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3318. Find X-Sum of All K-Long Subarrays I

EasyArrayHash 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 [2, 1, 3, 1, 1, 2] with x = 2:

  • Count occurrences: 1 appears 3 times, 2 appears 2 times, 3 appears 1 time
  • Top 2 most frequent: 1 (3 occurrences) and 2 (2 occurrences)
  • The x-sum would be: 3 × 1 + 2 × 2 = 7

The problem requires calculating the x-sum for each sliding window of size k as it moves through the array from left to right.

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Intuition

The key challenge is efficiently maintaining the top x most frequent elements as we slide the window of size k through the array.

When we move the sliding window by one position, only two elements change: one element leaves the window (from the left) and one element enters (from the right). This means we don't need to recalculate everything from scratch - we can update our data structures incrementally.

The main insight is to use two ordered sets to partition elements based on their frequency and value:

  • Set l contains the top x most frequent elements (these contribute to the x-sum)
  • Set r contains all other elements (these don't contribute to the x-sum)

Each element is stored as a tuple (frequency, value) so that ordering naturally prioritizes by frequency first, then by value if frequencies are equal.

When an element's frequency changes (either entering or leaving the window), we need to:

  1. Remove it from its current position in either l or r
  2. Update its frequency in our counter
  3. Re-insert it into the appropriate set based on its new frequency

The brilliance of this approach is maintaining the invariant that l always contains exactly x elements (or fewer if there aren't enough distinct elements). After each update, we rebalance:

  • If l has fewer than x elements and r is not empty, we move the largest element from r to l
  • If l has more than x elements, we move the smallest element from l to r

By tracking the sum s of elements in l incrementally (adding when elements move into l, subtracting when they leave), we can compute each window's x-sum in constant time after the rebalancing.

This sliding window technique with two ordered sets allows us to efficiently maintain the top x elements without repeatedly sorting or scanning the entire window, achieving O(log k) time per window movement.

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 the top x most frequent elements in each sliding window.

Data Structures:

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

Helper Functions:

  1. remove(v): Removes element v from its current position in either l or r

    • Creates tuple p = (cnt[v], v) with current frequency
    • If p is in l, subtracts its contribution from sum s and removes it
    • Otherwise, removes it from r

  2. add(v): Adds element v to the appropriate set based on its new frequency

    • Creates tuple p = (cnt[v], v) with updated frequency
    • If l is empty or p is greater than the smallest element in l, adds to l and updates sum
    • Otherwise, adds to r

Main Algorithm:

  1. Initialize the first window (elements 0 to k-1):

    • For each element, remove it from its old position, increment its count, then add it back
    • This builds up the initial frequency counts and populates the sets

  2. Process each window position:

    • After the window is fully formed (when j = i - k + 1 >= 0):

      a. Rebalance the sets to ensure l contains exactly x elements:

      • While l has fewer than x elements and r is not empty:
        • Pop the largest element from r and add to l
        • Update sum s accordingly
      • While l has more than x elements:
        • Pop the smallest element from l and add to r
        • Update sum s accordingly

      b. Record the x-sum: ans[j] = s

      c. Slide the window:

      • Remove the leftmost element nums[j] from the window
      • Call remove(nums[j]) to remove it from sets
      • Decrement cnt[nums[j]]
      • Call add(nums[j]) to re-add with updated frequency

The time complexity is O(n × log k) where n is the array length, as each element insertion/removal in the SortedList takes O(log k) time. The space complexity is O(n) for storing the counter and ordered sets.

This approach efficiently maintains the x-sum by only updating the elements that change, rather than recalculating from scratch for each window.

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Example Walkthrough

Let's walk through the solution with a concrete example:

  • nums = [1, 1, 2, 2, 3, 4, 2, 3], k = 6, x = 2

We need to find the x-sum for each window of size 6.

