Maximum Total from Optimal Activation Order - Solution & Explanation

Problem Overview: You are given an array where each element contributes to the total when it becomes activated. The order of activation changes the final score, so the goal is to choose an activation sequence that maximizes the total value. The challenge is identifying which elements should be activated earlier versus later to maximize cumulative gain.

Approach 1: Brute Force Permutation (O(n!))

The most direct method is to generate every possible activation order and compute the resulting total for each permutation. For every order, iterate through the array and simulate the activation step by step while accumulating the contribution. The maximum across all permutations becomes the answer. This approach clearly demonstrates how activation order affects the result but becomes infeasible very quickly because permutations grow factorially. Time complexity is O(n!) and space complexity is O(n) for recursion or permutation storage.

Approach 2: Greedy with Sorting (O(n log n))

The key insight is that elements that produce larger contributions should generally be activated earlier. By sorting the array based on the metric that determines contribution (often value or gain potential), you ensure high-impact activations happen first. After sorting, iterate through the array and accumulate the total while maintaining any running contribution needed by the scoring rule. Sorting dominates the runtime with O(n log n) time and O(1) or O(n) auxiliary space depending on the language implementation. This greedy ordering works because each step locally maximizes the gain without harming future choices.

Approach 3: Greedy with Heap / Priority Queue (O(n log n))

When the activation decision depends on dynamic conditions (for example, only some elements are eligible at each step), a heap (priority queue) becomes useful. First sort elements by the constraint that determines when they become available. Use two pointers to scan the sorted list while pushing eligible candidates into a max-heap. At each step, extract the element with the highest contribution and activate it next. The heap ensures the best candidate is always chosen among currently valid options. Each push and pop costs O(log n), giving an overall time complexity of O(n log n) and space complexity O(n). This pattern often appears in scheduling and resource allocation problems.

Recommended for interviews: Interviewers expect the greedy reasoning that leads to sorting or a priority queue. Starting with brute force shows you understand that order matters, but the real signal of problem-solving ability is recognizing that a greedy strategy with sorting or a greedy heap selection reduces the search space to O(n log n). That optimization is the typical production-ready solution.