Introduction

In this blog post, we will explore how to solve the LeetCode problem "1224" 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 nums of positive integers, return the longest possible length of an array prefix of nums, such that it is possible to remove exactly one element from this prefix so that every number that has appeared in it will have the same number of occurrences. If after removing one element there are no remaining elements, it's still considered that every appeared number has the same number of ocurrences (0). Example 1: Input: nums = [2,2,1,1,5,3,3,5] Output: 7 Explanation: For the subarray [2,2,1,1,5,3,3] of length 7, if we remove nums[4] = 5, we will get [2,2,1,1,3,3], so that each number will appear exactly twice. Example 2: Input: nums = [1,1,1,2,2,2,3,3,3,4,4,4,5] Output: 13 Constraints: 2 <= nums.length <= 105 1 <= nums[i] <= 105

Explanation

Here's the breakdown of the solution:

• Frequency Counting and Validation: Keep track of the frequency of each number encountered in the prefix. Then, check if removing one element can make all remaining numbers have the same frequency.
• Early Exit Conditions: Handle base cases and conditions that immediately satisfy the requirement (e.g., prefix of length 1 or all elements being the same).

• Efficient Validation Logic: Optimize the validation process by tracking the counts of frequencies and checking for specific scenarios where a single element removal can equalize frequencies.

• Runtime Complexity: O(n), where n is the length of the input array. Storage Complexity: O(n) due to the frequency maps.

Code

    def longest_prefix_equal_frequency(nums):
    """
    Finds the longest prefix of an array such that removing one element results
    in all remaining elements having the same frequency.
    """
    freq = {}  # Element -> Frequency
    freq_count = {}  # Frequency -> Count of elements with that frequency
    max_len = 0

    for i in range(len(nums)):
        num = nums[i]
        if num not in freq:
            freq[num] = 0

        old_freq = freq[num]
        freq[num] += 1
        new_freq = freq[num]

        if old_freq > 0:
            freq_count[old_freq] -= 1
            if freq_count[old_freq] == 0:
                del freq_count[old_freq]

        if new_freq not in freq_count:
            freq_count[new_freq] = 0
        freq_count[new_freq] += 1

        # Check if the current prefix satisfies the condition
        if len(freq) == 1: # all the same elements
            max_len = i + 1
            continue

        if 1 in freq_count and freq_count[1] == 1 and len(freq_count) == 2 and new_freq - 1 not in freq_count : # removing the element with frequency 1 gives all the rest the same frequency
             max_len = i + 1
             continue

        if new_freq == 1 and len(freq) == i+1: #all distinct
             max_len = i+1
             continue


        if len(freq_count) == 2:
            freqs = sorted(freq_count.keys())
            if freqs[1] == freqs[0] + 1 and freq_count[freqs[1]] == 1: #remove one element to match the other frequency
                max_len = i + 1
                continue

    return max_len