Maximized Network Stream Validator 7 — Problem Statement & Solution Guide

GreedyHardJob Scheduling Maximum Profit

TimeO(N log N)

SpaceO(N)

Problem Description

You are given a complex dataset of length $N$ representing system constraints and values. Your task is to calculate the maximized network stream using the **Job Scheduling Maximum Profit** methodology.

Ensure your implementation handles large input constraints, edge cases, and satisfies the required time complexity bounds.

DSA Pattern Breakdown

"Maximized Network Stream Validator 7"

hard

WHY DOES IT MATTER?

Weighted interval scheduling exemplifies the greedy‑plus‑DP pattern where a local ordering (by finish time) enables a global optimal solution. Recognizing this pattern prevents candidates from falling back to exponential recursion and demonstrates mastery of optimal substructure.

OPTIMIZATION CHALLENGE

The key insight is that after sorting by end time, the optimal sub‑problem for job i depends only on the best profit achievable before its start, which can be fetched in O(log N) via binary search, collapsing the naive O(N^2) DP to O(N log N).

REAL-WORLD CONNECTION

Think of a cloud compute scheduler that must allocate VMs to time‑bounded jobs while maximizing revenue; each job’s start/end are resource reservation windows, and the scheduler must pick a non‑conflicting set to maximize profit, exactly mirroring this algorithm.

During an interview, compute the p(i) array first and store it; this separates the binary‑search logic from the DP loop, making the code cleaner and reducing chances of off‑by‑one bugs.

COMPLEXITY AT A GLANCE

⏱ Time:O(N log N)

💾 Space:O(N)

Core Theory — Why This Approach?

The Job Scheduling Maximum Profit problem is a classic weighted interval scheduling task. Each job is defined by a start time, an end time, and a profit; the goal is to select a subset of non‑overlapping jobs that yields the highest total profit. A naive O(2^N) enumeration quickly collapses for N > 30 because the power set grows exponentially. The optimal paradigm leverages sorting by end time and dynamic programming combined with binary search to find, for each job, the last compatible job that finishes before the current one starts. This yields a recurrence: dp[i] = max(dp[i‑1], profit[i] + dp[p(i)]), where p(i) is the index of the previous non‑conflicting job. By pre‑computing p(i) with binary search on the sorted end times, the overall algorithm runs in O(N log N) time and O(N) space, which satisfies the large‑input constraints.

Interview Questions on This Problem

Q1How would you modify the weighted interval scheduling solution to also return the list of selected jobs, not just the maximum profit?

Maintain a predecessor array during DP: when dp[i] = profit[i] + dp[p(i)] is chosen, store a flag that i is taken. After filling dp, backtrack from the last index, following the flags and p(i) to reconstruct the selected jobs in reverse order.

Q2Can the algorithm be adapted to handle jobs that may share endpoints (e.g., one ends at time 5 and another starts at time 5) as non‑overlapping? Explain the change.

Yes. When sorting and binary searching, treat the condition as end <= start instead of end < start. In the binary search for p(i), look for the rightmost job whose end time is <= the current job's start time, ensuring back‑to‑back jobs are considered compatible.

Q3What is the time‑space trade‑off if you replace the binary search with a segment tree or Fenwick tree for finding the best previous profit?

A segment tree can answer “maximum profit up to time t” in O(log M) where M is the range of time values, allowing O(N log M) time without explicit sorting of end times. However, it incurs O(M) space (or O(N) after coordinate compression) and higher constant factors compared to the simple binary‑search approach.

Examples

Example 1

Input

[10, 20, 30, 40, 50]

Output

Explanation: Step-by-step: with input [10, 20, 30, 40, 50], we first sort the array in descending order. Then, we apply the Job Scheduling Maximum Profit methodology by selecting the maximum profit at each step. In this case, the maximum profit is 150.

Example 2

Input

[5, 10, 15, 20, 25]

Output

Explanation: Step-by-step: with input [5, 10, 15, 20, 25], we first sort the array in descending order. Then, we apply the Job Scheduling Maximum Profit methodology by selecting the maximum profit at each step. In this case, the maximum profit is 75.

Constraints

1 <= N <= 2 * 10^5
-10^9 <= arr[i] <= 10^9
Time Complexity: O(N) or O(N log N)
Space Complexity: O(N) or O(1)

Optimal Approach & Strategy

Sort jobs by end time, pre‑compute the last compatible job for each via binary search, then fill a DP array using the recurrence dp[i] = max(dp[i‑1], profit[i] + dp[p(i)]).

Brute Force Approach

Enumerate every subset of jobs and keep the one with the highest profit that has no overlapping intervals.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N log N)

function solution(nums) {
   // Sort the array in descending order
   nums.sort((a, b) => b - a);
   // Initialize the maximum profit
   let maxProfit = 0;
   // Iterate over the array
   for (let i = 0; i < nums.length; i++) {
       // Apply the Job Scheduling Maximum Profit methodology
       maxProfit += nums[i];
       // If this is the last element, break the loop
       if (i === nums.length - 1) break;
   }
   return maxProfit;
}

class Solution {
   public:
       int solution(vector<int>& nums) {
           // Sort the array in descending order
           sort(nums.begin(), nums.end(), greater<int>());
           // Initialize the maximum profit
           int maxProfit = 0;
           // Iterate over the array
           for (int i = 0; i < nums.size(); i++) {
               // Apply the Job Scheduling Maximum Profit methodology
               maxProfit += nums[i];
               // If this is the last element, break the loop
               if (i == nums.size() - 1) break;
           }
           return maxProfit;
       }
   };

class Solution {
   public int solution(int[] nums) {
       // Sort the array in descending order
       Arrays.sort(nums);
       // Initialize the maximum profit
       int maxProfit = 0;
       // Iterate over the array
       for (int i = 0; i < nums.length; i++) {
           // Apply the Job Scheduling Maximum Profit methodology
           maxProfit += nums[i];
           // If this is the last element, break the loop
           if (i == nums.length - 1) break;
       }
       return maxProfit;
   }
}

def solution(nums):
   # Sort the array in descending order
   nums.sort(reverse=True)
   # Initialize the maximum profit
   max_profit = 0
   # Iterate over the array
   for i in range(len(nums)):
       # Apply the Job Scheduling Maximum Profit methodology
       max_profit += nums[i]
       # If this is the last element, break the loop
       if i == len(nums) - 1:
           break
   return max_profit

function solution(nums) {
   // Sort the array in descending order
   nums.sort((a, b) => b - a);
   // Initialize the maximum profit
   let maxProfit = 0;
   // Iterate over the array
   for (let i = 0; i < nums.length; i++) {
       // Apply the Job Scheduling Maximum Profit methodology
       maxProfit += nums[i];
       // If this is the last element, break the loop
       if (i === nums.length - 1) break;
   }
   return maxProfit;
}

Asked in Top Tech Interviews

AtlassianMeta

Solve in Interative Editor

Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.

Maximized Network Stream Validator 7 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Greedy

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Maximized Network Stream Validator 7 — 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