Problem Description

You have an integer array nums and two positive integers m and k.

Your task is to find the maximum sum among all subarrays of length k that are almost unique. If no such subarray exists, return 0.

A subarray is considered almost unique if it contains at least m distinct elements.

A subarray is a contiguous sequence of elements from the array.

For example:

• If nums = [1, 2, 1, 3, 4], m = 3, and k = 4, you need to check all subarrays of length 4
• The subarray [2, 1, 3, 4] has 4 distinct elements (≥ 3), so it's almost unique with sum = 10
• The subarray [1, 2, 1, 3] has 3 distinct elements (≥ 3), so it's almost unique with sum = 7
• Among all almost unique subarrays of length 4, return the maximum sum

The solution uses a sliding window approach with a hash table to track distinct elements. It maintains a window of size k, counts the frequency of each element using Counter, and tracks the current sum. When the number of distinct elements is at least m, it updates the maximum sum. As the window slides, it adds the new element and removes the old one, updating both the counter and sum accordingly.

Intuition

When we need to find subarrays of a fixed length k, a sliding window is a natural choice since we're examining consecutive elements. The key insight is that as we slide the window one position to the right, we only need to remove one element from the left and add one element from the right - we don't need to recalculate everything from scratch.

For tracking distinct elements, we need to know not just which elements are present, but also their frequencies. Why? Because when an element leaves the window, we need to know if it was the last occurrence of that element (making it no longer distinct) or if there are still other occurrences remaining in the window.

This leads us to use a hash table (Counter) to track frequencies. When we add an element, we increment its count. When we remove an element, we decrement its count. If the count reaches zero, we remove it from the hash table entirely since it's no longer in the window.

The number of distinct elements at any point is simply the size of our hash table (number of keys with non-zero counts). We can check this against m to determine if the current window is "almost unique."

For the sum, instead of recalculating the sum of all k elements each time, we can maintain a running sum. When the window slides, we simply add the new element and subtract the old element: s = s + nums[i] - nums[i-k]. This optimization reduces the time complexity from O(n*k) to O(n).

By combining these techniques - sliding window for fixed-size subarrays, hash table for tracking distinct elements, and running sum for efficiency - we can solve the problem in a single pass through the array.

Learn more about Sliding Window patterns.

Solution Approach

The solution implements a sliding window technique combined with a hash table to efficiently find the maximum sum of almost unique subarrays.

Initial Window Setup: First, we initialize the window with the first k elements:

• Create a Counter object cnt to track the frequency of elements in the first k elements: cnt = Counter(nums[:k])
• Calculate the sum of the first k elements: s = sum(nums[:k])
• Check if this initial window is almost unique (has at least m distinct elements). If yes, set ans = s, otherwise ans = 0

Sliding the Window: We then slide the window through the rest of the array from position k to len(nums) - 1:

For each position i:

1. Add the new element to the window:
   • Increment the count of the new element: cnt[nums[i]] += 1
2. Remove the old element from the window:
   • Decrement the count of the element leaving the window: cnt[nums[i - k]] -= 1
   • If the count becomes 0, remove it from the counter entirely: if cnt[nums[i - k]] == 0: cnt.pop(nums[i - k])
3. Update the sum efficiently:
   • Instead of recalculating the entire sum, adjust it: s += nums[i] - nums[i - k]
4. Check if current window is almost unique:
   • If len(cnt) >= m (at least m distinct elements), update the maximum: ans = max(ans, s)

Why This Works:

• The Counter maintains exact frequencies, so len(cnt) gives us the number of distinct elements
• Removing elements with zero count ensures len(cnt) accurately reflects distinct elements in the current window
• The sliding window ensures we check all possible subarrays of length k
• The running sum optimization reduces time complexity from O(n*k) to O(n)

Time Complexity: O(n) where n is the length of the array Space Complexity: O(k) for storing at most k distinct elements in the hash table

Example Walkthrough

Let's walk through the solution with nums = [1, 2, 1, 3, 4], m = 3, and k = 4.

