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3321. Find X-Sum of All K-Long Subarrays II

HardArrayHash TableSliding WindowHeap (Priority Queue)
Leetcode Link

Problem Description

You are given an array nums of n integers and two integers k and x.

The x-sum of an array is calculated using the following procedure:

  1. Count the occurrences of all elements in the array
  2. Keep only the occurrences of the top x most frequent elements
    • If two elements have the same number of occurrences, the element with the bigger value is considered more frequent
  3. Calculate the sum of the resulting array

Note: If an array has less than x distinct elements, its x-sum is simply the sum of the entire array.

Your task is to return an integer array answer of length n - k + 1 where answer[i] is the x-sum of the subarray nums[i..i + k - 1].

For example, if we have a subarray and need to find its x-sum with x = 2:

  • First, count how many times each element appears
  • Select the top 2 most frequent elements (using value as tiebreaker if frequencies are equal)
  • Sum up all occurrences of these top 2 elements

The problem asks you to compute this x-sum for every sliding window of size k in the array, moving from left to right one position at a time.

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Intuition

The key insight is that we need to efficiently maintain the top x most frequent elements in a sliding window of size k. As the window slides, elements enter and leave, which changes their frequencies and potentially affects which elements belong in the top x.

Think of this problem as maintaining two groups of elements:

  • The "elite" group: the top x most frequent elements whose occurrences contribute to the x-sum
  • The "waiting" group: all other elements that might get promoted to the elite group

When we slide the window:

  1. A new element enters on the right - its frequency increases
  2. An old element leaves on the left - its frequency decreases

These frequency changes might cause elements to move between the elite and waiting groups. An element with increased frequency might deserve promotion to the elite group, while an element with decreased frequency might need demotion.

To handle the tie-breaking rule (larger values are preferred when frequencies are equal), we can use tuples (frequency, value) as our comparison key. This naturally ensures that when frequencies are equal, the larger value wins.

The challenge is efficiently maintaining these two groups and quickly calculating the x-sum. Using ordered data structures (like sorted lists) allows us to:

  • Always know the weakest member of the elite group (smallest frequency/value pair)
  • Always know the strongest member of the waiting group (largest frequency/value pair)
  • Efficiently transfer elements between groups when needed

By maintaining a running sum s of the elite group, we avoid recalculating the entire x-sum from scratch for each window. We only update s when elements move in or out of the elite group, making each window's calculation very efficient.

The pattern of "remove old occurrence count, update frequency, add new occurrence count" ensures we properly handle the frequency changes without double-counting or missing elements during transitions.

Learn more about Sliding Window and Heap (Priority Queue) patterns.

Solution Approach

The solution uses a hash table combined with two ordered sets to efficiently track and update the x-sum as we slide the window across the array.

Data Structures:

  • cnt: A Counter (hash table) to track the frequency of each element in the current window
  • l: A SortedList storing the top x most frequent elements as tuples (frequency, value)
  • r: A SortedList storing the remaining elements not in the top x
  • s: Running sum of all elements in l (the x-sum)

Helper Functions:

The remove(v) function handles removing an element's old state before updating its frequency:

  • Creates a tuple p = (cnt[v], v) representing the element's current state
  • If p is in the elite set l, removes it and subtracts its contribution from s
  • Otherwise, removes it from the waiting set r

The add(v) function handles adding an element's new state after updating its frequency:

  • Creates a tuple p = (cnt[v], v) representing the element's new state
  • If l is not empty and p is greater than the smallest element in l, it belongs in the elite group
    • Adds p to l and updates the sum s
  • Otherwise, adds p to the waiting set r

Main Algorithm:

  1. Initialize the first window: Add the first k elements to build the initial window

    • For each element, remove its old state, increment its count, then add its new state

  2. Balance the sets: After the window is built, ensure l contains exactly x elements (or all distinct elements if less than x)

    • While l has fewer than x elements and r is not empty:
      • Move the largest element from r to l and update s
    • While l has more than x elements:
      • Move the smallest element from l to r and update s

  3. Record the x-sum: The current value of s is the x-sum for this window position

  4. Slide the window: Remove the leftmost element and continue to the next position

    • Remove the old state of the departing element
    • Decrement its count in cnt
    • Add its new state (which might be with count 0)

The process repeats for each window position, giving us an x-sum for each of the n - k + 1 possible windows.

Time Complexity: O(n × log k) - Each element is added and removed once, with each operation taking O(log k) time due to the sorted lists.