Initial Setup:

  • cnt = {} (frequency counter)
  • l = [] (top x elements)
  • r = [] (remaining elements)
  • s = 0 (sum of elements in l)

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

Process each element:

  1. nums[0] = 1:

    • Remove 1 (not present), update cnt[1] = 1, add (1, 1) to l
    • l = [(1, 1)], s = 1

  2. nums[1] = 1:

    • Remove (1, 1) from l, update cnt[1] = 2, add (2, 1) to l
    • l = [(2, 1)], s = 3

  3. nums[2] = 2:

    • Remove 2 (not present), update cnt[2] = 1, add (1, 2) to l
    • l = [(1, 2), (2, 1)], s = 5

  4. nums[3] = 2:

    • Remove (1, 2) from l, update cnt[2] = 2, add (2, 2) to l
    • l = [(2, 1), (2, 2)], s = 7

  5. nums[4] = 3:

    • Remove 3 (not present), update cnt[3] = 1, add (1, 3) to r (since l has 2 elements and (1, 3) < (2, 1))
    • l = [(2, 1), (2, 2)], r = [(1, 3)], s = 7

  6. nums[5] = 4:

    • Remove 4 (not present), update cnt[4] = 1, add (1, 4) to r
    • l = [(2, 1), (2, 2)], r = [(1, 3), (1, 4)], s = 7

Window 1 Complete:

  • Current frequencies: {1: 2, 2: 2, 3: 1, 4: 1}
  • l = [(2, 1), (2, 2)] contains the top 2 most frequent
  • x-sum = 7 (2×1 + 2×2)
  • answer[0] = 7

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

Remove nums[0] = 1 (the element leaving):

  • Remove (2, 1) from l, s = 4
  • Update cnt[1] = 1
  • Add (1, 1) to r since (1, 1) < (2, 2)
  • l = [(2, 2)], r = [(1, 1), (1, 3), (1, 4)]

Process nums[6] = 2 (the element entering):

  • Remove (2, 2) from l, s = 0
  • Update cnt[2] = 3
  • Add (3, 2) to l
  • l = [(3, 2)], r = [(1, 1), (1, 3), (1, 4)], s = 6

Rebalance: Since |l| = 1 < x = 2, move the largest from r to l:

  • Move (1, 4) from r to l
  • l = [(1, 4), (3, 2)], r = [(1, 1), (1, 3)], s = 10

Window 2 Complete:

  • Current frequencies: {1: 1, 2: 3, 3: 1, 4: 1}
  • l = [(1, 4), (3, 2)] contains top 2 (2 with freq 3, and 4 with freq 1)
  • x-sum = 10 (1×4 + 3×2)
  • answer[1] = 10

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

Remove nums[1] = 1:

  • Remove (1, 1) from r
  • Update cnt[1] = 0

Process nums[7] = 3:

  • Remove (1, 3) from r
  • Update cnt[3] = 2
  • Add (2, 3) to l since (2, 3) > (1, 4)
  • l = [(1, 4), (2, 3), (3, 2)], s = 16

Rebalance: Since |l| = 3 > x = 2, move the smallest from l to r:

  • Move (1, 4) from l to r
  • l = [(2, 3), (3, 2)], r = [(1, 4)], s = 12

Window 3 Complete:

  • Current frequencies: {2: 3, 3: 2, 4: 1}
  • l = [(2, 3), (3, 2)] contains top 2 (2 with freq 3, and 3 with freq 2)
  • x-sum = 12 (2×3 + 3×2)
  • answer[2] = 12

Final Answer: [7, 10, 12]

This example demonstrates how the algorithm efficiently maintains the top x elements using two ordered sets, rebalancing after each window slide to ensure we always track the correct x-sum.

