single 4, only 5 is left. Example 2: Input: arr = [4,3,1,1,3,3,2], k = 3 Output: 2 Explanation: Remove 4, 2 and either one of the two 1s or three 3s. 1 and 3 will be left. Constraints: 1 <= arr.length <= 10^5 1 <= arr[i] <= 10^9 0 <= k <= arr.length

Explanation

Here's the breakdown of the solution:

• Frequency Counting: Count the frequency of each number in the input array.
• Greedy Removal: Sort the frequencies in ascending order. Greedily remove elements with the lowest frequencies until k elements have been removed.

• Count Remaining Unique: Count the number of unique integers remaining after the removals.

• Runtime Complexity: O(n log n), where n is the length of the input array (due to sorting). Storage Complexity: O(n) (to store the frequencies).

Code

    from collections import Counter

def find_least_num_of_unique_ints(arr, k):
    """
    Finds the least number of unique integers after removing exactly k elements.

    Args:
        arr: A list of integers.
        k: The number of elements to remove.

    Returns:
        The least number of unique integers after removing k elements.
    """

    counts = Counter(arr)
    frequencies = sorted(counts.values())

    removed_count = 0
    unique_count = len(frequencies)

    for freq in frequencies:
        if removed_count + freq <= k:
            removed_count += freq
            unique_count -= 1
        else:
            break

    return unique_count