Space Complexity: O(n) - For storing the counter and sorted lists.

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Example Walkthrough

Let's walk through a concrete example with nums = [1, 1, 2, 2, 3, 4, 2, 3], k = 6, and x = 2.

Initial Setup:

  • We need to find the x-sum for each window of size 6
  • The x-sum includes only the top 2 most frequent elements

Window 1: [1, 1, 2, 2, 3, 4]

Building the window element by element:

  • Add 1: cnt = {1:1}, move (1,1) to l, s = 1
  • Add 1: cnt = {1:2}, update to (2,1) in l, s = 2
  • Add 2: cnt = {1:2, 2:1}, add (1,2) to l, s = 3
  • Add 2: cnt = {1:2, 2:2}, update to (2,2) in l, s = 5
  • Add 3: cnt = {1:2, 2:2, 3:1}, add (1,3) to r (since we already have 2 elements in l)
  • Add 4: cnt = {1:2, 2:2, 3:1, 4:1}, add (1,4) to r

Current state:

  • l = [(2,1), (2,2)] - top 2 frequent elements (both appear twice)
  • r = [(1,3), (1,4)] - remaining elements
  • s = 2×1 + 2×2 = 6 (x-sum for this window)

Sliding to Window 2: [1, 2, 2, 3, 4, 2]

Remove leftmost element (1):

  • Remove (2,1) from l, s = 4
  • Update cnt = {1:1, 2:2, 3:1, 4:1}
  • Add (1,1) back, but now it goes to r since (1,1) < (2,2)

Add rightmost element (2):

  • Remove (2,2) from l, s = 0
  • Update cnt = {1:1, 2:3, 3:1, 4:1}
  • Add (3,2) to l, s = 6

Balance sets (move highest from r to l until we have 2 elements):

  • Move (1,4) from r to l, s = 7

Current state:

  • l = [(1,4), (3,2)] - element 2 appears 3 times, element 4 appears once
  • r = [(1,1), (1,3)]
  • s = 1×4 + 3×2 = 10 (x-sum for this window)

Sliding to Window 3: [2, 2, 3, 4, 2, 3]

Remove leftmost element (1):

  • Remove (1,1) from r
  • Update cnt = {2:3, 3:1, 4:1} (element 1 is removed entirely)
  • No need to add anything back since count is 0

Add rightmost element (3):

  • Remove (1,3) from r
  • Update cnt = {2:3, 3:2, 4:1}
  • Add (2,3) to l, s = 12

Balance sets (we now have 3 elements in l, need to demote one):

  • Move (1,4) from l to r, s = 11

Current state:

  • l = [(2,3), (3,2)] - element 3 appears 2 times, element 2 appears 3 times
  • r = [(1,4)]
  • s = 2×3 + 3×2 = 12 (x-sum for this window)

Final Result: answer = [6, 10, 12]

This walkthrough demonstrates how:

  1. The sorted lists maintain the top x elements dynamically
  2. The sum s is updated incrementally as elements move between sets
  3. The tie-breaking rule favors larger values when frequencies are equal
  4. Elements smoothly transition between the elite and waiting groups as their frequencies change