Solution Implementation

1class Solution:
2    def findXSum(self, nums: List[int], k: int, x: int) -> List[int]:
3        """
4        Find the x-sum for each sliding window of size k.
5        The x-sum is the sum of the top x most frequent elements (frequency * value).
6      
7        Args:
8            nums: Input array of integers
9            k: Size of the sliding window
10            x: Number of top frequent elements to consider
11          
12        Returns:
13            List of x-sums for each window
14        """
15      
16        def add_element_to_sets(value: int) -> None:
17            """Add an element to either top_x_elements or remaining_elements based on priority."""
18            if element_count[value] == 0:
19                return
20          
21            # Create priority tuple (frequency, value) for comparison
22            priority_tuple = (element_count[value], value)
23          
24            # If top set is empty or new element has higher priority than minimum in top set
25            if top_x_elements and priority_tuple > top_x_elements[0]:
26                nonlocal current_sum
27                current_sum += priority_tuple[0] * priority_tuple[1]  # frequency * value
28                top_x_elements.add(priority_tuple)
29            else:
30                remaining_elements.add(priority_tuple)
31      
32        def remove_element_from_sets(value: int) -> None:
33            """Remove an element from either top_x_elements or remaining_elements."""
34            if element_count[value] == 0:
35                return
36          
37            priority_tuple = (element_count[value], value)
38          
39            if priority_tuple in top_x_elements:
40                nonlocal current_sum
41                current_sum -= priority_tuple[0] * priority_tuple[1]  # frequency * value
42                top_x_elements.remove(priority_tuple)
43            else:
44                remaining_elements.remove(priority_tuple)
45      
46        # Initialize data structures
47        top_x_elements = SortedList()      # Stores top x frequent elements
48        remaining_elements = SortedList()   # Stores remaining elements
49        element_count = Counter()           # Tracks frequency of each element
50        current_sum = 0                     # Current x-sum
51        n = len(nums)
52        result = [0] * (n - k + 1)         # Result array for all windows
53      
54        # Process each element
55        for i, current_value in enumerate(nums):
56            # Update element in the window
57            remove_element_from_sets(current_value)
58            element_count[current_value] += 1
59            add_element_to_sets(current_value)
60          
61            # Calculate window start index
62            window_start = i - k + 1
63          
64            # Skip if window is not complete yet
65            if window_start < 0:
66                continue
67          
68            # Balance the sets: ensure top_x_elements has exactly x elements
69            # Move elements from remaining_elements to top_x_elements if needed
70            while remaining_elements and len(top_x_elements) < x:
71                element_to_promote = remaining_elements.pop()
72                top_x_elements.add(element_to_promote)
73                current_sum += element_to_promote[0] * element_to_promote[1]
74          
75            # Move excess elements from top_x_elements to remaining_elements
76            while len(top_x_elements) > x:
77                element_to_demote = top_x_elements.pop(0)
78                current_sum -= element_to_demote[0] * element_to_demote[1]
79                remaining_elements.add(element_to_demote)
80          
81            # Store the x-sum for current window
82            result[window_start] = current_sum
83          
84            # Remove the leftmost element from window for next iteration
85            leftmost_element = nums[window_start]
86            remove_element_from_sets(leftmost_element)
87            element_count[leftmost_element] -= 1
88            if element_count[leftmost_element] > 0:
89                add_element_to_sets(leftmost_element)
90      
91        return result
92
1class Solution {
2    // TreeSet to maintain top x 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 (same comparator)
7    private TreeSet<int[]> remainingElements = new TreeSet<>(topElements.comparator());
8  
9    // Map to track frequency count of each number
10    private Map<Integer, Integer> frequencyMap = new HashMap<>();
11  
12    // Current sum of top x elements
13    private int currentSum;
14
15    /**
16     * Finds the x-sum for each k-length sliding window in the array.
17     * X-sum is the sum of (value * frequency) for the x most frequent elements.
18     * 
19     * @param nums The input array
20     * @param k The window size
21     * @param x The number of top frequent elements to consider
22     * @return Array containing x-sum for each window
23     */
24    public int[] findXSum(int[] nums, int k, int x) {
25        int n = nums.length;
26        int[] result = new int[n - k + 1];
27      
28        for (int i = 0; i < n; ++i) {
29            int currentValue = nums[i];
30          
31            // Remove old frequency entry before updating
32            removeFromSets(currentValue);
33          
34            // Update frequency count
35            frequencyMap.merge(currentValue, 1, Integer::sum);
36          
37            // Add back with new frequency
38            addToSets(currentValue);
39          
40            // Calculate window starting index
