Sequence Optimization Problem — Problem Statement & Solution Guide

QueueHardMatrix Chain Multiplication

TimeO(n log k)

SpaceO(k)

Problem Description

Given a sequence of operations with specific execution times and a constraint on the number of parallel processors, determine the optimal sequence of operations that minimizes the total execution time.

DSA Pattern Breakdown

"Sequence Optimization Problem"

hard

WHY DOES IT MATTER?

Efficiently balancing work across limited resources is a cornerstone of high‑throughput systems, from web servers handling request bursts to GPU kernels scheduling kernels. Mastering this pattern demonstrates a candidate’s ability to translate abstract scheduling constraints into concrete, performant code.

OPTIMIZATION CHALLENGE

The key insight is to avoid recomputing the load of every processor for each operation. By storing only the next free timestamp in a priority queue, each assignment becomes a O(log k) heap operation, collapsing a potential O(n k) brute‑force scan into a scalable solution.

REAL-WORLD CONNECTION

Think of a call center with a fixed number of agents. Each incoming call has an estimated handling time, and the goal is to keep the overall waiting time minimal. Assigning each call to the agent who becomes free first mirrors the min‑heap scheduling algorithm used in distributed job queues like Apache Kafka's consumer groups.

During an interview, implement the heap manually or use the language’s built‑in priority queue. Initialize the heap with k zeros (all processors idle), then loop through the tasks, updating the heap in place. Remember to return the max value after the loop—track it on the fly to avoid a second O(k) scan.

COMPLEXITY AT A GLANCE

⏱ Time:O(n log k)

💾 Space:O(k)

Core Theory — Why This Approach?

The Sequence Optimization Problem is a classic instance of scheduling jobs on identical parallel machines with a hard bound on the number of concurrent processors. The naive solution—trying every permutation of the operation sequence—has factorial time complexity and quickly becomes infeasible even for modest input sizes. The underlying optimal paradigm leverages the greedy principle combined with a min‑heap (priority queue) to always assign the next operation to the processor that will become idle the soonest. This approach respects the original order of operations (if required) while guaranteeing the minimal possible makespan because any deviation would only increase the load on the most‑loaded processor. The proof of optimality follows from an exchange argument: swapping two consecutive assignments cannot reduce the maximum completion time, thus the greedy schedule is locally optimal and, by induction, globally optimal for the ordered‑job variant. When the order can be rearranged, the problem becomes NP‑hard (the P||Cmax problem), and approximation schemes such as Longest‑Processing‑Time (LPT) are used, but the min‑heap greedy remains the de‑facto optimal for the ordered case.

In practice, the algorithm maintains a priority queue of size equal to the number of processors. For each operation, it extracts the smallest current finish time, adds the operation’s execution time, and pushes the updated finish time back. After processing all operations, the largest value in the heap represents the total execution time. This method reduces the exponential search space to a linear scan with logarithmic updates, yielding O(n log k) time where n is the number of operations and k is the processor count. Space usage is O(k) for the heap plus O(1) auxiliary storage. The elegance of this solution lies in its simplicity: a single data structure captures the dynamic state of all processors, and each decision is locally optimal yet provably leads to a globally optimal schedule under the given constraints.

Interview Questions on This Problem

Q1How would you schedule a list of tasks with known execution times on 4 identical CPUs to minimize total completion time while preserving the original task order?

Use a min‑heap of size 4 to track each CPU's next free time. Iterate through the tasks, pop the smallest free time, add the task's duration, and push the new free time back. The maximum value in the heap after all tasks are processed is the minimal makespan.

Q2Why does the greedy min‑heap approach guarantee the optimal makespan for ordered tasks, whereas the same greedy rule fails when tasks can be reordered arbitrarily?

When order is fixed, each decision only affects the current earliest‑available processor; any alternative assignment would leave that processor idle longer, increasing the makespan. An exchange argument shows swapping two consecutive assignments cannot improve the maximum load, proving optimality. If tasks can be reordered, the problem becomes NP‑hard, and the same greedy rule may produce sub‑optimal schedules because the global distribution of long and short tasks matters.

Q3Explain how you would modify the algorithm if each processor has a different speed factor (heterogeneous machines).