Solution Implementation

1class Solution:
2    def findXSum(self, nums: List[int], k: int, x: int) -> List[int]:
3        """
4        Find the sum of top x frequent elements for each sliding window of size k.
5        Elements are ranked by (frequency, value) in descending order.
6      
7        Args:
8            nums: Input array
9            k: Window size
10            x: Number of top frequent elements to sum
11      
12        Returns:
13            List of sums for each window
14        """
15      
16        def add_to_sets(value: int) -> None:
17            """Add element to either top_x_set or remaining_set based on its priority."""
18            if frequency_map[value] == 0:
19                return
20          
21            priority_tuple = (frequency_map[value], value)
22          
23            # If top_x_set is empty or this element has higher priority than the minimum in top_x_set
24            if top_x_set and priority_tuple > top_x_set[0]:
25                nonlocal current_sum
26                current_sum += priority_tuple[0] * priority_tuple[1]  # frequency * value
27                top_x_set.add(priority_tuple)
28            else:
29                remaining_set.add(priority_tuple)
30      
31        def remove_from_sets(value: int) -> None:
32            """Remove element from whichever set it belongs to."""
33            if frequency_map[value] == 0:
34                return
35          
36            priority_tuple = (frequency_map[value], value)
37          
38            if priority_tuple in top_x_set:
39                nonlocal current_sum
40                current_sum -= priority_tuple[0] * priority_tuple[1]  # frequency * value
41                top_x_set.remove(priority_tuple)
42            else:
43                remaining_set.remove(priority_tuple)
44      
45        # Initialize data structures
46        top_x_set = SortedList()      # Maintains top x elements by (frequency, value)
47        remaining_set = SortedList()   # Maintains remaining elements
48        frequency_map = Counter()       # Tracks frequency of each element in current window
49        current_sum = 0                 # Sum of top x elements
50        n = len(nums)
51        result = [0] * (n - k + 1)     # Result array for each window
52      
53        # Process each element
54        for i, current_value in enumerate(nums):
55            # Update frequency for current element
56            remove_from_sets(current_value)
57            frequency_map[current_value] += 1
58            add_to_sets(current_value)
59          
60            # Calculate window start index
61            window_start = i - k + 1
62          
63            # Skip if we haven't formed a complete window yet
64            if window_start < 0:
65                continue
66          
67            # Balance the sets: ensure top_x_set has exactly x elements
68            # Move elements from remaining_set to top_x_set if needed
69            while remaining_set and len(top_x_set) < x:
70                element = remaining_set.pop()
71                top_x_set.add(element)
72                current_sum += element[0] * element[1]
73          
74            # Move excess elements from top_x_set to remaining_set
75            while len(top_x_set) > x:
76                element = top_x_set.pop(0)  # Remove smallest element
77                current_sum -= element[0] * element[1]
78                remaining_set.add(element)
79          
80            # Store result for current window
81            result[window_start] = current_sum
82          
83            # Remove the leftmost element of the window for next iteration
84            leftmost_element = nums[window_start]
85            remove_from_sets(leftmost_element)
86            frequency_map[leftmost_element] -= 1
87            if frequency_map[leftmost_element] > 0:
88                add_to_sets(leftmost_element)
89      
90        return result
91
1class Solution {
2    // TreeSet to maintain top X frequent elements (sorted by frequency desc, then value desc)
3    private TreeSet<int[]> topElements = new TreeSet<>((a, b) -> 
4        a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
5  
6    // TreeSet to maintain remaining elements not in top X
7    private TreeSet<int[]> remainingElements = new TreeSet<>(topElements.comparator());
8  
9    // Map to track frequency count of each number in current window
10    private Map<Integer, Integer> frequencyMap = new HashMap<>();
11  
12    // Running sum of (value * frequency) for top X elements
13    private long currentSum;
14
15    public long[] findXSum(int[] nums, int k, int x) {
16        int n = nums.length;
17        long[] result = new long[n - k + 1];
18      
19        for (int i = 0; i < n; i++) {
20            int currentValue = nums[i];
21          
22            // Remove old frequency entry before updating
23            removeFromSets(currentValue);
24          
25            // Update frequency count for current value
26            frequencyMap.merge(currentValue, 1, Integer::sum);
27          
28            // Add back with new frequency
29            addToSets(currentValue);
30          
31            // Check if we have a complete window of size k
32            int windowStartIndex = i - k + 1;
33            if (windowStartIndex < 0) {
34                continue;
35            }
36          
37            // Balance the sets to ensure topElements has exactly x elements (if possible)
38            // Move elements from remainingElements to topElements if needed
39            while (!remainingElements.isEmpty() && topElements.size() < x) {
40                int[] element = remainingElements.pollLast();
41                currentSum += 1L * element[0] * element[1];  // frequency * value