41            int windowStart = i - k + 1;
42          
43            // Skip if window is not complete yet
44            if (windowStart < 0) {
45                continue;
46            }
47          
48            // Move elements from remaining to top if needed
49            while (!remainingElements.isEmpty() && topElements.size() < x) {
50                int[] element = remainingElements.pollLast();
51                currentSum += element[0] * element[1];  // frequency * value
52                topElements.add(element);
53            }
54          
55            // Move excess elements from top to remaining
56            while (topElements.size() > x) {
57                int[] element = topElements.pollFirst();
58                currentSum -= element[0] * element[1];  // frequency * value
59                remainingElements.add(element);
60            }
61          
62            // Store the x-sum for current window
63            result[windowStart] = currentSum;
64          
65            // Remove the element going out of window
66            removeFromSets(nums[windowStart]);
67            frequencyMap.merge(nums[windowStart], -1, Integer::sum);
68            addToSets(nums[windowStart]);
69        }
70      
71        return result;
72    }
73
74    /**
75     * Removes an element's frequency entry from the appropriate set.
76     * 
77     * @param value The value to remove
78     */
79    private void removeFromSets(int value) {
80        // Skip if value doesn't exist in frequency map
81        if (!frequencyMap.containsKey(value)) {
82            return;
83        }
84      
85        // Create array with [frequency, value]
86        int[] element = new int[] {frequencyMap.get(value), value};
87      
88        // Remove from appropriate set
89        if (topElements.contains(element)) {
90            topElements.remove(element);
91            currentSum -= element[0] * element[1];  // frequency * value
92        } else {
93            remainingElements.remove(element);
94        }
95    }
96
97    /**
98     * Adds an element's frequency entry to the appropriate set.
99     * 
100     * @param value The value to add
101     */
102    private void addToSets(int value) {
103        // Skip if value doesn't exist in frequency map
104        if (!frequencyMap.containsKey(value)) {
105            return;
106        }
107      
108        // Create array with [frequency, value]
109        int[] element = new int[] {frequencyMap.get(value), value};
110      
111        // Add to top set if it's better than the minimum in top set
112        if (!topElements.isEmpty() && 
113            topElements.comparator().compare(topElements.first(), element) < 0) {
114            topElements.add(element);
115            currentSum += element[0] * element[1];  // frequency * value
116        } else {
117            remainingElements.add(element);
118        }
119    }
120}
121
1class Solution {
2public:
3    vector<int> findXSum(vector<int>& nums, int k, int x) {
4        using pii = pair<int, int>;  // pair of (frequency, value)
5      
6        // Two sets to maintain top x elements and remaining elements
7        set<pii> topX;        // Stores the x most frequent elements
8        set<pii> remaining;   // Stores elements not in top x
9        int xSum = 0;         // Sum of (frequency * value) for top x elements
10      
11        // Frequency map to track count of each element
12        unordered_map<int, int> frequency;
13      
14        // Lambda function to add an element to the appropriate set
15        auto addElement = [&](int value) {
16            if (frequency[value] == 0) {
17                return;
18            }
19          
20            pii element = {frequency[value], value};
21          
22            // If topX is empty or current element should be in topX
23            if (!topX.empty() && element > *topX.begin()) {
24                xSum += element.first * element.second;  // Add to sum
25                topX.insert(element);
26            } else {
27                remaining.insert(element);
28            }
29        };
30      
31        // Lambda function to remove an element from its current set
32        auto removeElement = [&](int value) {
33            if (frequency[value] == 0) {
34                return;
35            }
36          
37            pii element = {frequency[value], value};
38          
39            // Try to find and remove from topX set
40            auto it = topX.find(element);
41            if (it != topX.end()) {
42                xSum -= element.first * element.second;  // Subtract from sum
43                topX.erase(it);
44            } else {
45                // Remove from remaining set
46                remaining.erase(element);
47            }
48        };
49      
50        vector<int> result;
51      
52        // Sliding window approach
53        for (int i = 0; i < nums.size(); ++i) {
54            // Update frequency for current element
55            removeElement(nums[i]);      // Remove old frequency entry
56            ++frequency[nums[i]];         // Increment frequency
57            addElement(nums[i]);          // Add with new frequency
58          
59            // Calculate window start index
60            int windowStart = i - k + 1;
61            if (windowStart < 0) {
62                continue;  // Window not yet complete
63            }
64          
65            // Balance the sets: move elements from remaining to topX if needed
66            while (!remaining.empty() && topX.size() < x) {
67                pii largestRemaining = *remaining.rbegin();
