Find X-Sum of All K-Long Subarrays I

Editorial

Core insight

Find X-Sum of All K-Long Subarrays I · Sliding Window

Core Insight for Find X-Sum of All K-Long Subarrays I

The key intuition is that two consecutive windows, nums[i...i+k-1] and nums[i+1...i+k], share k-1 elements. The only difference is that nums[i] leaves the window and nums[i+k] enters the window.

Instead of rebuilding the frequency map from scratch, we can maintain a single frequency map. As the window slides one position to the right:

Decrement the count of the element leaving the window.
Increment the count of the element entering the window.

After this $O(1)$ update to the frequency map, the map correctly reflects the state of the new window. We then perform the selection logic (finding the top x elements) on this updated map.

Because the constraints for this specific problem are small ( $N \le 50$ ), we can simply sort the distinct elements in the map at every step to find the top x. This hybrid approach—Sliding Window for state maintenance and Sorting for selection—is optimal for these constraints and prepares the groundwork for the more complex "Part II" version of the problem.

Deep dive

Full breakdown

Full reasoning, pitfalls, and implementation flow

LeetCode 3318: Find X-Sum of All K-Long Subarrays I Solution & Explanation

Problem Overview

TL;DR: Use a fixed-size sliding window to maintain element frequencies, updating the count of only the entering and exiting elements at each step to efficiently calculate the sum of the top x frequent elements.

The problem asks us to process an array nums using a window of size k. For every contiguous subarray of length k, we must calculate a specific metric called the "x-sum." The x-sum is defined as the sum of elements corresponding to the top x most frequent values in that subarray. If there is a tie in frequency, the larger value takes precedence. We need to return an array containing these x-sums for every valid window position.

This is a popular interview question that tests the ability to manage state within a moving window, specifically combining frequency counting with dynamic sorting or selection logic.

Brute Force Approach for Find X-Sum of All K-Long Subarrays I

The naive approach involves iterating through every possible subarray of length k, calculating the frequency of its elements from scratch, sorting them, and computing the sum.

Loop from index i = 0 to n - k.
Extract the subarray nums[i...i+k-1].
Create a frequency map for the current subarray.
Convert the map entries into a list of pairs (frequency, value).
Sort the list first by frequency (descending), then by value (descending).
Sum the contributions (frequency $\times$ value) of the top x elements.

python

1# Pseudo-code for Brute Force
2results = []
3for i from 0 to n - k:
4    subarray = nums[i : i+k]
5    freq_map = count_frequencies(subarray)
6    sorted_elements = sort(freq_map, key=(freq desc, value desc))
7    
8    current_x_sum = 0
9    count = 0
10    for element in sorted_elements:
11        if count < x:
12            current_x_sum += element.freq * element.value
13            count += 1
14    results.append(current_x_sum)
15return results

Time Complexity: $O((N-K) \cdot K \log K)$ . For each of the $N-K+1$ windows, we iterate $K$ elements to build the map and sort them. Why it fails: While this approach passes the constraints for LeetCode 3318 ( $N \le 50$ ), it is computationally redundant. It re-processes $K-1$ overlapping elements for every step. In a scenario with larger constraints (like LeetCode 3319), this leads to a Time Limit Exceeded (TLE).

Algorithm Strategy: Sliding Window Patterns

Initialization:
- Create a frequency map (Hash Map or Array).
- Iterate through the first k elements (nums[0] to nums[k-1]) and populate the map.
- Calculate the x-sum for this initial window by sorting the unique elements based on the criteria (frequency desc, value desc) and summing the top x.
Sliding Phase:
- Iterate i from k to n - 1.
- Remove Old: Identify nums[i - k]. Decrement its count in the map. If the count drops to 0, remove the key to keep the map size proportional to distinct elements.
- Add New: Identify nums[i]. Increment its count in the map.
- Calculate: Extract all (frequency, value) pairs from the map, sort them, and compute the x-sum for the current window.
- Append the result to the answer list.
Return: The list of x-sums.

