Problem Description

You have an integer array nums (0-indexed) and an integer key that appears in the array.

Your task is to find which number appears most frequently immediately after key in the array.

Specifically, for each unique integer target in nums, count how many times target appears right after key. This means counting the number of valid indices i where:

• i is between 0 and nums.length - 2 (inclusive)
• nums[i] == key
• nums[i + 1] == target

Return the target value that has the maximum count. The problem guarantees that there will be exactly one target with the maximum count (no ties).

For example, if nums = [1, 100, 200, 1, 100] and key = 1, you would check what comes after each occurrence of 1:

• At index 0: 1 is followed by 100
• At index 3: 1 is followed by 100

So 100 appears 2 times after key = 1, making it the answer.

Intuition

The problem asks us to find the most frequent number that appears immediately after a specific key value. This is essentially a counting problem where we need to track consecutive pairs.

The key insight is that we need to examine every consecutive pair (nums[i], nums[i+1]) in the array. Whenever the first element of a pair equals key, we should count the second element as a potential answer.

To solve this efficiently, we can use a hash table (or Counter in Python) to keep track of how many times each number appears after key. As we traverse the array:

• Check each consecutive pair
• When we find nums[i] == key, increment the count for nums[i+1]
• Keep track of the maximum count seen so far and which number has that count

This approach works because:

1. We only need one pass through the array
2. The hash table allows us to count occurrences of each target in O(1) time
3. By maintaining the maximum count during traversal, we avoid needing a second pass to find the answer

The pairwise function in Python makes this even cleaner by automatically giving us consecutive pairs (a, b) from the array, where we can simply check if a == key and then count b.

Solution Approach

The solution uses a hash table to count occurrences and a single traversal to find the answer.

Data Structures Used:

• Counter (hash table): To store the count of each target that appears after key
• Two variables: mx to track the maximum count, and ans to store the target with maximum count

Implementation Steps:

1. Initialize the data structures:

   • Create a Counter object cnt to store frequency counts
   • Set ans = 0 (to store the final answer) and mx = 0 (to track maximum frequency)

2. Traverse consecutive pairs:

   • Use pairwise(nums) to iterate through all consecutive pairs (a, b) in the array
   • This gives us pairs like (nums[0], nums[1]), (nums[1], nums[2]), etc.

3. Count targets after key:

   • For each pair (a, b):
     • If a == key, then b is a target that follows key
     • Increment the count: cnt[b] += 1

4. Update maximum on the fly:

   • After incrementing cnt[b], check if this new count exceeds mx
   • If mx < cnt[b]:
     • Update mx = cnt[b] (new maximum count)
     • Update ans = b (the target with this maximum count)

5. Return the result:

   • After traversing all pairs, ans contains the target with the maximum count

Time Complexity: O(n) where n is the length of nums, as we traverse the array once.

Space Complexity: O(k) where k is the number of unique targets that appear after key, for storing the counts in the hash table.

Example Walkthrough

Let's walk through the solution with nums = [2, 2, 2, 2, 3] and key = 2.

Step 1: Initialize data structures

• cnt = {} (empty counter)
• ans = 0 (will store our answer)
• mx = 0 (tracks maximum frequency)

Step 2: Process consecutive pairs

We'll examine each consecutive pair (a, b):

Pair 1: (2, 2) at indices (0, 1)

• a = 2 equals key = 2 ✓
• So b = 2 is a target that follows key
• Update: cnt[2] = 1
• Check maximum: cnt[2] = 1 > mx = 0 ✓
• Update: mx = 1, ans = 2

Pair 2: (2, 2) at indices (1, 2)

• a = 2 equals key = 2 ✓
• So b = 2 is a target that follows key
• Update: cnt[2] = 2
• Check maximum: cnt[2] = 2 > mx = 1 ✓
• Update: mx = 2, ans = 2

Pair 3: (2, 2) at indices (2, 3)

• a = 2 equals key = 2 ✓
• So b = 2 is a target that follows key
• Update: cnt[2] = 3
• Check maximum: cnt[2] = 3 > mx = 2 ✓
• Update: mx = 3, ans = 2

Pair 4: (2, 3) at indices (3, 4)

• a = 2 equals key = 2 ✓
• So b = 3 is a target that follows key
• Update: cnt[3] = 1
• Check maximum: cnt[3] = 1 > mx = 3? ✗
• No update needed (3 still has the maximum count)

Step 3: Return the result

• Final counter: cnt = {2: 3, 3: 1}
• The number 2 appears 3 times after key = 2
• The number 3 appears 1 time after key = 2
• Return ans = 2

The answer is 2 because it appears most frequently (3 times) immediately after the key value 2.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. The algorithm iterates through the array once using pairwise(nums), which generates n-1 pairs. For each pair, it performs constant-time operations: checking if a == key, updating the counter, and comparing/updating the maximum frequency.

The space complexity is O(M), where M represents the number of distinct elements that appear immediately after the key in the array. In the worst case, this could be all unique values that follow the key, which is bounded by the maximum value of elements in nums if we consider the counter stores at most M distinct integers. The Counter object stores the frequency of each target element that appears after key, and the additional variables (ans, mx) use O(1) space.

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

Common Pitfalls

1. Off-by-One Error in Loop Range

A common mistake is iterating through the entire array without considering that we need to check pairs. If you iterate up to len(nums) instead of len(nums) - 1, you'll get an index out of bounds error when accessing nums[i + 1].

Incorrect:

for i in range(len(nums)):  # Goes up to len(nums) - 1
    if nums[i] == key:
        target = nums[i + 1]  # IndexError when i = len(nums) - 1

Correct:

for i in range(len(nums) - 1):  # Stops at len(nums) - 2
    if nums[i] == key:
        target = nums[i + 1]  # Safe access

2. Not Handling Empty or Single-Element Arrays

The code assumes there are at least two elements to form a pair. Edge cases with empty arrays or arrays with only one element could cause issues.

Solution: Add a guard clause at the beginning:

if len(nums) < 2:
    return 0  # or handle according to problem constraints

3. Incorrect Maximum Tracking

Some might try to find all counts first, then search for the maximum in a separate pass. This works but is less efficient and more error-prone.

Less Efficient Approach:

# First pass: count all
for i in range(len(nums) - 1):
    if nums[i] == key:
        count_map[nums[i + 1]] = count_map.get(nums[i + 1], 0) + 1

# Second pass: find maximum
result = 0
max_freq = 0
for target, count in count_map.items():
    if count > max_freq:
        max_freq = count
        result = target

Better Approach (as shown in solution): Update the maximum during the counting process itself, eliminating the need for a second pass.

4. Using Wrong Comparison Operator

When updating the maximum, using <= instead of < could potentially return a different target if multiple targets have the same maximum count (though the problem guarantees uniqueness).

Potentially Problematic:

if max_frequency <= count_map[next_element]:  # Uses <=
    max_frequency = count_map[next_element]
    result = next_element

Correct:

if max_frequency < count_map[next_element]:  # Uses <
    max_frequency = count_map[next_element]
    result = next_element

5. Forgetting to Initialize Dictionary Values

When using a regular dictionary instead of defaultdict or Counter, forgetting to initialize new keys leads to KeyError.

Incorrect:

count_map[next_element] += 1  # KeyError if next_element not in count_map

Correct:

if next_element not in count_map:
    count_map[next_element] = 0
count_map[next_element] += 1

Or use defaultdict:

from collections import defaultdict
count_map = defaultdict(int)
count_map[next_element] += 1  # Automatically initializes to 0