42                topElements.add(element);
43            }
44          
45            // Move excess elements from topElements to remainingElements
46            while (topElements.size() > x) {
47                int[] element = topElements.pollFirst();
48                currentSum -= 1L * element[0] * element[1];  // frequency * value
49                remainingElements.add(element);
50            }
51          
52            // Store the sum for current window
53            result[windowStartIndex] = currentSum;
54          
55            // Remove the element going out of window
56            removeFromSets(nums[windowStartIndex]);
57          
58            // Decrease frequency count for element leaving the window
59            frequencyMap.merge(nums[windowStartIndex], -1, Integer::sum);
60          
61            // Add back with updated frequency (or not at all if frequency becomes 0)
62            addToSets(nums[windowStartIndex]);
63        }
64      
65        return result;
66    }
67
68    /**
69     * Removes an element from either topElements or remainingElements set
70     * based on its current frequency
71     */
72    private void removeFromSets(int value) {
73        if (!frequencyMap.containsKey(value)) {
74            return;
75        }
76      
77        // Create pair [frequency, value] to search in sets
78        int[] elementPair = new int[] {frequencyMap.get(value), value};
79      
80        // Remove from appropriate set and update sum if needed
81        if (topElements.contains(elementPair)) {
82            topElements.remove(elementPair);
83            currentSum -= 1L * elementPair[0] * elementPair[1];  // frequency * value
84        } else {
85            remainingElements.remove(elementPair);
86        }
87    }
88
89    /**
90     * Adds an element to either topElements or remainingElements set
91     * based on comparison with current minimum in topElements
92     */
93    private void addToSets(int value) {
94        if (!frequencyMap.containsKey(value)) {
95            return;
96        }
97      
98        // Create pair [frequency, value] for the element
99        int[] elementPair = new int[] {frequencyMap.get(value), value};
100      
101        // Add to topElements if it's better than the minimum element there
102        if (!topElements.isEmpty() && 
103            topElements.comparator().compare(topElements.first(), elementPair) < 0) {
104            topElements.add(elementPair);
105            currentSum += 1L * elementPair[0] * elementPair[1];  // frequency * value
106        } else {
107            remainingElements.add(elementPair);
108        }
109    }
110}
111
1class Solution {
2public:
3    vector<long long> findXSum(vector<int>& nums, int k, int x) {
4        using FreqValuePair = pair<int, int>; // {frequency, value}
5      
6        // topX: maintains the top x elements (by frequency, then by value)
7        // remaining: maintains elements not in top x
8        set<FreqValuePair> topX, remaining;
9        long long currentSum = 0; // Sum of (frequency * value) for elements in topX
10        unordered_map<int, int> frequency; // Maps value to its frequency in current window
11      
12        // Helper function to add an element to the appropriate set after its frequency changes
13        auto addToSets = [&](int value) {
14            if (frequency[value] == 0) {
15                return; // Element not in window, nothing to add
16            }
17          
18            FreqValuePair element = {frequency[value], value};
19          
20            // If topX is empty or element is better than worst in topX, add to topX
21            if (!topX.empty() && element > *topX.begin()) {
22                currentSum += 1LL * element.first * element.second; // frequency * value
23                topX.insert(element);
24            } else {
25                remaining.insert(element);
26            }
27        };
28      
29        // Helper function to remove an element from sets before its frequency changes
30        auto removeFromSets = [&](int value) {
31            if (frequency[value] == 0) {
32                return; // Element not in window, nothing to remove
33            }
34          
35            FreqValuePair element = {frequency[value], value};
36          
37            // Check if element is in topX
38            auto it = topX.find(element);
39            if (it != topX.end()) {
40                currentSum -= 1LL * element.first * element.second; // frequency * value
41                topX.erase(it);
42            } else {
43                remaining.erase(element);
44            }
45        };
46      
47        vector<long long> result;
48      
49        // Process each element in the array
50        for (int i = 0; i < nums.size(); ++i) {
51            // Update frequency for new element entering window
52            removeFromSets(nums[i]); // Remove old frequency entry
53            ++frequency[nums[i]];     // Update frequency
54            addToSets(nums[i]);       // Add new frequency entry
55          
56            // Calculate window start position
57            int windowStart = i - k + 1;
58            if (windowStart < 0) {
59                continue; // Window not yet complete
60            }
61          
62            // Rebalance sets to maintain exactly x elements in topX (if possible)
63            // Move elements from remaining to topX if needed
64            while (!remaining.empty() && topX.size() < x) {
65                FreqValuePair best = *remaining.rbegin(); // Get best from remaining
66                currentSum += 1LL * best.first * best.second;
67                remaining.erase(best);
68                topX.insert(best);