Execution Flow

Consider nums = [1, 1, 2, 2, 3, 4], k = 4, x = 2.

Initialize Window (Indices 0-3): [1, 1, 2, 2]
- Map: {1: 2, 2: 2}
- Pairs: (2, 1), (2, 2)
- Sort (Freq desc, Val desc): (2, 2), (2, 1)
- Top x=2: Both are kept.
- Sum: $(2 \times 2) + (2 \times 1) = 6$ .
- Answer: [6]
Slide Window to Index 4: [1, 2, 2, 3]
- Remove nums[0] (value 1). Map: {1: 1, 2: 2}
- Add nums[4] (value 3). Map: {1: 1, 2: 2, 3: 1}
- Pairs: (1, 1), (2, 2), (1, 3)
- Sort:
  1. (2, 2) (Freq 2)
  2. (1, 3) (Freq 1, Val 3 > 1)
  3. (1, 1) (Freq 1, Val 1)
- Top x=2: (2, 2) and (1, 3).
- Sum: $(2 \times 2) + (1 \times 3) = 7$ .
- Answer: [6, 7]
Slide Window to Index 5: [2, 2, 3, 4]
- Remove nums[1] (value 1). Map: {1: 0, 2: 2, 3: 1} $\to$ {2: 2, 3: 1} (1 removed).
- Add nums[5] (value 4). Map: {2: 2, 3: 1, 4: 1}
- Sort: (2, 2), (1, 4), (1, 3)
- Top x=2: (2, 2) and (1, 4).
- Sum: $(2 \times 2) + (1 \times 4) = 8$ .
- Answer: [6, 7, 8]

Proof of Correctness

The correctness relies on the invariant that the frequency map always accurately reflects the counts of elements in the current window nums[i-k+1...i].

Base Case: The map is correctly built for the first k elements.
Inductive Step: When moving from window W_i to W_{i+1}, we decrement the count of the element leaving W_i and increment the count of the element entering W_{i+1}. Since addition and subtraction are inverse operations, the map accurately transitions to represent W_{i+1}.
Selection: Since we sort the entire set of distinct elements present in the window based on the problem's criteria (frequency, then value) at each step, the summation of the top x elements is guaranteed to be the correct x-sum.

Pattern Reuse Notes

The Sliding Window - Fixed Size pattern is fundamental for array processing. The logic used here (updating a state incrementally as the window slides) is identical to:

LeetCode 346: Moving Average from Data Stream (Maintains a running sum).
LeetCode 643: Maximum Average Subarray I (Maintains a running sum to find max average).
LeetCode 2985: Calculate Compressed Mean (Similar statistical calculation over a window).
LeetCode 3254: Find the Power of K-Size Subarrays I (Checks properties of subarrays like sorted order).

In all these cases, the $O(1)$ update step is the key to efficiency.

Feynman coach · Before you peek at the code

Can you explain the core insight in your own words?

If you can explain it before reading code, you’re far more likely to recall it under interview pressure.

Loading your attempt…

Reference Implementation

C++ Solution for LeetCode 3318

cpp

1#include <vector>
2#include <unordered_map>
3#include <algorithm>
4
5using namespace std;
6
7class Solution {
8public:
9    vector<int> findXSum(vector<int>& nums, int k, int x) {
10        int n = nums.size();
11        vector<int> answer;
12        unordered_map<int, int> freqMap;
13
14        // Initialize first window
15        for (int i = 0; i < k; ++i) {
16            freqMap[nums[i]]++;
17        }
18
19        // Helper lambda to calculate x-sum from current map
20        auto calculateXSum = [&]() -> int {
21            vector<pair<int, int>> elements;
22            for (auto const& [val, count] : freqMap) {
23                elements.push_back({count, val});
24            }
25            // Sort by frequency desc, then value desc
26            sort(elements.begin(), elements.end(), [](const pair<int, int>& a, const pair<int, int>& b) {
27                if (a.first != b.first) {
28                    return a.first > b.first;
29                }
30                return a.second > b.second;
31            });
32
33            long long currentSum = 0;
34            int taken = 0;
35            for (const auto& p : elements) {
36                if (taken >= x) break;
37                currentSum += (long long)p.first * p.second;
38                taken++;
39            }
40            return (int)currentSum;
41        };
42
43        answer.push_back(calculateXSum());
44
45        // Slide the window
46        for (int i = k; i < n; ++i) {
47            // Remove outgoing element
48            int out = nums[i - k];
49            freqMap[out]--;
50            if (freqMap[out] == 0) {
51                freqMap.erase(out);
52            }
53
54            // Add incoming element
55            int in = nums[i];
56            freqMap[in]++;
57
58            answer.push_back(calculateXSum());
59        }
60
61        return answer;
62    }
63};

