Problem Description

In this problem, we are given an array nums consisting of positive integers. The goal is to find the length of the longest prefix (an initial segment of the array) with the property that if you remove exactly one element from it, all the remaining numbers in the prefix will have the same frequency of occurrence.

This means that in the longest prefix, for all the numbers that have appeared thus far, either all numbers appear the same number of times, or removing one instance of a number can achieve this state of equal frequency.

An important part of the description specifies that if we end up removing one element from the prefix, and there are no elements left, this also counts as having all numbers appearing with the same frequency (since they all appear zero times).

Intuition

To solve this problem, we need to keep track of two things at each step:

1. The frequency of each number in the prefix (cnt).
2. The count of frequencies that we've seen (ccnt).

Using these two counters, we can determine at each step if the prefix can have equal frequency after removing a single element, and hence calculate the longest such prefix.

The thinking process is to iterate through the array, and at each step, update our two counters: one for the frequency of the current element (cnt[v]) and another for the frequency of this frequency (ccnt[cnt[v]]). While doing so, we maintain a variable mx that stores the maximum frequency of any number in the prefix so far.

We consider three cases where the prefix could potentially have all numbers appearing the same number of times after removing one element:

1. If the maximum frequency (mx) is 1, which means all elements are unique, any prefix is valid by removing any one of them.
2. If the highest frequency number (mx) appears exactly once (ccnt[mx] == 1) and the rest of the numbers have one less frequency (mx - 1), then by removing the single occurrence of the highest frequency number, all numbers would have the same frequency.
3. Likewise, if all the numbers have the same frequency (ccnt[mx] * mx), except for one number that is alone (ccnt[1] == 1), we can remove this single occurrence to match all the others.

We repeat these checks at each step and update the answer everytime we find a longer valid prefix. After iterating through the array, the value in ans will be the length of the longest possible prefix satisfying the problem condition.

Solution Approach

The solution uses a couple of key data structures along with some logical checks that help execute the strategy described in the intuition effectively.

Data Structures

1. Counter cnt: This is a dictionary that keeps track of the occurrences of each element in the prefix of the array nums. If v is an element in nums, then cnt[v] will tell us how many times v has appeared so far.

2. Counter ccnt: This is a dictionary that keeps track of the count of the frequencies themselves. It tells us how many numbers have a certain frequency. If f is a frequency, then ccnt[f] will tell us how many numbers have appeared exactly f times.

Algorithm

• Initialize the cnt and ccnt dictionaries (imported from collections class in Python). They are both initially empty.

• Start iterating through each element v in nums, keeping track of the index i (starting from 1 instead of 0).

• At each iteration, increase the count for v in cnt. If v has appeared before, decrease the counter for its old frequency in ccnt - since the frequency of v is about to increase by one.

• Update ccnt for the new frequency of v.

• Keep track of the maximum frequency seen so far in variable mx. This is updated to the maximum of its current value and the frequency of the current number cnt[v].

• Perform the checks to see if we could make all frequencies equal by removing one element:

  • Check if mx is 1, meaning all numbers are unique, and any prefix is valid.
  • Check if ccnt[mx] is 1 and the total count of elements with frequency mx or mx - 1 equals the current length of the array (i). This means we could remove the one element with frequency mx to even out the counts.
  • Check if the total count of elements with frequency mx plus 1 equals the current length of the array and there is only one number with the frequency of 1 (ccnt[1] == 1). We could remove the single occurrence of an element to achieve equal frequency.

• If any of the above conditions are true, then the current prefix (up to the index i) can be made to have equal frequencies by removing one element. We thus update ans to store the maximum prefix length satisfying the condition.

• After the loop terminates, ans will contain the length of the longest prefix which can achieve equal frequency by removing one element, and we return this value.

Using a combination of logical checks along with efficient tracking of frequencies of the numbers and their counts using hashmaps (cnt and ccnt), the solution is able to identify the longest prefix satisfying the problem's conditions without having to check each possible prefix repeatedly.

Example Walkthrough

Let's take an array nums with the values [1, 2, 2, 3, 3, 3, 4, 4, 4, 4] and walk through the solution step by step to understand how we apply the solution approach described above.

• We initialize our cnt and ccnt dictionaries empty and a variable ans for the answer.

• Start with i = 1, processing the first element v = 1.

  • Update cnt[1] from 0 to 1. Since 1 has not appeared before, there's no old frequency count to decrease in ccnt.
  • Update ccnt[1] from 0 to 1, since cnt[1] is now 1.
  • Set mx to 1, as it's the first and thus the highest frequency so far.
  • Since mx is 1, we update ans to 1, which is the length of the prefix now.

• Moving to i = 2 with the element v = 2.

  • cnt[2] gets updated from 0 to 1.
  • ccnt[1] gets updated to 2 now, as there are two numbers with a frequency of 1.
  • mx is still 1, and thus the prefix of length 2 also satisfies the condition, so ans is updated to 2.

• Continue with i = 3 with the element v = 2.

  • cnt[2] increases from 1 to 2.
  • Decrease ccnt[1] from 2 to 1 since the frequency of 2 has changed.
  • Update ccnt[2] from 0 to 1.
  • mx updates to 2.
  • Now, ccnt[mx] * mx + ccnt[1] is 3, which matches the length of the prefix, so ans is updated to 3.

In the interest of brevity, let’s jump to i = 10 with the element v = 4.

• We should have cnt showing {1: 1, 2: 2, 3: 3, 4: 4}, and ccnt showing {1: 1, 2: 1, 3: 1, 4: 1}.

• The mx now is 4 after processing all elements up to index 10.

  • Check if mx is 1 (false in this case).
  • Check if ccnt[mx] is 1 and the total count of elements with frequency mx or mx - 1 equals the current index (10). This is false because ccnt[4] is 1, but ccnt[3] * 3 + ccnt[4] * 4 is 9, not 10.
  • Check if ccnt[mx] * mx + ccnt[1] equals the current index (10) and ccnt[1] is 1. This is false because ccnt[4] * 4 + ccnt[1] is 17, not 10.

• Since no condition is met, we do not update ans.

After processing all elements, ans remains 3, which is the length of the longest prefix where one can remove exactly one element to make the frequency of the remaining elements the same. Therefore, the answer for our example array [1, 2, 2, 3, 3, 3, 4, 4, 4, 4] is 3.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code is O(N) where N is the length of the nums list. This is because the code iterates through each element of the nums list exactly once, performing a constant amount of work for each element concerning updating the cnt and ccnt Counters, and computing the maximum frequency mx. Every operation inside the loop—such as checking if v is in cnt, updating counters, and the conditional checks—is done in constant time, assuming that the counter operations are O(1) on average due to the hashing mechanism used.

Space Complexity

The space complexity of the provided code is O(N) as well. The cnt and ccnt Counters will each store elements proportional to the number of distinct elements in nums. In the worst case, if all elements of nums are unique, the space taken by these counters will be linear in terms of the size of the input. Additionally, nums, ans, mx, and other variables use a constant amount of extra space, but this does not change the overall space complexity.

Learn more about how to find time and space complexity quickly using problem constraints.