69            }
70          
71            // Move excess elements from topX to remaining if needed
72            while (topX.size() > x) {
73                FreqValuePair worst = *topX.begin(); // Get worst from topX
74                currentSum -= 1LL * worst.first * worst.second;
75                topX.erase(worst);
76                remaining.insert(worst);
77            }
78          
79            // Add current window's x-sum to result
80            result.push_back(currentSum);
81          
82            // Remove element leaving the window
83            removeFromSets(nums[windowStart]); // Remove old frequency entry
84            --frequency[nums[windowStart]];     // Update frequency
85            addToSets(nums[windowStart]);       // Add new frequency entry (or nothing if freq becomes 0)
86        }
87      
88        return result;
89    }
90};
91
1function findXSum(nums: number[], k: number, x: number): number[] {
2    // Custom comparator for frequency-value pairs
3    // Sorts by frequency (ascending), then by value (ascending)
4    const compareFreqValuePair = (a: [number, number], b: [number, number]): number => {
5        if (a[0] !== b[0]) return a[0] - b[0]; // Compare by frequency
6        return a[1] - b[1]; // Compare by value if frequency is same
7    };
8  
9    // topX: maintains the top x elements (by frequency, then by value)
10    // Using arrays to simulate sorted sets
11    let topX: [number, number][] = []; // [frequency, value] pairs
12    let remaining: [number, number][] = []; // [frequency, value] pairs
13    let currentSum: number = 0; // Sum of (frequency * value) for elements in topX
14    const frequency: Map<number, number> = new Map(); // Maps value to its frequency in current window
15  
16    // Helper function to add an element to the appropriate set after its frequency changes
17    const addToSets = (value: number): void => {
18        const freq = frequency.get(value) || 0;
19        if (freq === 0) {
20            return; // Element not in window, nothing to add
21        }
22      
23        const element: [number, number] = [freq, value];
24      
25        // If topX is empty or element is better than worst in topX, add to topX
26        if (topX.length > 0 && compareFreqValuePair(element, topX[0]) > 0) {
27            currentSum += freq * value; // frequency * value
28            topX.push(element);
29            topX.sort(compareFreqValuePair); // Keep sorted
30        } else {
31            remaining.push(element);
32            remaining.sort(compareFreqValuePair); // Keep sorted
33        }
34    };
35  
36    // Helper function to remove an element from sets before its frequency changes
37    const removeFromSets = (value: number): void => {
38        const freq = frequency.get(value) || 0;
39        if (freq === 0) {
40            return; // Element not in window, nothing to remove
41        }
42      
43        const element: [number, number] = [freq, value];
44      
45        // Check if element is in topX
46        const topXIndex = topX.findIndex(item => item[0] === element[0] && item[1] === element[1]);
47        if (topXIndex !== -1) {
48            currentSum -= freq * value; // frequency * value
49            topX.splice(topXIndex, 1);
50        } else {
51            // Remove from remaining
52            const remainingIndex = remaining.findIndex(item => item[0] === element[0] && item[1] === element[1]);
53            if (remainingIndex !== -1) {
54                remaining.splice(remainingIndex, 1);
55            }
56        }
57    };
58  
59    const result: number[] = [];
60  
61    // Process each element in the array
62    for (let i = 0; i < nums.length; i++) {
63        // Update frequency for new element entering window
64        removeFromSets(nums[i]); // Remove old frequency entry
65        frequency.set(nums[i], (frequency.get(nums[i]) || 0) + 1); // Update frequency
66        addToSets(nums[i]); // Add new frequency entry
67      
68        // Calculate window start position
69        const windowStart = i - k + 1;
70        if (windowStart < 0) {
71            continue; // Window not yet complete
72        }
73      
74        // Rebalance sets to maintain exactly x elements in topX (if possible)
75        // Move elements from remaining to topX if needed
76        while (remaining.length > 0 && topX.length < x) {
77            const best = remaining[remaining.length - 1]; // Get best from remaining (last element)
78            currentSum += best[0] * best[1]; // frequency * value
79            remaining.pop();
80            topX.push(best);
81            topX.sort(compareFreqValuePair); // Keep sorted
82        }
83      
84        // Move excess elements from topX to remaining if needed
85        while (topX.length > x) {
86            const worst = topX[0]; // Get worst from topX (first element)
87            currentSum -= worst[0] * worst[1]; // frequency * value
88            topX.shift();
89            remaining.push(worst);
90            remaining.sort(compareFreqValuePair); // Keep sorted
91        }
92      
93        // Add current window's x-sum to result
94        result.push(currentSum);
95      
96        // Remove element leaving the window
97        removeFromSets(nums[windowStart]); // Remove old frequency entry
98        const currentFreq = frequency.get(nums[windowStart]) || 0;
99        if (currentFreq > 1) {
100            frequency.set(nums[windowStart], currentFreq - 1); // Update frequency
101        } else {
102            frequency.delete(nums[windowStart]); // Remove if frequency becomes 0
103        }
104        addToSets(nums[windowStart]); // Add new frequency entry (or nothing if freq becomes 0)
105    }
106  
107    return result;
108}
109

