Problem Description

You are given an integer array nums and an integer k.

An integer x is called almost missing if it appears in exactly one subarray of size k within nums. A subarray is a contiguous sequence of elements from the array.

Your task is to find the largest almost missing integer from nums. If no such integer exists, return -1.

Let's break down what this means:

• You need to look at all possible subarrays of size k in the array
• For each unique integer in these subarrays, count how many subarrays it appears in
• An integer is "almost missing" if it appears in exactly one of these subarrays
• Among all almost missing integers, return the largest one

For example, if nums = [1, 2, 3, 4] and k = 2:

• The subarrays of size 2 are: [1, 2], [2, 3], [3, 4]
• The integer 1 appears only in [1, 2] (exactly one subarray) → almost missing
• The integer 2 appears in [1, 2] and [2, 3] (two subarrays) → not almost missing
• The integer 3 appears in [2, 3] and [3, 4] (two subarrays) → not almost missing
• The integer 4 appears only in [3, 4] (exactly one subarray) → almost missing
• The almost missing integers are 1 and 4, so the answer would be 4 (the largest)

The solution handles three cases:

• When k = 1: Each element forms its own subarray, so we need elements that appear exactly once in the entire array
• When k = len(nums): There's only one subarray (the entire array), so all elements in it are almost missing, and we return the maximum
• When 1 < k < len(nums): Only the first and last elements can be almost missing (if they don't appear elsewhere in the array)

Intuition

The key insight is to understand which elements can possibly be "almost missing" based on the value of k.

Let's think about when an element can appear in exactly one subarray of size k:

1. When k = 1: Each element forms its own individual subarray. An element appears in exactly one subarray if and only if it appears exactly once in the entire array. This is straightforward - we just need to count occurrences.

2. When k = n (where n is the array length): There's only one possible subarray - the entire array itself. Every element in the array appears in this single subarray, making all elements "almost missing". We simply return the maximum element.

3. When 1 < k < n: This is the interesting case. Let's think about which elements can appear in exactly one subarray:

   • For an element in the middle of the array, it will appear in multiple overlapping subarrays of size k. For instance, nums[i] where k-1 ≤ i ≤ n-k will appear in at least k different subarrays.
   • The only elements that have a chance of appearing in exactly one subarray are those at the extreme positions: nums[0] and nums[n-1].
   • nums[0] only appears in the first subarray [0, k-1]
   • nums[n-1] only appears in the last subarray [n-k, n-1]
   • However, if nums[0] appears anywhere else in the array (positions 1 to n-1), it will appear in more than one subarray
   • Similarly, if nums[n-1] appears anywhere else in the array (positions 0 to n-2), it will appear in more than one subarray

This observation drastically simplifies our problem. Instead of checking all possible elements and all possible subarrays (which would be computationally expensive), we only need to:

• Check if nums[0] appears elsewhere in the array
• Check if nums[n-1] appears elsewhere in the array
• Return the maximum among the valid "almost missing" elements

The solution elegantly handles all three cases with minimal computation, avoiding the need to generate and check all subarrays explicitly.

Solution Approach

Based on the case analysis, we implement the solution as follows:

Case 1: When k = 1

• Each element forms its own subarray of size 1
• We need to find elements that appear exactly once in the entire array
• Use a Counter to count the frequency of each element
• Filter elements with frequency equal to 1
• Return the maximum among these elements, or -1 if none exist

if k == 1:
    cnt = Counter(nums)
    return max((x for x, v in cnt.items() if v == 1), default=-1)

Case 2: When k = len(nums)

• The entire array is the only subarray of size k
• All elements in the array appear in exactly one subarray
• Simply return the maximum element in the array

if k == len(nums):
    return max(nums)

Case 3: When 1 < k < len(nums)

• Only nums[0] and nums[n-1] can potentially be almost missing
• Define a helper function f(k) to check if nums[k] is almost missing:
  • Iterate through the array
  • If nums[k] appears at any other position (i.e., nums[i] == nums[k] where i != k), it's not almost missing, return -1
  • Otherwise, return nums[k]
• Check both nums[0] (by calling f(0)) and nums[n-1] (by calling f(len(nums) - 1))
• Return the maximum of the two results

def f(k: int) -> int:
    for i, x in enumerate(nums):
        if i != k and x == nums[k]:
            return -1
    return nums[k]

return max(f(0), f(len(nums) - 1))

The solution efficiently handles all cases with:

• Time Complexity: O(n) where n is the length of the array
  • For k = 1: O(n) to count frequencies
  • For k = n: O(n) to find the maximum
  • For 1 < k < n: O(n) to check if first/last elements appear elsewhere
• Space Complexity: O(n) for the Counter in case k = 1, O(1) for other cases

This approach avoids the brute force method of generating all subarrays and checking each element, which would have been O(n * k) time complexity.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example: nums = [2, 3, 5, 3, 1], k = 3

Since 1 < k < len(nums) (specifically, 1 < 3 < 5), we're in Case 3.