Java Solution for LeetCode 3318

java

1import java.util.*;
2
3class Solution {
4    public int[] findXSum(int[] nums, int k, int x) {
5        int n = nums.length;
6        int[] answer = new int[n - k + 1];
7        Map<Integer, Integer> freqMap = new HashMap<>();
8
9        // Initialize first window
10        for (int i = 0; i < k; i++) {
11            freqMap.put(nums[i], freqMap.getOrDefault(nums[i], 0) + 1);
12        }
13
14        answer[0] = calculateXSum(freqMap, x);
15
16        // Slide the window
17        for (int i = k; i < n; i++) {
18            // Remove outgoing element
19            int out = nums[i - k];
20            if (freqMap.get(out) == 1) {
21                freqMap.remove(out);
22            } else {
23                freqMap.put(out, freqMap.get(out) - 1);
24            }
25
26            // Add incoming element
27            int in = nums[i];
28            freqMap.put(in, freqMap.getOrDefault(in, 0) + 1);
29
30            answer[i - k + 1] = calculateXSum(freqMap, x);
31        }
32
33        return answer;
34    }
35
36    private int calculateXSum(Map<Integer, Integer> map, int x) {
37        List<int[]> elements = new ArrayList<>();
38        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
39            elements.add(new int[]{entry.getValue(), entry.getKey()});
40        }
41
42        // Sort: Freq descending, then Value descending
43        Collections.sort(elements, (a, b) -> {
44            if (a[0] != b[0]) return b[0] - a[0];
45            return b[1] - a[1];
46        });
47
48        int sum = 0;
49        int count = 0;
50        for (int[] p : elements) {
51            if (count >= x) break;
52            sum += p[0] * p[1];
53            count++;
54        }
55        return sum;
56    }
57}

Python Solution for LeetCode 3318

python

1from typing import List
2from collections import Counter
3
4class Solution:
5    def findXSum(self, nums: List[int], k: int, x: int) -> List[int]:
6        n = len(nums)
7        # Frequency map
8        freq_map = Counter()
9        
10        # Initialize first window
11        for i in range(k):
12            freq_map[nums[i]] += 1
13            
14        def get_x_sum():
15            # Convert map to list of (freq, value) tuples
16            elements = []
17            for val, count in freq_map.items():
18                elements.append((count, val))
19            
20            # Sort: Primary key freq (desc), Secondary key val (desc)
21            elements.sort(key=lambda p: (p[0], p[1]), reverse=True)
22            
23            current_sum = 0
24            taken = 0
25            for count, val in elements:
26                if taken >= x:
27                    break
28                current_sum += count * val
29                taken += 1
30            return current_sum
31
32        ans = []
33        ans.append(get_x_sum())
34        
35        # Slide the window
36        for i in range(k, n):
37            # Remove outgoing element
38            out_elem = nums[i - k]
39            freq_map[out_elem] -= 1
40            if freq_map[out_elem] == 0:
41                del freq_map[out_elem]
42            
43            # Add incoming element
44            in_elem = nums[i]
45            freq_map[in_elem] += 1
46            
47            ans.append(get_x_sum())
48            
49        return ans

Editorial

Core insight

Find X-Sum of All K-Long Subarrays I · Sliding Window

Core Insight for Find X-Sum of All K-Long Subarrays I

Instead of rebuilding the frequency map from scratch, we can maintain a single frequency map. As the window slides one position to the right:

Decrement the count of the element leaving the window.
Increment the count of the element entering the window.

