Problem Description

The problem presents us with an array of integers named nums and an integer key, which is guaranteed to be found within nums. Our objective is to identify the integer in nums that follows key the most frequently. To clarify, we are looking for the value target that appears exactly one position after key (i.e., nums[i + 1] == target when nums[i] == key) the greatest number of times throughout the array.

We should iterate through the array, checking pairs of consecutive numbers (nums[i] and nums[i + 1]). We need to keep track of how many times each target comes after key, and then return the target number that has the highest frequency of appearing immediately after key. If multiple target numbers have the same frequency, we only return the one with the maximum count, which per the problem description, is unique.

Intuition

To solve the problem, one efficient approach is to traverse through nums and use a data structure to keep a tally of the frequencies of each target that follows key. A common data structure that can be used for this purpose is a Counter (available in Python's collections module), which will hold target integers as keys and their counts as values.

We start by initializing an empty Counter object. As we loop through the array, we examine each pair of adjacent elements (nums[i] and nums[i + 1]). When we find an occurrence of key, we increment the count for the following number (nums[i + 1]). While doing this, we keep track of both the number with the maximum count encountered so far, and the current maximum count. If at any point we find a target with a higher count than the previous maximum, we update both the ans (answer) with the new target and mx (maximum count) with the new count.

Finally, once we've passed through the array, the variable ans will hold the target with the highest frequency of occurrence immediately following key, and that is what we return.

Solution Approach

The implementation of the solution uses a Counter from Python's collections module to track the frequency of each integer appearing immediately after the key. Also, it leverages the pairwise iterator from Python's itertools module to efficiently iterate over the array in adjacent pairs. Here is a step-by-step explanation of the code:

1. Initialize a Counter object named cnt to hold the frequencies, and two variables ans and mx to keep track of the answer (the most frequented target number) and the maximum frequency encountered so far, respectively.

2. Use pairwise(nums) to create an iterator that returns consecutive pairs of elements in nums. This will look like (nums[0], nums[1]), (nums[1], nums[2]), ..., (nums[n-2], nums[n-1]).

3. Iterate over these pairs of numbers a (current key) and b (potential target):

   • If a (the current number) is equal to key, it means b is immediately following key. Then increment the counter for b by 1: cnt[b] += 1.

   • After incrementing, check if the count for b exceeds the current maximum (mx). If it does, update mx to the new count for b, and set ans to b because b is now the new target with the maximum count following key.

4. Once the loop concludes, ans will hold the value of the most frequent target after key, and we return ans as the solution.

The use of a Counter object is crucial in this approach for efficient frequency tracking, which allows for the update and query operations to happen in constant time (O(1)). The pairwise utility simplifies the process of inspecting adjacent elements without having to manage index values manually. Together, these strategies offer a straightforward and efficient solution for the given problem description.

Example Walkthrough

Let's illustrate the solution approach with a small example:

nums = [1, 2, 3, 2, 2, 3, 4, 2, 5]
key = 2

Here, our task is to find the number that most frequently appears immediately after the key, which in this case is the integer 2.

1. We initialize the Counter object cnt as empty, ans as None, and mx (maximum frequency) as 0.

2. We create pairwise pairs from nums, which would look like this:

   (1, 2), (2, 3), (3, 2), (2, 2), (2, 3), (3, 4), (4, 2), (2, 5)

   Notice that the number 2, our key, appears in the first slot of several pairs.

3. We iterate over each pair (a, b):

   • The first interesting pair is (2, 3). Since a is 2 (our key), we increment the counter for b, which is 3: cnt[3] becomes 1. mx was 0, so mx is updated to 1, and ans is set to 3.
   • The next occurrence of 2 is in (2, 2). We increment the counter for 2: cnt[2] becomes 1. Since mx is 1 and cnt[2] equals mx, nothing changes with mx and ans.
   • We encounter (2, 3) again. Incrementing the counter for 3: cnt[3] becomes 2. mx is then updated to 2, and ans becomes 3.
   • Finally, we see (2, 5). We increment the counter for 5: cnt[5] becomes 1. Since mx is still higher (2), we do nothing.

4. After completion, the Counter object cnt contains:

   cnt = {3: 2, 2: 1, 5: 1}

   ans holds 3, and mx holds 2. Thus, the most frequent number following 2 is 3.

Therefore, the solution would return 3 as the target number that most frequently appears immediately after the key in the array nums.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code is O(n) where n is the length of the input list nums. This is because we are iterating through the list exactly once with pairwise(nums), and the operations within the loop are performed in constant time.

The space complexity is O(u) where u is the number of unique elements that come immediately after key in the list nums. The worst case for space complexity occurs when every element following key is unique, in which case the Counter will store each unique element. However, on average, the space used by the Counter would probably be less than n because not all elements in nums would be unique or follow the key.

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