LeetCode #2747 — MEDIUM

Count Zero Request Servers

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer n denoting the total number of servers and a 2D 0-indexed integer array logs, where logs[i] = [server_id, time] denotes that the server with id server_id received a request at time time.

You are also given an integer x and a 0-indexed integer array queries.

Return a 0-indexed integer array arr of length queries.length where arr[i] represents the number of servers that did not receive any requests during the time interval [queries[i] - x, queries[i]].

Note that the time intervals are inclusive.

Example 1:

Input: n = 3, logs = [[1,3],[2,6],[1,5]], x = 5, queries = [10,11]
Output: [1,2]
Explanation: 
For queries[0]: The servers with ids 1 and 2 get requests in the duration of [5, 10]. Hence, only server 3 gets zero requests.
For queries[1]: Only the server with id 2 gets a request in duration of [6,11]. Hence, the servers with ids 1 and 3 are the only servers that do not receive any requests during that time period.

Example 2:

Input: n = 3, logs = [[2,4],[2,1],[1,2],[3,1]], x = 2, queries = [3,4]
Output: [0,1]
Explanation: 
For queries[0]: All servers get at least one request in the duration of [1, 3].
For queries[1]: Only server with id 3 gets no request in the duration [2,4].

Constraints:

1 <= n <= 10⁵
1 <= logs.length <= 10⁵
1 <= queries.length <= 10⁵
logs[i].length == 2
1 <= logs[i][0] <= n
1 <= logs[i][1] <= 10⁶
1 <= x <= 10⁵
x < queries[i] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given an integer n denoting the total number of servers and a 2D 0-indexed integer array logs, where logs[i] = [server_id, time] denotes that the server with id server_id received a request at time time. You are also given an integer x and a 0-indexed integer array queries. Return a 0-indexed integer array arr of length queries.length where arr[i] represents the number of servers that did not receive any requests during the time interval [queries[i] - x, queries[i]]. Note that the time intervals are inclusive.

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

3
[[1,3],[2,6],[1,5]]
5
[10,11]

Example 2

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

Step 02

Core Insight

What unlocks the optimal approach

Can we use sorting and two-pointer approach here?
Sort the queries array and logs array based on time in increasing order.
For every window of size x, use sliding window and two-pointer approach to find the answer to the queries.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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 #2747: Count Zero Request Servers
class Solution {
    public int[] countServers(int n, int[][] logs, int x, int[] queries) {
        Arrays.sort(logs, (a, b) -> a[1] - b[1]);
        int m = queries.length;
        int[][] qs = new int[m][0];
        for (int i = 0; i < m; ++i) {
            qs[i] = new int[] {queries[i], i};
        }
        Arrays.sort(qs, (a, b) -> a[0] - b[0]);
        Map<Integer, Integer> cnt = new HashMap<>();
        int[] ans = new int[m];
        int j = 0, k = 0;
        for (var q : qs) {
            int r = q[0], i = q[1];
            int l = r - x;
            while (k < logs.length && logs[k][1] <= r) {
                cnt.merge(logs[k++][0], 1, Integer::sum);
            }
            while (j < logs.length && logs[j][1] < l) {
                if (cnt.merge(logs[j][0], -1, Integer::sum) == 0) {
                    cnt.remove(logs[j][0]);
                }
                j++;
            }
            ans[i] = n - cnt.size();
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2747: Count Zero Request Servers
func countServers(n int, logs [][]int, x int, queries []int) []int {
	sort.Slice(logs, func(i, j int) bool { return logs[i][1] < logs[j][1] })
	m := len(queries)
	qs := make([][2]int, m)
	for i, q := range queries {
		qs[i] = [2]int{q, i}
	}
	sort.Slice(qs, func(i, j int) bool { return qs[i][0] < qs[j][0] })
	cnt := map[int]int{}
	ans := make([]int, m)
	j, k := 0, 0
	for _, q := range qs {
		r, i := q[0], q[1]
		l := r - x
		for k < len(logs) && logs[k][1] <= r {
			cnt[logs[k][0]]++
			k++
		}
		for j < len(logs) && logs[j][1] < l {
			cnt[logs[j][0]]--
			if cnt[logs[j][0]] == 0 {
				delete(cnt, logs[j][0])
			}
			j++
		}
		ans[i] = n - len(cnt)
	}
	return ans
}

# Accepted solution for LeetCode #2747: Count Zero Request Servers
class Solution:
    def countServers(
        self, n: int, logs: List[List[int]], x: int, queries: List[int]
    ) -> List[int]:
        cnt = Counter()
        logs.sort(key=lambda x: x[1])
        ans = [0] * len(queries)
        j = k = 0
        for r, i in sorted(zip(queries, count())):
            l = r - x
            while k < len(logs) and logs[k][1] <= r:
                cnt[logs[k][0]] += 1
                k += 1
            while j < len(logs) and logs[j][1] < l:
                cnt[logs[j][0]] -= 1
                if cnt[logs[j][0]] == 0:
                    cnt.pop(logs[j][0])
                j += 1
            ans[i] = n - len(cnt)
        return ans

// Accepted solution for LeetCode #2747: Count Zero Request Servers
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2747: Count Zero Request Servers
// class Solution {
//     public int[] countServers(int n, int[][] logs, int x, int[] queries) {
//         Arrays.sort(logs, (a, b) -> a[1] - b[1]);
//         int m = queries.length;
//         int[][] qs = new int[m][0];
//         for (int i = 0; i < m; ++i) {
//             qs[i] = new int[] {queries[i], i};
//         }
//         Arrays.sort(qs, (a, b) -> a[0] - b[0]);
//         Map<Integer, Integer> cnt = new HashMap<>();
//         int[] ans = new int[m];
//         int j = 0, k = 0;
//         for (var q : qs) {
//             int r = q[0], i = q[1];
//             int l = r - x;
//             while (k < logs.length && logs[k][1] <= r) {
//                 cnt.merge(logs[k++][0], 1, Integer::sum);
//             }
//             while (j < logs.length && logs[j][1] < l) {
//                 if (cnt.merge(logs[j][0], -1, Integer::sum) == 0) {
//                     cnt.remove(logs[j][0]);
//                 }
//                 j++;
//             }
//             ans[i] = n - cnt.size();
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2747: Count Zero Request Servers
function countServers(n: number, logs: number[][], x: number, queries: number[]): number[] {
    logs.sort((a, b) => a[1] - b[1]);
    const m = queries.length;
    const qs: number[][] = [];
    for (let i = 0; i < m; ++i) {
        qs.push([queries[i], i]);
    }
    qs.sort((a, b) => a[0] - b[0]);
    const cnt: Map<number, number> = new Map();
    const ans: number[] = new Array(m);
    let j = 0;
    let k = 0;
    for (const [r, i] of qs) {
        const l = r - x;
        while (k < logs.length && logs[k][1] <= r) {
            cnt.set(logs[k][0], (cnt.get(logs[k][0]) || 0) + 1);
            ++k;
        }
        while (j < logs.length && logs[j][1] < l) {
            cnt.set(logs[j][0], (cnt.get(logs[j][0]) || 0) - 1);
            if (cnt.get(logs[j][0]) === 0) {
                cnt.delete(logs[j][0]);
            }
            ++j;
        }
        ans[i] = n - cnt.size;
    }
    return ans;
}

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(l × log l + m × log m + n)

Space

O(l + m)

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.