Maintain a min‑heap keyed by the projected finish time, which is the current load divided by the processor's speed plus the new task's time divided by the same speed. When assigning a task, pop the processor with the smallest projected finish, update its load accordingly, and push it back. This adapts the greedy rule to heterogeneous speeds while still running in O(n log k).

Examples

Example 1

Input

[[5, 15], [25, 35], [45, 55], [65, 75], [85, 95]]

Output

Explanation: Step-by-step: Given a sequence of operations with specific execution times and a constraint on the number of parallel processors, we can distribute the operations across processors in a way that each processor gets as close to equal time as possible. For the input [[5, 15], [25, 35], [45, 55], [65, 75], [85, 95]], we can distribute the operations as follows: (5+15)+(25+35)+(45+55)+(65+75)+(85+95) = 155 units.

Example 2

Input

[[5, 25], [35, 45], [55, 65], [75, 85]]

Output

Explanation: Step-by-step: Given a sequence of operations with specific execution times and a constraint on the number of parallel processors, we can distribute the operations across processors in a way that each processor gets as close to equal time as possible. For the input [[5, 25], [35, 45], [55, 65], [75, 85]], we can distribute the operations as follows: (5+25)+(35+45)+(55+65)+(75+85) = 105 units.

Constraints

The number of parallel processors is limited and should be a non-negative integer.
Each operation's execution time is a non-negative integer.
The input sequence contains at least one operation and is a list of non-negative integers.

Optimal Approach & Strategy

Use a min‑heap to always assign the next operation to the processor that becomes free earliest, updating finish times in O(log k) per operation.

Brute Force Approach

Try every possible assignment of operations to processors, compute the makespan for each, and pick the minimum.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n log k)

function solution(operations, processors) {
   operations.sort((a, b) => a[0] - b[0]);
   let start = 0;
   let end = processors - 1;
   let time = 0;
   while (start < operations.length) {
       let maxTime = 0;
       for (let i = start; i <= end && i < operations.length; i++) {
           maxTime = Math.max(maxTime, operations[i][0]);
       }
       time += maxTime;
       start = end + 1;
       end = Math.min(end + processors - 1, operations.length - 1);
   }
   return time;
}

class Solution {
   public:
       int solution(vector<vector<int>>& operations, int processors) {
           sort(operations.begin(), operations.end(), [](const vector<int>& a, const vector<int>& b) {
               return a[0] < b[0];
           });
           int start = 0;
           int end = processors - 1;
           int time = 0;
           while (start < operations.size()) {
               int maxTime = 0;
               for (int i = start; i <= end && i < operations.size(); i++) {
                   maxTime = max(maxTime, operations[i][0]);
               }
               time += maxTime;
               start = end + 1;
               end = min(end + processors - 1, operations.size() - 1);
           }
           return time;
       }
   };

class Solution {
   public int solution(int[][] operations, int processors) {
       Arrays.sort(operations, (a, b) -> a[0] - b[0]);
       int start = 0;
       int end = processors - 1;
       int time = 0;
       while (start < operations.length) {
           int maxTime = 0;
           for (int i = start; i <= end && i < operations.length; i++) {
               maxTime = Math.max(maxTime, operations[i][0]);
           }
           time += maxTime;
           start = end + 1;
           end = Math.min(end + processors - 1, operations.length - 1);
       }
       return time;
   }
}

def solution(operations, processors):
   operations.sort(key=lambda x: x[0])
   start = 0
   end = processors - 1
   time = 0
   while start < len(operations):
       max_time = 0
       for i in range(start, end + 1):
           max_time = max(max_time, operations[i][0])
       time += max_time
       start = end + 1
       end = min(end + processors - 1, len(operations) - 1)
   return time

function solution(operations, processors) {
   operations.sort((a, b) => a[0] - b[0]);
   let start = 0;
   let end = processors - 1;
   let time = 0;
   while (start < operations.length) {
       let maxTime = 0;
       for (let i = start; i <= end && i < operations.length; i++) {
           maxTime = Math.max(maxTime, operations[i][0]);
       }
       time += maxTime;
       start = end + 1;
       end = Math.min(end + processors - 1, operations.length - 1);
   }
   return time;
}

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Sequence Optimization Problem Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Queue

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Sequence Optimization Problem — 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

Solve in Interative Editor