Time and Space Complexity

Time Complexity: O(n * k * log(k))

The algorithm uses a sliding window approach with two SortedLists to maintain the top x elements by frequency and value. Breaking down the operations:

  • The outer loop runs n times (once for each element in nums)
  • For each iteration:
    • remove(v) operation: Removing from a SortedList takes O(log(k)) time in the worst case, as the SortedList can contain at most k distinct elements
    • add(v) operation: Adding to a SortedList takes O(log(k)) time
    • The while loops for balancing between l and r SortedLists:
      • Each element can move between l and r at most once per window
      • Pop and add operations each take O(log(k)) time
    • For the window at position j:
      • Remove operation for nums[j]: O(log(k))
      • Add operation for nums[j]: O(log(k))

Since we perform a constant number of O(log(k)) operations per element, and we process n elements total, with potential rebalancing operations that could involve up to k elements in the worst case, the overall time complexity is O(n * k * log(k)).

Space Complexity: O(k)

The space usage consists of:

  • l and r SortedLists: Combined they store at most k distinct elements (one entry per distinct value in the current window), so O(k) space
  • cnt Counter: Stores at most k distinct elements from the current window, so O(k) space
  • ans array: Stores n - k + 1 results, which is O(n) space
  • Other variables (s, i, j, etc.): O(1) space

The dominant space complexity excluding the output array is O(k). Including the output array, the total space complexity is O(n).

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

Common Pitfalls

1. Incorrect Handling of Element Priority When Frequencies Are Equal

A critical pitfall is misunderstanding how elements should be prioritized when they have the same frequency. The problem states that when frequencies are equal, the element with the bigger value should be considered more frequent. This means the comparison tuple should be (frequency, value) in that order, which naturally gives us the correct sorting behavior.

Incorrect approach:

# Wrong: Using (value, frequency) would prioritize value over frequency
priority_tuple = (value, frequency_map[value])

Correct approach:

# Correct: Using (frequency, value) prioritizes frequency first, then value as tiebreaker
priority_tuple = (frequency_map[value], value)

2. Not Handling Zero-Frequency Elements Properly

When an element's frequency becomes 0 (after being removed from the window), attempting to add it to the sorted sets creates unnecessary overhead and potential bugs. The element with frequency 0 shouldn't contribute to any calculations.

Problem:

def add_to_sets(value: int) -> None:
    # Missing check for zero frequency
    priority_tuple = (frequency_map[value], value)
    # This would add (0, value) to the sets unnecessarily

Solution:

def add_to_sets(value: int) -> None:
    if frequency_map[value] == 0:
        return  # Skip elements with zero frequency
    priority_tuple = (frequency_map[value], value)
    # ... rest of the logic

3. Incorrect Sum Calculation

A common mistake is forgetting that the sum should include all occurrences of the top x elements, not just counting each unique element once.

Incorrect:

# Wrong: Only adds the value once
current_sum += priority_tuple[1]  # Just adds the value

Correct:

# Correct: Multiplies value by its frequency
current_sum += priority_tuple[0] * priority_tuple[1]  # frequency * value

4. Improper Set Balancing After Window Updates

After adding or removing elements, the sets might become unbalanced. The top_x_set should always contain exactly x elements (or all distinct elements if fewer than x exist). Failing to rebalance after each window update leads to incorrect results.

Issue: Not moving elements between sets when needed:

# After updating the window, forgetting to rebalance:
result[window_start] = current_sum  # Wrong if sets are unbalanced

Solution: Always rebalance before calculating the result:

# Move elements from remaining_set to top_x_set if needed
while remaining_set and len(top_x_set) < x:
    element = remaining_set.pop()
    top_x_set.add(element)
    current_sum += element[0] * element[1]

# Move excess elements from top_x_set to remaining_set
while len(top_x_set) > x:
    element = top_x_set.pop(0)
    current_sum -= element[0] * element[1]
    remaining_set.add(element)

# Now calculate the result
result[window_start] = current_sum

5. Edge Case: When Window Has Fewer Than x Distinct Elements

The problem states that if a window has fewer than x distinct elements, the x-sum is the sum of the entire array. The balancing logic handles this automatically by keeping all elements in top_x_set when there are fewer than x distinct elements, but it's important to understand this behavior.

Key insight: The while loop condition while remaining_set and len(top_x_set) < x ensures we only move elements if they exist in remaining_set, naturally handling the case where we have fewer than x distinct elements.

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