LeetCode #3318 — EASY

Find X-Sum of All K-Long Subarrays I

Build confidence with an intuition-first walkthrough focused on array fundamentals.

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The Problem

Problem Statement

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

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

  • Count the occurrences of all elements in the array.
  • 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.
  • Calculate the sum of the resulting array.

Note that if an array has less than x distinct elements, its x-sum is the sum of the array.

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].

Example 1:

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

Output: [6,10,12]

Explanation:

  • For subarray [1, 1, 2, 2, 3, 4], only elements 1 and 2 will be kept in the resulting array. Hence, answer[0] = 1 + 1 + 2 + 2.
  • For subarray [1, 2, 2, 3, 4, 2], only elements 2 and 4 will be kept in the resulting array. Hence, answer[1] = 2 + 2 + 2 + 4. Note that 4 is kept in the array since it is bigger than 3 and 1 which occur the same number of times.
  • For subarray [2, 2, 3, 4, 2, 3], only elements 2 and 3 are kept in the resulting array. Hence, answer[2] = 2 + 2 + 2 + 3 + 3.

Example 2:

Input: nums = [3,8,7,8,7,5], k = 2, x = 2

Output: [11,15,15,15,12]

Explanation:

Since k == x, answer[i] is equal to the sum of the subarray nums[i..i + k - 1].

Constraints:

  • 1 <= n == nums.length <= 50
  • 1 <= nums[i] <= 50
  • 1 <= x <= k <= nums.length
Patterns Used
🪟Sliding Window→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an array nums of n integers and two integers k and x. The x-sum of an array is calculated by the following procedure: Count the occurrences of all elements in the array. 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. Calculate the sum of the resulting array. Note that if an array has less than x distinct elements, its x-sum is the sum of the array. 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].

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Sliding Window

Example 1

[1,1,2,2,3,4,2,3]
6
2

Example 2

[3,8,7,8,7,5]
2
2
Step 02

Core Insight

What unlocks the optimal approach

  • Implement the x-sum function. Then, run x-sum on every subarray of <code>nums</code> of size <code>k</code>.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3318: Find X-Sum of All K-Long Subarrays I
class Solution {
    private TreeSet<int[]> l = new TreeSet<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
    private TreeSet<int[]> r = new TreeSet<>(l.comparator());
    private Map<Integer, Integer> cnt = new HashMap<>();
    private int s;

    public int[] findXSum(int[] nums, int k, int x) {
        int n = nums.length;
        int[] ans = new int[n - k + 1];
        for (int i = 0; i < n; ++i) {
            int v = nums[i];
            remove(v);
            cnt.merge(v, 1, Integer::sum);
            add(v);
            int j = i - k + 1;
            if (j < 0) {
                continue;
            }
            while (!r.isEmpty() && l.size() < x) {
                var p = r.pollLast();
                s += p[0] * p[1];
                l.add(p);
            }
            while (l.size() > x) {
                var p = l.pollFirst();
                s -= p[0] * p[1];
                r.add(p);
            }
            ans[j] = s;

            remove(nums[j]);
            cnt.merge(nums[j], -1, Integer::sum);
            add(nums[j]);
        }
        return ans;
    }

    private void remove(int v) {
        if (!cnt.containsKey(v)) {
            return;
        }
        var p = new int[] {cnt.get(v), v};
        if (l.contains(p)) {
            l.remove(p);
            s -= p[0] * p[1];
        } else {
            r.remove(p);
        }
    }

    private void add(int v) {
        if (!cnt.containsKey(v)) {
            return;
        }
        var p = new int[] {cnt.get(v), v};
        if (!l.isEmpty() && l.comparator().compare(l.first(), p) < 0) {
            l.add(p);
            s += p[0] * p[1];
        } else {
            r.add(p);
        }
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × log k)
Space
O(n)

Approach Breakdown

BRUTE FORCE
O(n × k) time
O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW
O(n) time
O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.

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