Maximum Temporal Node Connections — Problem Statement & Solution Guide

Binary SearchMediumMixed

TimeO(n log n)

SpaceO(n)

Problem Description

In a distributed event logging system, nodes are recorded with integer timestamps indicating their activation sequence. A valid connection chain is defined as a subsequence of these timestamps where each subsequent timestamp is strictly greater than the previous one, representing a causal progression of events. Your task is to determine the maximum length of such a strictly increasing temporal chain that can be extracted from the given sequence.

Given an integer array timestamps of length n, where timestamps[i] denotes the activation time of the i-th node, compute the length of the longest strictly increasing subsequence (LIS). The subsequence does not need to be contiguous, but the relative order of elements must be preserved, and each element in the subsequence must be strictly larger than its predecessor.

Return an integer representing the maximum number of nodes that can be connected in this strictly increasing temporal order.

DSA Pattern Breakdown

"Maximum Temporal Node Connections"

medium

WHY DOES IT MATTER?

The LIS pattern captures the essence of ordering constraints in many real‑world streams, such as event causality, version upgrades, and stock price trends. Mastering this pattern equips engineers to reason about optimal subsequences under monotonicity, a frequent requirement in performance‑critical code.

OPTIMIZATION CHALLENGE

The key insight is to replace the exponential enumeration of subsequences with a compact representation of potential tails, and to locate the correct tail to update via binary search. This reduces both time from O(2^n) to O(n log n) and space from O(2^n) to O(n).

REAL-WORLD CONNECTION

In distributed systems, timestamps represent causal dependencies; the longest increasing chain corresponds to the deepest causal path, akin to the critical path in a DAG of events. Identifying this path helps in latency analysis and bottleneck detection.

During an interview, implement the O(n log n) solution first, then discuss how to reconstruct the actual sequence. Keep the binary search clean—use std::lower_bound in C++ or bisect_left in Python—to avoid off‑by‑one bugs.

COMPLEXITY AT A GLANCE

⏱ Time:O(n log n)

💾 Space:O(n)

Core Theory — Why This Approach?

The problem of finding the longest strictly increasing subsequence (LIS) of timestamps maps directly to the classic LIS problem. A naïve solution enumerates all possible subsequences, leading to exponential time (O(2^n)) and quickly becomes infeasible for large logs that can contain millions of entries. The optimal paradigm leverages the fact that the LIS can be built incrementally: for each timestamp we maintain the smallest possible tail value for all increasing subsequences of a given length. By using a binary‑searchable structure (typically a dynamic array), we can update these tails in O(log n) time per element, yielding an overall O(n log n) algorithm.

The underlying theory rests on the patience‑sorting analogy: each tail represents the top card of a pile, and placing a new timestamp on the leftmost pile whose top is greater or equal preserves optimality. This greedy‑binary‑search combination guarantees that the length of the tail array equals the length of the LIS, while the actual subsequence can be reconstructed with additional predecessor tracking if needed. The approach scales linearly with input size in memory and logarithmically in time, making it ideal for high‑throughput distributed logging systems.

Interview Questions on This Problem

Q1How would you modify the LIS algorithm to also output one actual longest increasing subsequence, not just its length?

Maintain an additional array 'prev' that stores the index of the predecessor for each element when it updates a tail. Also keep an array 'tailIndices' that stores the index of the element that ends each length. After processing, backtrack from the index stored for the longest length using 'prev' to reconstruct the sequence.

Q2If timestamps can repeat, how does the definition of a valid connection chain change and how would you adapt the algorithm?

When repeats are allowed, the chain must be strictly increasing, so equal timestamps cannot be part of the same chain. The binary search should use lower_bound (first element >= current) to replace, ensuring duplicates do not extend the length. The O(n log n) algorithm remains unchanged, only the comparison operator matters.

Q3Explain how you could solve the problem in O(n) time if the timestamps are guaranteed to be a permutation of 1..n.

When the input is a permutation of 1..n, the LIS length equals the length of the longest increasing contiguous segment after mapping each value to its position. By scanning once and counting consecutive increasing positions, we obtain the answer in O(n) time and O(1) extra space.