After this $O(1)$ update to the frequency map, the map correctly reflects the state of the new window. We then perform the selection logic (finding the top x elements) on this updated map.

Deep dive

Full breakdown

Full reasoning, pitfalls, and implementation flow

LeetCode 3318: Find X-Sum of All K-Long Subarrays I Solution & Explanation

Problem Overview

This is a popular interview question that tests the ability to manage state within a moving window, specifically combining frequency counting with dynamic sorting or selection logic.

Brute Force Approach for Find X-Sum of All K-Long Subarrays I

The naive approach involves iterating through every possible subarray of length k, calculating the frequency of its elements from scratch, sorting them, and computing the sum.

Loop from index i = 0 to n - k.
Extract the subarray nums[i...i+k-1].
Create a frequency map for the current subarray.
Convert the map entries into a list of pairs (frequency, value).
Sort the list first by frequency (descending), then by value (descending).
Sum the contributions (frequency $\times$ value) of the top x elements.

python

1# Pseudo-code for Brute Force
2results = []
3for i from 0 to n - k:
4    subarray = nums[i : i+k]
5    freq_map = count_frequencies(subarray)
6    sorted_elements = sort(freq_map, key=(freq desc, value desc))
7    
8    current_x_sum = 0
9    count = 0
10    for element in sorted_elements:
11        if count < x:
12            current_x_sum += element.freq * element.value
13            count += 1
14    results.append(current_x_sum)
15return results

Algorithm Strategy: Sliding Window Patterns

Initialization:
- Create a frequency map (Hash Map or Array).
- Iterate through the first k elements (nums[0] to nums[k-1]) and populate the map.
- Calculate the x-sum for this initial window by sorting the unique elements based on the criteria (frequency desc, value desc) and summing the top x.
Sliding Phase:
- Iterate i from k to n - 1.
- Remove Old: Identify nums[i - k]. Decrement its count in the map. If the count drops to 0, remove the key to keep the map size proportional to distinct elements.
- Add New: Identify nums[i]. Increment its count in the map.
- Calculate: Extract all (frequency, value) pairs from the map, sort them, and compute the x-sum for the current window.
- Append the result to the answer list.
Return: The list of x-sums.

Execution Flow

Consider nums = [1, 1, 2, 2, 3, 4], k = 4, x = 2.

Initialize Window (Indices 0-3): [1, 1, 2, 2]
- Map: {1: 2, 2: 2}
- Pairs: (2, 1), (2, 2)
- Sort (Freq desc, Val desc): (2, 2), (2, 1)
- Top x=2: Both are kept.
- Sum: $(2 \times 2) + (2 \times 1) = 6$ .
- Answer: [6]
Slide Window to Index 4: [1, 2, 2, 3]
- Remove nums[0] (value 1). Map: {1: 1, 2: 2}
- Add nums[4] (value 3). Map: {1: 1, 2: 2, 3: 1}
- Pairs: (1, 1), (2, 2), (1, 3)
- Sort:
  1. (2, 2) (Freq 2)
  2. (1, 3) (Freq 1, Val 3 > 1)
  3. (1, 1) (Freq 1, Val 1)
- Top x=2: (2, 2) and (1, 3).
- Sum: $(2 \times 2) + (1 \times 3) = 7$ .
- Answer: [6, 7]
Slide Window to Index 5: [2, 2, 3, 4]
- Remove nums[1] (value 1). Map: {1: 0, 2: 2, 3: 1} $\to$ {2: 2, 3: 1} (1 removed).
- Add nums[5] (value 4). Map: {2: 2, 3: 1, 4: 1}
- Sort: (2, 2), (1, 4), (1, 3)
- Top x=2: (2, 2) and (1, 4).
- Sum: $(2 \times 2) + (1 \times 4) = 8$ .
- Answer: [6, 7, 8]

Proof of Correctness

The correctness relies on the invariant that the frequency map always accurately reflects the counts of elements in the current window nums[i-k+1...i].