Step 1: Initialize the first window [1, 2, 1, 3]

• Create counter: cnt = {1: 2, 2: 1, 3: 1} (element 1 appears twice)
• Calculate sum: s = 1 + 2 + 1 + 3 = 7
• Check distinct elements: len(cnt) = 3 (elements are 1, 2, 3)
• Since 3 ≥ 3, this window is almost unique, so ans = 7

Step 2: Slide window to position i = 4, new window [2, 1, 3, 4]

Add new element (nums[4] = 4):

• cnt[4] += 1 → cnt = {1: 2, 2: 1, 3: 1, 4: 1}

Remove old element (nums[0] = 1):

• cnt[1] -= 1 → cnt[1] = 1 (still one occurrence of 1 remaining)
• Since count isn't 0, keep it in counter: cnt = {1: 1, 2: 1, 3: 1, 4: 1}

Update sum:

• s = 7 + 4 - 1 = 10

Check if almost unique:

• len(cnt) = 4 distinct elements (1, 2, 3, 4)
• Since 4 ≥ 3, window is almost unique
• Update maximum: ans = max(7, 10) = 10

Result: The maximum sum among all almost unique subarrays of length 4 is 10.

This example demonstrates how the sliding window efficiently moves through the array, maintaining the counter and sum with minimal operations - we only update for the one element entering and one element leaving, rather than recalculating everything from scratch.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the input array nums.

The algorithm performs a single pass through the array starting from index k to n-1. In the initial setup, we process the first k elements to build the counter and calculate the initial sum, which takes O(k) time. Then, the main loop iterates n-k times. Within each iteration, all operations are constant time:

• Adding and removing elements from the counter: O(1)
• Updating the sum: O(1)
• Checking and removing zero-count entries: O(1)
• Checking the counter size and updating the maximum: O(1)

Since k ≤ n, the total time complexity is O(k) + O(n-k) = O(n).

Space Complexity: O(k), where k is the size of the sliding window.

The counter (cnt) stores at most k distinct elements at any given time, since it represents the elements within a sliding window of size k. Even though elements are added and removed as the window slides, the maximum number of unique elements in the counter cannot exceed k. The remaining variables (s, ans, i) use constant space O(1). Therefore, the overall space complexity is O(k).

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

Common Pitfalls

1. Not Handling Edge Cases Where k > len(nums)

The code assumes that the array has at least k elements. If k > len(nums), the slicing operation nums[:k] will only return the entire array, and the loop range(k, len(nums)) won't execute at all, potentially returning an incorrect result.

Solution: Add a check at the beginning to handle this edge case:

def maxSum(self, nums: List[int], m: int, k: int) -> int:
    if k > len(nums):
        return 0
    # rest of the code...

2. Forgetting to Remove Elements with Zero Count from Counter

A critical mistake is forgetting to remove elements with zero count from the counter. If you only decrement without removing zero-count elements, len(window_counter) will incorrectly include elements that are no longer in the window, leading to wrong distinct element counts.

Incorrect approach:

# Only decrementing without removing
window_counter[nums[i - k]] -= 1
# len(window_counter) might still count this element even if its count is 0

Correct approach:

window_counter[nums[i - k]] -= 1
if window_counter[nums[i - k]] == 0:
    window_counter.pop(nums[i - k])

3. Incorrect Order of Operations When Sliding the Window

The order matters when updating the counter and sum. A common mistake is removing the old element before adding the new one, which could lead to issues if the same element is being added and removed.

Problematic order (if same element entering and leaving):

# If nums[i] == nums[i-k], this could cause issues
window_counter[nums[i - k]] -= 1
if window_counter[nums[i - k]] == 0:
    window_counter.pop(nums[i - k])  # Might remove the element completely
window_counter[nums[i]] += 1  # Then have to re-add it

Better approach (as shown in the solution):

# Add first, then remove
window_counter[nums[i]] += 1
window_counter[nums[i - k]] -= 1
if window_counter[nums[i - k]] == 0:
    window_counter.pop(nums[i - k])

4. Recalculating the Entire Sum Instead of Using Incremental Updates

While not incorrect, recalculating the sum of the entire window at each step is inefficient and defeats the purpose of the sliding window optimization.

Inefficient approach:

# O(k) operation for each window position
window_sum = sum(nums[i-k+1:i+1])

Efficient approach:

# O(1) operation
window_sum += nums[i] - nums[i - k]

5. Not Initializing max_sum Correctly When No Valid Window Exists

If the first window doesn't meet the "almost unique" criteria (has fewer than m distinct elements), and no subsequent windows do either, the function should return 0, not leave max_sum uninitialized or with an incorrect value.

Correct initialization ensures proper handling:

max_sum = window_sum if len(window_counter) >= m else 0