Examples

Example 1

Input

timestamps = [10, 9, 2, 5, 3, 7, 101, 18]

Output

Explanation: The strictly increasing subsequences include [2, 3, 7, 18] and [2, 5, 7, 101]. Both have a length of 4. No subsequence of length 5 exists because the sequence drops significantly after 101. Thus, the maximum number of connected nodes is 4.

Example 2

Input

timestamps = [0, 1, 0, 3, 2, 3]

Output

Explanation: One valid longest strictly increasing subsequence is [0, 1, 2, 3] (indices 0, 1, 4, 5). Another is [0, 1, 3] (indices 0, 1, 3) which is shorter. The subsequence [0, 1, 2, 3] has length 4. Note that [0, 0, 3] is invalid because 0 is not strictly greater than 0. The maximum length is 4.

Example 3

Input

timestamps = [7, 7, 7, 7]

Output

Explanation: Since all elements are identical, no two elements can form a strictly increasing pair. The longest strictly increasing subsequence consists of a single element. Therefore, the maximum number of connected nodes is 1.

Example 4

Input

timestamps = [1, 2, 3, 4, 5]

Output

Explanation: The entire array is already strictly increasing. The subsequence [1, 2, 3, 4, 5] has length 5, which is the maximum possible length for an array of size 5.

Constraints

1 <= timestamps.length <= 10^5
-10^9 <= timestamps[i] <= 10^9

Optimal Approach & Strategy

Maintain an array of minimal tail values for increasing subsequences of each length and update it with binary search for each timestamp, achieving O(n log n) time.

Brute Force Approach

Generate every possible subsequence and check if it is strictly increasing, tracking the longest length. This exhaustive search runs in exponential time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n log n)

/**
 * @param {number[]} nums
 * @return {number}
 */
var findLengthOfLIS = function(nums) {
    const n = nums.length;
    if (n === 0) return 0;
    const dp = new Array(n).fill(1);
    let maxLen = 1;
    for (let i = 1; i < n; i++) {
        for (let j = 0; j < i; j++) {
            if (nums[j] < nums[i]) {
                dp[i] = Math.max(dp[i], dp[j] + 1);
            }
        }
        maxLen = Math.max(maxLen, dp[i]);
    }
    return maxLen;
};

class Solution {
public:
    int findLengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        if (n == 0) return 0;
        vector<int> dp(n, 1);
        int maxLen = 1;
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[j] < nums[i]) {
                    dp[i] = max(dp[i], dp[j] + 1);
                }
            }
            maxLen = max(maxLen, dp[i]);
        }
        return maxLen;
    }
};

class Solution {
    public int findLengthOfLIS(int[] nums) {
        int n = nums.length;
        if (n == 0) return 0;
        int[] dp = new int[n];
        Arrays.fill(dp, 1);
        int maxLen = 1;
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[j] < nums[i]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
            maxLen = Math.max(maxLen, dp[i]);
        }
        return maxLen;
    }
}

class Solution:
    def findLengthOfLIS(self, nums: List[int]) -> int:
        n = len(nums)
        if n == 0:
            return 0
        dp = [1] * n
        max_len = 1
        for i in range(1, n):
            for j in range(i):
                if nums[j] < nums[i]:
                    dp[i] = max(dp[i], dp[j] + 1)
            max_len = max(max_len, dp[i])
        return max_len

/**
 * @param {number[]} nums
 * @return {number}
 */
var findLengthOfLIS = function(nums) {
    const n = nums.length;
    if (n === 0) return 0;
    const dp = new Array(n).fill(1);
    let maxLen = 1;
    for (let i = 1; i < n; i++) {
        for (let j = 0; j < i; j++) {
            if (nums[j] < nums[i]) {
                dp[i] = Math.max(dp[i], dp[j] + 1);
            }
        }
        maxLen = Math.max(maxLen, dp[i]);
    }
    return maxLen;
};

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Maximum Temporal Node Connections Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Binary Search

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Maximum Temporal Node Connections — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor