Monotonic Envelope Engine 6 — Problem Statement & Solution Guide

StringsMediumSubsequence Verification

TimeO(n log n)

SpaceO(n)

Problem Description

You are tasked with analyzing a sequence of integer values representing signal amplitudes in a telemetry stream. The goal is to determine the 'Monotonic Envelope' by identifying the longest subsequence that is strictly monotonically increasing. A subsequence is derived by deleting zero or more elements from the original sequence without changing the relative order of the remaining elements. Strictly monotonically increasing means that for every pair of consecutive elements in the subsequence, the later element must be strictly greater than the previous one.

Given an array of integers nums, return the length of the longest strictly increasing subsequence. If the array is empty or contains only one element, the length is 0 or 1, respectively. The solution must efficiently handle large input sizes, implying that a brute-force approach is insufficient.

Input: An array nums of integers.

Output: An integer representing the length of the longest strictly increasing subsequence.

DSA Pattern Breakdown

"Monotonic Envelope Engine 6"

medium

WHY DOES IT MATTER?

LIS exemplifies the ‘patience sorting’ pattern, where a greedy structure combined with binary search compresses combinatorial explosion. Mastery of this pattern unlocks efficient solutions for a wide class of ordering and scheduling problems.

OPTIMIZATION CHALLENGE

The key insight is that only the minimal possible tail for each subsequence length matters; by discarding larger tails we reduce the state from O(2^n) possibilities to O(n) and enable O(log n) updates via binary search.

REAL-WORLD CONNECTION

Think of a telemetry buffer that must keep the longest stretch of rising signal strength; the tail array acts like a set of stacked plates where each plate represents the smallest possible top value for a stack of a given height, mirroring load‑balancing in distributed queues.

During an interview, code the binary‑search update loop first and verify it with a few hand‑crafted cases; if you get stuck, fall back to the classic ‘patience sorting’ analogy—it often clarifies why replacement is safe.

COMPLEXITY AT A GLANCE

⏱ Time:O(n log n)

💾 Space:O(n)

Core Theory — Why This Approach?

The Monotonic Envelope problem is a classic formulation of the Longest Increasing Subsequence (LIS). The naive solution examines every possible subsequence, leading to exponential time, which quickly becomes infeasible for streams with tens of thousands of amplitudes. The optimal paradigm leverages a greedy‑plus‑binary‑search technique: we maintain a dynamic list where the i‑th element stores the smallest possible tail value of an increasing subsequence of length i+1 seen so far. For each new amplitude we locate its position in this list using binary search; if it extends the list we append, otherwise we replace the existing tail, guaranteeing that the list always represents the best (smallest‑ending) subsequences for each length.

Why this works stems from the observation that a smaller tail gives more flexibility for future elements, never hurting the length of any subsequence we could build later. By repeatedly applying this replacement rule, we compress the exponential state space into a linear‑size structure while preserving the ability to reconstruct the length of the longest monotonic envelope. The resulting algorithm runs in O(n log n) time and O(n) space, which is optimal for the comparison‑based model.

In practice, the LIS formulation also underpins many variations—strictly increasing, non‑decreasing, or with constraints on gaps—making the greedy‑binary‑search skeleton a versatile tool in a senior engineer’s algorithmic toolbox.

Interview Questions on This Problem

Q1How would you modify the LIS algorithm to handle a non‑strictly increasing (i.e., allowing equal values) envelope?

Replace the binary search condition from lower_bound (first element >= x) to upper_bound (first element > x) when updating the tail array, so equal values extend the current subsequence rather than replace it.

Q2Given a stream of amplitudes that cannot be stored entirely in memory, how can you compute the length of the monotonic envelope using O(k) space where k << n?

Maintain the tail array of size at most k (the current LIS length) and process the stream element‑by‑element; the algorithm only needs the tail values, not the full input, thus achieving O(k) additional space.

Q3Explain why the greedy replacement of tails does not affect the final LIS length, and provide a proof sketch.

Because any subsequence ending with a larger tail can be replaced by one ending with a smaller tail without reducing its length; the smaller tail can only increase future extension possibilities. By induction on subsequence length, the minimal tail for each length is sufficient to preserve the optimal length.

Examples

Example 1

Input

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

Output

Explanation: The longest strictly increasing subsequence is [2, 3, 7, 101] or [2, 5, 7, 101] or [2, 3, 7, 18] or [2, 5, 7, 18]. All have length 4. Note that [10, 9, 2, 5, 3, 7, 101, 18] is not increasing. The subsequence [2, 5, 7, 101] is valid because 2 < 5 < 7 < 101.

Example 2

Input

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

Output

Explanation: The longest strictly increasing subsequence is [0, 1, 2, 3] or [0, 1, 3, 3] is invalid because 3 is not strictly greater than 3. Valid subsequences of length 4 include [0, 1, 2, 3] (indices 0, 1, 4, 5) or [0, 1, 3, 3] is invalid. Another valid one is [0, 1, 2, 3] using indices 0, 1, 4, 5. Wait, [0, 1, 3, 3] is not strictly increasing. Let's check [0, 1, 2, 3]: 0<1, 1<2, 2<3. Yes. Length is 4.

Example 3

Input

nums = [7, 7, 7, 7, 7]

Output

Explanation: Since all elements are equal, no two consecutive elements in any subsequence can satisfy the strictly increasing condition (a < b). Therefore, the longest strictly increasing subsequence has length 1 (any single element).

Example 4

Input

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

Output

Explanation: The entire array is strictly increasing. Thus, the longest strictly increasing subsequence is the array itself, with length 5.

Constraints

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

Optimal Approach & Strategy

Iterate once, maintaining a tail array updated via binary search to keep the smallest possible ending values for each subsequence length.

Brute Force Approach

Generate all subsets of the sequence, filter those that are strictly increasing, and track the longest length.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n log n)

function monotonicEnvelope(nums) {
      if (nums.length === 0 || nums.length === 1) return nums;
      nums.sort((a, b) => a - b);
      let monotonicEnvelope = [nums[0]];
      for (let i = 1; i < nums.length; i++) {
         if (nums[i] !== nums[i - 1]) monotonicEnvelope.push(nums[i]);
      }
      return monotonicEnvelope;
   }

class Solution {
   public:
      int* monotonicEnvelope(int* nums, int numsSize) {
         if (numsSize == 0 || numsSize == 1) return nums;
         sort(nums, nums + numsSize);
         int* monotonicEnvelope = new int[numsSize];
         monotonicEnvelope[0] = nums[0];
         for (int i = 1; i < numsSize; i++) {
            if (nums[i] != nums[i - 1]) monotonicEnvelope[i] = nums[i];
         }
         return monotonicEnvelope;
      }
   };

class Solution {
      public int[] monotonicEnvelope(int[] nums) {
         if (nums.length == 0 || nums.length == 1) return nums;
         Arrays.sort(nums);
         int[] monotonicEnvelope = new int[nums.length];
         monotonicEnvelope[0] = nums[0];
         for (int i = 1; i < nums.length; i++) {
            if (nums[i] != nums[i - 1]) monotonicEnvelope[i] = nums[i];
         }
         return monotonicEnvelope;
      }
   }

def monotonic_envelope(nums):
      if len(nums) == 0 or len(nums) == 1:
         return nums
      nums.sort()
      monotonic_envelope = [nums[0]]
      for i in range(1, len(nums)):
         if nums[i] != nums[i - 1]:
            monotonic_envelope.append(nums[i])
      return monotonic_envelope

function monotonicEnvelope(nums) {
      if (nums.length === 0 || nums.length === 1) return nums;
      nums.sort((a, b) => a - b);
      let monotonicEnvelope = [nums[0]];
      for (let i = 1; i < nums.length; i++) {
         if (nums[i] !== nums[i - 1]) monotonicEnvelope.push(nums[i]);
      }
      return monotonicEnvelope;
   }

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Monotonic Envelope Engine 6 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Strings

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Monotonic Envelope Engine 6 — 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