68                xSum += largestRemaining.first * largestRemaining.second;
69                remaining.erase(largestRemaining);
70                topX.insert(largestRemaining);
71            }
72          
73            // Balance the sets: move excess elements from topX to remaining
74            while (topX.size() > x) {
75                pii smallestTop = *topX.begin();
76                xSum -= smallestTop.first * smallestTop.second;
77                topX.erase(smallestTop);
78                remaining.insert(smallestTop);
79            }
80          
81            // Add current window's x-sum to result
82            result.push_back(xSum);
83          
84            // Remove the element going out of window
85            removeElement(nums[windowStart]);      // Remove old frequency entry
86            --frequency[nums[windowStart]];        // Decrement frequency
87            addElement(nums[windowStart]);          // Add with new frequency
88        }
89      
90        return result;
91    }
92};
93
1function findXSum(nums: number[], k: number, x: number): number[] {
2    // Custom comparator for sorting pairs by frequency (ascending) then value (ascending)
3    const comparePairs = (a: [number, number], b: [number, number]): number => {
4        if (a[0] !== b[0]) return a[0] - b[0];
5        return a[1] - b[1];
6    };
7  
8    // Two sorted sets to maintain top x elements and remaining elements
9    // Using arrays with manual sorting to simulate C++ set behavior
10    let topX: [number, number][] = [];        // Stores the x most frequent elements [frequency, value]
11    let remaining: [number, number][] = [];   // Stores elements not in top x
12    let xSum = 0;                             // Sum of (frequency * value) for top x elements
13  
14    // Frequency map to track count of each element
15    const frequency: Map<number, number> = new Map();
16  
17    // Helper function to add an element to the appropriate set
18    const addElement = (value: number): void => {
19        const freq = frequency.get(value) || 0;
20        if (freq === 0) {
21            return;
22        }
23      
24        const element: [number, number] = [freq, value];
25      
26        // Determine which set to add the element to
27        if (topX.length > 0 && comparePairs(element, topX[0]) > 0) {
28            // Element should be in topX
29            xSum += element[0] * element[1];  // Add to sum
30            topX.push(element);
31            topX.sort(comparePairs);  // Keep sorted
32        } else {
33            // Element should be in remaining
34            remaining.push(element);
35            remaining.sort(comparePairs);  // Keep sorted
36        }
37    };
38  
39    // Helper function to remove an element from its current set
40    const removeElement = (value: number): void => {
41        const freq = frequency.get(value) || 0;
42        if (freq === 0) {
43            return;
44        }
45      
46        const element: [number, number] = [freq, value];
47      
48        // Try to find and remove from topX set
49        const topXIndex = topX.findIndex(e => e[0] === element[0] && e[1] === element[1]);
50        if (topXIndex !== -1) {
51            xSum -= element[0] * element[1];  // Subtract from sum
52            topX.splice(topXIndex, 1);
53        } else {
54            // Try to remove from remaining set
55            const remainingIndex = remaining.findIndex(e => e[0] === element[0] && e[1] === element[1]);
56            if (remainingIndex !== -1) {
57                remaining.splice(remainingIndex, 1);
58            }
59        }
60    };
61  
62    const result: number[] = [];
63  
64    // Process each element using sliding window approach
65    for (let i = 0; i < nums.length; i++) {
66        // Update frequency for current element
67        removeElement(nums[i]);                           // Remove old frequency entry
68        frequency.set(nums[i], (frequency.get(nums[i]) || 0) + 1);  // Increment frequency
69        addElement(nums[i]);                              // Add with new frequency
70      
71        // Calculate window start index
72        const windowStart = i - k + 1;
73        if (windowStart < 0) {
74            continue;  // Window not yet complete
75        }
76      
77        // Balance the sets: move elements from remaining to topX if needed
78        while (remaining.length > 0 && topX.length < x) {
79            const largestRemaining = remaining[remaining.length - 1];
80            xSum += largestRemaining[0] * largestRemaining[1];
81            remaining.pop();
82            topX.push(largestRemaining);
83            topX.sort(comparePairs);  // Keep sorted
84        }
85      
86        // Balance the sets: move excess elements from topX to remaining
87        while (topX.length > x) {
88            const smallestTop = topX[0];
89            xSum -= smallestTop[0] * smallestTop[1];
90            topX.shift();
91            remaining.push(smallestTop);
92            remaining.sort(comparePairs);  // Keep sorted
93        }
94      
95        // Add current window's x-sum to result
96        result.push(xSum);
97      
98        // Remove the element going out of window
99        removeElement(nums[windowStart]);                 // Remove old frequency entry
100        const currentFreq = frequency.get(nums[windowStart]) || 0;
101        frequency.set(nums[windowStart], currentFreq - 1);  // Decrement frequency
102        addElement(nums[windowStart]);                    // Add with new frequency
103    }
104  
105    return result;
106}
107