Base Case: The map is correctly built for the first k elements.
Inductive Step: When moving from window W_i to W_{i+1}, we decrement the count of the element leaving W_i and increment the count of the element entering W_{i+1}. Since addition and subtraction are inverse operations, the map accurately transitions to represent W_{i+1}.
Selection: Since we sort the entire set of distinct elements present in the window based on the problem's criteria (frequency, then value) at each step, the summation of the top x elements is guaranteed to be the correct x-sum.

Pattern Reuse Notes

The Sliding Window - Fixed Size pattern is fundamental for array processing. The logic used here (updating a state incrementally as the window slides) is identical to:

LeetCode 346: Moving Average from Data Stream (Maintains a running sum).
LeetCode 643: Maximum Average Subarray I (Maintains a running sum to find max average).
LeetCode 2985: Calculate Compressed Mean (Similar statistical calculation over a window).
LeetCode 3254: Find the Power of K-Size Subarrays I (Checks properties of subarrays like sorted order).

In all these cases, the $O(1)$ update step is the key to efficiency.

Feynman coach · Before you peek at the code

Can you explain the core insight in your own words?

If you can explain it before reading code, you’re far more likely to recall it under interview pressure.

Loading your attempt…

Reference Implementation

C++ Solution for LeetCode 3318

cpp

1#include <vector>
2#include <unordered_map>
3#include <algorithm>
4
5using namespace std;
6
7class Solution {
8public:
9    vector<int> findXSum(vector<int>& nums, int k, int x) {
10        int n = nums.size();
11        vector<int> answer;
12        unordered_map<int, int> freqMap;
13
14        // Initialize first window
15        for (int i = 0; i < k; ++i) {
16            freqMap[nums[i]]++;
17        }
18
19        // Helper lambda to calculate x-sum from current map
20        auto calculateXSum = [&]() -> int {
21            vector<pair<int, int>> elements;
22            for (auto const& [val, count] : freqMap) {
23                elements.push_back({count, val});
24            }
25            // Sort by frequency desc, then value desc
26            sort(elements.begin(), elements.end(), [](const pair<int, int>& a, const pair<int, int>& b) {
27                if (a.first != b.first) {
28                    return a.first > b.first;
29                }
30                return a.second > b.second;
31            });
32
33            long long currentSum = 0;
34            int taken = 0;
35            for (const auto& p : elements) {
36                if (taken >= x) break;
37                currentSum += (long long)p.first * p.second;
38                taken++;
39            }
40            return (int)currentSum;
41        };
42
43        answer.push_back(calculateXSum());
44
45        // Slide the window
46        for (int i = k; i < n; ++i) {
47            // Remove outgoing element
48            int out = nums[i - k];
49            freqMap[out]--;
50            if (freqMap[out] == 0) {
51                freqMap.erase(out);
52            }
53
54            // Add incoming element
55            int in = nums[i];
56            freqMap[in]++;
57
58            answer.push_back(calculateXSum());
59        }
60
61        return answer;
62    }
63};

Java Solution for LeetCode 3318

java

1import java.util.*;
2
3class Solution {
4    public int[] findXSum(int[] nums, int k, int x) {
5        int n = nums.length;
6        int[] answer = new int[n - k + 1];
7        Map<Integer, Integer> freqMap = new HashMap<>();
8
9        // Initialize first window
10        for (int i = 0; i < k; i++) {
11            freqMap.put(nums[i], freqMap.getOrDefault(nums[i], 0) + 1);
12        }
13
14        answer[0] = calculateXSum(freqMap, x);
15
16        // Slide the window
17        for (int i = k; i < n; i++) {
18            // Remove outgoing element
19            int out = nums[i - k];
20            if (freqMap.get(out) == 1) {
21                freqMap.remove(out);
22            } else {
23                freqMap.put(out, freqMap.get(out) - 1);
24            }
25
26            // Add incoming element
27            int in = nums[i];
28            freqMap.put(in, freqMap.getOrDefault(in, 0) + 1);
29
30            answer[i - k + 1] = calculateXSum(freqMap, x);
31        }
32
33        return answer;
34    }
35
36    private int calculateXSum(Map<Integer, Integer> map, int x) {
37        List<int[]> elements = new ArrayList<>();
38        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
39            elements.add(new int[]{entry.getValue(), entry.getKey()});
40        }
41
42        // Sort: Freq descending, then Value descending
43        Collections.sort(elements, (a, b) -> {
44            if (a[0] != b[0]) return b[0] - a[0];
45            return b[1] - a[1];
46        });
47
48        int sum = 0;
49        int count = 0;
50        for (int[] p : elements) {
51            if (count >= x) break;
52            sum += p[0] * p[1];
53            count++;
54        }
55        return sum;
56    }
57}

Python Solution for LeetCode 3318

python

1from typing import List
2from collections import Counter
3
4class Solution:
5    def findXSum(self, nums: List[int], k: int, x: int) -> List[int]:
6        n = len(nums)
7        # Frequency map
8        freq_map = Counter()
9        
10        # Initialize first window
11        for i in range(k):
12            freq_map[nums[i]] += 1
13            
14        def get_x_sum():
15            # Convert map to list of (freq, value) tuples
16            elements = []
17            for val, count in freq_map.items():
18                elements.append((count, val))
19            
20            # Sort: Primary key freq (desc), Secondary key val (desc)
21            elements.sort(key=lambda p: (p[0], p[1]), reverse=True)
22            
23            current_sum = 0
24            taken = 0
25            for count, val in elements:
26                if taken >= x:
27                    break
28                current_sum += count * val
29                taken += 1
30            return current_sum
31
32        ans = []
33        ans.append(get_x_sum())
34        
35        # Slide the window
36        for i in range(k, n):
37            # Remove outgoing element
38            out_elem = nums[i - k]
39            freq_map[out_elem] -= 1
40            if freq_map[out_elem] == 0:
41                del freq_map[out_elem]
42            
43            # Add incoming element
44            in_elem = nums[i]
45            freq_map[in_elem] += 1
46            
47            ans.append(get_x_sum())
48            
49        return ans

Core insight

Core Insight for Find X-Sum of All K-Long Subarrays I

Full breakdown

LeetCode 3318: Find X-Sum of All K-Long Subarrays I Solution & Explanation

Problem Overview

Brute Force Approach for Find X-Sum of All K-Long Subarrays I

Algorithm Strategy: Sliding Window Patterns

Execution Flow

Proof of Correctness

Pattern Reuse Notes

Can you explain the core insight in your own words?

Reference Implementation

C++ Solution for LeetCode 3318

Java Solution for LeetCode 3318

Python Solution for LeetCode 3318

Practise Find X-Sum of All K-Long Subarrays I and the problems around it

Related problems

Find X-Sum of All K-Long Subarrays I

Core insight

Core Insight for Find X-Sum of All K-Long Subarrays I

Full breakdown

LeetCode 3318: Find X-Sum of All K-Long Subarrays I Solution & Explanation

Problem Overview

Brute Force Approach for Find X-Sum of All K-Long Subarrays I

Algorithm Strategy: Sliding Window Patterns

Execution Flow

Proof of Correctness

Pattern Reuse Notes

Can you explain the core insight in your own words?

Reference Implementation

C++ Solution for LeetCode 3318

Java Solution for LeetCode 3318

Python Solution for LeetCode 3318

Practise Find X-Sum of All K-Long Subarrays I and the problems around it

Related problems