Maximized Network Stream Analyzer 6 — Problem Statement & Solution Guide
Problem Description
You are given a collection of N independent tasks. Each task i is described by three integers: start_i, end_i and profit_i, where start_i < end_i. A task occupies the half‑open interval [start_i, end_i) and yields profit_i if it is selected. No two selected tasks may overlap in time. Determine the maximum total profit obtainable by selecting a subset of non‑overlapping tasks.
Input:
The first line contains a single integer N, the number of tasks. Each of the next N lines contains three space‑separated integers start_i, end_i and profit_i.
Output:
Print a single integer – the maximum achievable profit.
Constraints guarantee that a solution exists and fits within 64‑bit signed integer range.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Maximized Network Stream Analyzer 6"
WHY DOES IT MATTER?
Weighted interval scheduling exemplifies the power of combining sorting, binary search, and DP to handle optimization under conflict constraints, a pattern that recurs in scheduling, resource allocation, and revenue maximization problems.
OPTIMIZATION CHALLENGE
The key insight is that after sorting by end time, the optimal profit up to any interval depends only on the best profit of the latest non‑overlapping interval, which can be located in O(log N) via binary search, collapsing an exponential search space into linear DP.
REAL-WORLD CONNECTION
Think of a cloud compute scheduler that must allocate VMs to time‑bounded jobs with different revenue potentials; the algorithm decides which jobs to accept to maximize total revenue while ensuring no two jobs occupy the same physical machine at overlapping times.
When coding, pre‑compute an array of predecessor indices using binary search once, then reuse it in the DP loop; this avoids repeated scans and keeps the implementation clean and fast.
COMPLEXITY AT A GLANCE
O(N log N)O(N)Core Theory — Why This Approach?
The problem is a classic instance of the Weighted Interval Scheduling problem, which can be solved optimally using dynamic programming combined with binary search. After sorting the tasks by their end times, each task i either contributes its profit plus the best profit achievable from the latest non‑overlapping task before i, or it is skipped in favor of the best profit up to i‑1. This recurrence captures the optimal substructure: the optimal solution for the first i tasks depends only on optimal solutions of smaller prefixes.
A naive exhaustive search would try every subset of tasks, leading to O(2^N) time, which quickly becomes infeasible even for moderate N (e.g., N=30). The greedy approach that works for unweighted interval scheduling (select the earliest finishing job) fails here because profits differ; picking the earliest finishing low‑profit job can block a later high‑profit interval. The DP‑with‑binary‑search paradigm reduces the complexity to O(N log N) by reusing previously computed optimal profits and locating the predecessor interval in logarithmic time.
Interview Questions on This Problem
Q1How would you modify the solution if tasks could share endpoints, i.e., intervals are closed [start, end] and overlapping at a single point is allowed?
Treat intervals as half‑open by incrementing the end time of each task by one unit (or adjust the binary search condition to use <= instead of <). This ensures that tasks ending exactly when another starts are considered non‑overlapping, preserving the DP recurrence.
Q2Can you solve the problem in O(N) time after sorting? If not, why is O(N log N) a lower bound?
After sorting by end time, finding the predecessor for each interval requires a search among earlier intervals. Without additional constraints (e.g., bounded time range), this search cannot be done in constant time for all intervals, making O(N log N) the optimal bound due to the need for binary search or a balanced tree.
Q3How would you extend the algorithm to also return the actual set of selected tasks, not just the maximum profit?
Maintain a predecessor array during DP. After computing the DP table, backtrack from the last interval: if DP[i] == profit_i + DP[pred[i]] include i and move to pred[i]; otherwise move to i‑1. This reconstructs the optimal subset in O(N) additional time.
Examples
Input
3 1 3 50 3 5 20 6 19 100
Output
150
Explanation: The tasks are: 1) [1,3) profit 50 2) [3,5) profit 20 3) [6,19) profit 100 Tasks 1 and 2 do not overlap, and task 3 also does not overlap with either of them. Selecting tasks 1 and 3 yields profit 50 + 100 = 150, which is larger than any other feasible combination.
Input
4 1 2 20 2 4 30 3 5 25 4 6 15
Output
45
Explanation: Possible non‑overlapping selections: - Tasks 1 and 3: profit 20 + 25 = 45 - Tasks 2 and 4: profit 30 + 15 = 45 All other combinations either overlap or give a smaller total. Hence the maximum profit is 45.
Input
5 1 4 10 2 6 20 5 7 30 6 9 25 8 10 15
Output
55
Explanation: Consider the following schedule: - Task 1: [1,4) profit 10 - Task 3: [5,7) profit 30 - Task 5: [8,10) profit 15 These three tasks are pairwise non‑overlapping, giving total profit 10 + 30 + 15 = 55. Any schedule that includes task 2 or task 4 reduces the total because they conflict with higher‑profit tasks. Therefore 55 is optimal.
Constraints
- 1 <= N <= 200000
- 0 <= start_i < end_i <= 10^9
- 1 <= profit_i <= 10^9
- All input values are integers and the answer fits in a 64‑bit signed integer.
Optimal Approach & Strategy
Sort tasks by end time, pre‑compute the predecessor for each task with binary search, then fill a DP array using the recurrence dp[i] = max(dp[i‑1], profit_i + dp[pred[i]]).
Brute Force Approach
Enumerate every subset of tasks, check if the subset is non‑overlapping, and compute its total profit, keeping the maximum.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
if (typeof num === 'number') {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
return sum(num for num in nums if isinstance(num, (int, float)))function solution(nums) {
let sum = 0;
for (let num of nums) {
if (typeof num === 'number') {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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