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. For each position in the array:

  • The add() and remove() operations are called, each performing operations on SortedList which take O(log(k)) time in the worst case (since the window size is k, there are at most k distinct elements)
  • The while loops for balancing between l and r SortedLists run at most O(k) times in total across all iterations, with each operation taking O(log(k)) time
  • For each of the n - k + 1 windows, we perform:
    • One remove() call: O(log(k))
    • One add() call: O(log(k))
    • Balancing operations: amortized O(log(k))

Therefore, the overall time complexity is O((n - k + 1) * log(k)) which simplifies to O(n * log(k)) for the main operations. However, in the worst case where rebalancing happens frequently, it could be O(n * k * log(k)).

Space Complexity: O(k)

The space usage includes:

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

The dominant space complexity is O(n) for the answer array, but if we only consider auxiliary space (excluding the output), it would be O(k).

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

Common Pitfalls

1. Incorrect Priority Comparison When Adding Elements

The most critical pitfall in this solution is the logic error in the add_element_to_sets function. The current code has:

if top_x_elements and priority_tuple > top_x_elements[0]:
    # Add to top_x_elements

This condition is checking if the new element is greater than the minimum element in top_x_elements, but it's also requiring top_x_elements to be non-empty. This creates a problem: when top_x_elements is empty, the element will always go to remaining_elements, even if it should be in the top set.

Solution: The condition should allow elements to be added to top_x_elements when it's empty OR when the element has higher priority than the minimum:

if not top_x_elements or priority_tuple > top_x_elements[0]:
    # Add to top_x_elements

2. Not Handling the Case Where x > Number of Distinct Elements

When the window has fewer than x distinct elements, all elements should contribute to the x-sum. The current balancing logic might not handle this edge case properly if not carefully implemented.

Solution: Ensure the balancing logic correctly handles this by only enforcing the size limit when there are more than x distinct elements:

# Only remove excess elements if we have more than x elements
while len(top_x_elements) > x and len(top_x_elements) > 0:
    element_to_demote = top_x_elements.pop(0)
    current_sum -= element_to_demote[0] * element_to_demote[1]
    remaining_elements.add(element_to_demote)

3. Inefficient Element Removal and Re-addition

When sliding the window, the code removes an element and then potentially re-adds it if its count is still positive. This can cause unnecessary operations if the element appears multiple times in the window.

Solution: Only perform removal/addition when the frequency actually changes meaningfully:

# Before removing from window
old_count = element_count[leftmost_element]
if old_count > 1:
    # Element still exists in window with reduced frequency
    remove_element_from_sets(leftmost_element)
    element_count[leftmost_element] -= 1
    add_element_to_sets(leftmost_element)
else:
    # Element is completely removed from window
    remove_element_from_sets(leftmost_element)
    element_count[leftmost_element] = 0

4. Missing Zero-Count Check in Helper Functions

The helper functions check if element_count[value] == 0 and return early, but this might not catch all edge cases, especially after decrementing counts.

Solution: Add more robust checking and consider removing elements from the Counter when their count reaches zero:

def remove_element_from_sets(value: int) -> None:
    if value not in element_count or element_count[value] == 0:
        return
    # ... rest of the function

# After decrementing
element_count[leftmost_element] -= 1
if element_count[leftmost_element] == 0:
    del element_count[leftmost_element]

5. Tie-Breaking Rule Confusion

The problem states that when two elements have the same frequency, the element with the bigger value is considered more frequent. The tuple comparison (frequency, value) handles this correctly, but it's easy to mistakenly reverse this order or misunderstand the priority.

Solution: Add clear comments and consider using a named tuple or class for clarity:

from collections import namedtuple

ElementPriority = namedtuple('ElementPriority', ['frequency', 'value'])
# This ensures that elements are compared first by frequency (ascending),
# then by value (ascending) when frequencies are equal
Discover Your Strengths and Weaknesses: Take Our 5-Minute Quiz to Tailor Your Study Plan:

Which data structure is used to implement recursion?

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