Step 1: Identify which elements can be almost missing

First, let's see all subarrays of size 3:

• Subarray 1: [2, 3, 5] (indices 0-2)
• Subarray 2: [3, 5, 3] (indices 1-3)
• Subarray 3: [5, 3, 1] (indices 2-4)

According to our solution insight, only nums[0] = 2 and nums[4] = 1 can potentially be almost missing because:

• Elements in middle positions will appear in multiple overlapping subarrays
• For example, nums[2] = 5 appears in all three subarrays
• nums[1] = 3 and nums[3] = 3 each appear in at least two subarrays

Step 2: Check if nums[0] = 2 is almost missing

Call f(0):

• Iterate through the array checking if 2 appears at any position other than index 0
• Index 0: nums[0] = 2 (this is k itself, skip)
• Index 1: nums[1] = 3 (not equal to 2)
• Index 2: nums[2] = 5 (not equal to 2)
• Index 3: nums[3] = 3 (not equal to 2)
• Index 4: nums[4] = 1 (not equal to 2)
• Since 2 doesn't appear anywhere else, return 2

Step 3: Check if nums[4] = 1 is almost missing

Call f(4):

• Iterate through the array checking if 1 appears at any position other than index 4
• Index 0: nums[0] = 2 (not equal to 1)
• Index 1: nums[1] = 3 (not equal to 1)
• Index 2: nums[2] = 5 (not equal to 1)
• Index 3: nums[3] = 3 (not equal to 1)
• Index 4: nums[4] = 1 (this is k itself, skip)
• Since 1 doesn't appear anywhere else, return 1

Step 4: Return the maximum

Both 2 and 1 are almost missing. Return max(2, 1) = 2.

Let's verify this is correct:

• 2 appears only in subarray [2, 3, 5] ✓
• 1 appears only in subarray [5, 3, 1] ✓
• The largest almost missing integer is indeed 2

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums.

Time Complexity Analysis:

• When k == 1: The Counter(nums) operation takes O(n) time to count all elements. The subsequent iteration through the counter items and finding the maximum takes at most O(n) time.
• When k == len(nums): The max(nums) operation takes O(n) time.
• Otherwise: The function f() is called at most twice (for indices 0 and len(nums) - 1). Each call to f() iterates through the array once, taking O(n) time. Therefore, the total time is O(2n) = O(n).

The space complexity is O(n).

Space Complexity Analysis:

• When k == 1: The Counter(nums) creates a dictionary that can contain up to n unique elements in the worst case, requiring O(n) space.
• When k == len(nums): Only O(1) extra space is used for the max operation.
• Otherwise: The function f() uses O(1) extra space for iteration variables, but no additional data structures are created.
• Overall, the worst-case space complexity is O(n) due to the Counter when k == 1.

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

Common Pitfalls

1. Misunderstanding the Problem Definition

Pitfall: Many developers initially misinterpret what "almost missing" means. They might think it refers to integers that are missing from the array but close in value, or that it means appearing in "almost all" subarrays.

Solution: Carefully read that "almost missing" means appearing in exactly one subarray of size k. Draw out examples with small arrays to verify understanding before coding.

2. Incorrect Handling of Edge Cases

Pitfall: Failing to properly handle the special cases when k = 1 or k = len(nums). Some might try to use the general case logic for all scenarios, which leads to incorrect results.

Solution: Explicitly check and handle these special cases first:

• When k = 1: Look for elements appearing exactly once in the entire array
• When k = len(nums): All elements are almost missing (they all appear in exactly one subarray)

3. Off-by-One Errors in Boundary Checking

Pitfall: When checking if boundary elements (nums[0] and nums[n-1]) appear elsewhere, using incorrect index ranges or conditions like i > 0 instead of i != k.

Solution: Use clear condition i != k when checking if an element at position k appears elsewhere:

for i, x in enumerate(nums):
    if i != k and x == nums[k]:  # Correct: excludes the position we're checking
        return -1

4. Forgetting to Handle Empty Results

Pitfall: When no almost missing integers exist (especially in the k = 1 case), not returning -1 properly. Using max() on an empty sequence causes a ValueError.

Solution: Always use default=-1 parameter with max() when the sequence might be empty:

return max((x for x, v in cnt.items() if v == 1), default=-1)

5. Inefficient Brute Force Approach

Pitfall: Attempting to generate all subarrays of size k and count occurrences for each element, resulting in O(n*k) time complexity.

Solution: Recognize the pattern that only boundary elements can be almost missing when 1 < k < n, reducing the problem to O(n) time complexity by checking just two elements.

6. Import Statement Placement

Pitfall: Importing Counter inside the function only when k = 1, which can cause NameError if the import statement is moved or forgotten.

Solution: Import at the beginning of the solution or at the module level:

from collections import Counter

class Solution:
    def largestInteger(self, nums: List[int], k: int) -> int:
        # Rest of the code