Problem Description

You have two arrays value and limit, both of length n. Each element starts as inactive and can be activated in any order you choose.

The activation process follows these rules:

1. Activation Condition: To activate element at index i, the number of currently active elements must be strictly less than limit[i]. For example, if limit[i] = 3, you can only activate element i when there are 0, 1, or 2 active elements.

2. Value Addition: When you activate element at index i, you add value[i] to your total score.

3. Deactivation Rule: After each activation, if the number of active elements becomes x, then all elements j where limit[j] ≤ x become permanently inactive. This happens even if those elements were already active - they still become permanently unusable.

The goal is to find the maximum total value you can achieve by choosing the optimal activation order.

Example walkthrough:

• If you have value = [2, 3, 4] and limit = [2, 1, 3]
• Element at index 1 has limit[1] = 1, so it can only be activated when 0 elements are active
• Element at index 0 has limit[0] = 2, so it can be activated when 0 or 1 elements are active
• Element at index 2 has limit[2] = 3, so it can be activated when 0, 1, or 2 elements are active
• After activating any element, check if any elements should be permanently deactivated based on the current count of active elements

The solution groups elements by their limit values and greedily selects the highest-value elements from each group, taking at most lim elements from each group with limit lim (since after activating lim elements, all elements with that limit become permanently inactive).

Intuition

The key insight is understanding what happens when we activate elements. Once we have x active elements, all elements with limit[j] ≤ x become permanently inactive. This means elements with the same limit value form a critical threshold - once we reach that many active elements, they all become unusable.

Think about elements with limit = k. We can activate them as long as we have fewer than k active elements. But once we reach k active elements total, all elements with limit = k become permanently inactive. This creates a natural grouping: elements with the same limit value compete with each other because activating k of them means none of the remaining ones in that group can ever be used.

Since elements within the same limit group compete for activation slots, and we can activate at most k elements before all elements with limit = k become inactive, we should greedily choose the most valuable ones from each group.

The strategy becomes clear:

1. Group all elements by their limit value
2. For each group with limit = k, we can activate at most k elements before they all become inactive
3. To maximize our total, we should pick the k most valuable elements from this group (or all of them if there are fewer than k)

This greedy approach works because the groups don't interfere with each other's selection - we're simply maximizing the value we extract from each limit threshold before it becomes unreachable. By sorting each group and taking the top values, we ensure we're getting the maximum possible contribution from each limit level.

The beauty of this solution is that it transforms a complex activation ordering problem into a simple selection problem: for each limit value lim, select up to lim elements with the highest values.

Learn more about Greedy, Two Pointers, Sorting and Heap (Priority Queue) patterns.

Solution Approach

The implementation follows the greedy strategy we identified:

1. Group elements by limit value: We use a defaultdict(list) to create a dictionary g where each key is a limit value and the corresponding value is a list of all element values that have that limit.

   g = defaultdict(list)
   for v, lim in zip(value, limit):
       g[lim].append(v)

   This groups elements efficiently - for example, if we have value = [2, 5, 3, 4] and limit = [2, 3, 2, 3], we get g = {2: [2, 3], 3: [5, 4]}.

2. Sort and select top values from each group: For each limit group, we sort the values in ascending order and then select the last lim elements (which are the largest).

   for lim, vs in g.items():
       vs.sort()
       ans += sum(vs[-lim:])

   The key insight here is using Python's negative indexing: vs[-lim:] gives us the last lim elements from the sorted list, which are the lim largest values.

   • If a group has more than lim elements, we take exactly the lim largest ones
   • If a group has fewer than lim elements, the slice vs[-lim:] will just return all elements in the group

3. Calculate the total: We accumulate the sum of selected values across all groups to get our maximum possible total.

Time Complexity: O(n log n) where n is the length of the input arrays. The dominant operation is sorting the values within each group.

Space Complexity: O(n) for storing the grouped elements in the dictionary.

The elegance of this solution lies in recognizing that the complex activation rules simplify to a straightforward selection problem when elements are grouped by their limit values.

Example Walkthrough

Let's walk through a concrete example with value = [6, 1, 5, 8] and limit = [2, 3, 2, 3].

Step 1: Group elements by limit value

• Elements with limit = 2: indices 0 and 2, with values [6, 5]
• Elements with limit = 3: indices 1 and 3, with values [1, 8]

This gives us: g = {2: [6, 5], 3: [1, 8]}

Step 2: Process each limit group

For limit = 2 group with values [6, 5]:

• We can activate at most 2 elements before all elements with limit = 2 become permanently inactive
• Sort the values: [5, 6]
• Take the last 2 elements (the 2 largest): [5, 6]
• Contribution to total: 5 + 6 = 11

For limit = 3 group with values [1, 8]:

• We can activate at most 3 elements before all elements with limit = 3 become permanently inactive
• Sort the values: [1, 8]
• Take the last 3 elements, but we only have 2, so take all: [1, 8]
• Contribution to total: 1 + 8 = 9

Step 3: Calculate final result Total maximum value = 11 + 9 = 20

Why this works:

• When we activate 2 elements total, all elements with limit ≤ 2 become permanently inactive
• When we activate 3 elements total, all elements with limit ≤ 3 become permanently inactive
• By taking the most valuable elements from each group before they become inactive, we maximize our total score
• The actual activation order doesn't matter as long as we pick the right elements from each group

Solution Implementation

Time and Space Complexity

Time Complexity: O(n + k * m log m)

• Creating the dictionary and grouping values by limit takes O(n) time, where n is the length of the input arrays
• Iterating through each unique limit group takes O(k) iterations, where k is the number of unique limits
• For each group with m values:
  • Sorting takes O(m log m) time
  • Slicing and summing the last lim elements takes O(min(lim, m)) time
• In the worst case where all values have the same limit, we have one group with n values, giving us O(n log n)
• In the best case where all values have different limits, we have n groups each with 1 value, giving us O(n)
• The overall time complexity is O(n + k * m log m) where m is the average group size, which simplifies to O(n log n) in the worst case

Space Complexity: O(n)

• The defaultdict g stores all n values from the input, requiring O(n) space
• The sorted lists are created in-place modifications of the lists already in the dictionary
• The slicing operation vs[-lim:] creates a temporary list of at most lim elements, but this doesn't exceed O(n) in total
• Overall space complexity is O(n) for storing all input values in the dictionary structure

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Misunderstanding the Deactivation Rule

A critical pitfall is misinterpreting when elements become permanently inactive. Many might think that elements with limit[j] ≤ x become inactive only if they haven't been activated yet. However, the rule states that all such elements become permanently inactive, even those already active. This means they can't contribute to future activations or be activated if not already active.

Wrong interpretation: "Only unactivated elements with limit[j] ≤ x become unavailable"
Correct interpretation: "All elements with limit[j] ≤ x become permanently inactive, regardless of their current state"

2. Incorrect Greedy Selection Within Groups

When selecting elements from each group, a common mistake is to take the first lim elements instead of the largest lim elements. This happens when forgetting to sort or when using incorrect slice notation.

Incorrect approach:

# Taking first lim elements (wrong!)
total_sum += sum(value_list[:limit_value])

Correct approach:

# Sort first, then take the last (largest) lim elements
value_list.sort()
total_sum += sum(value_list[-limit_value:])

3. Off-by-One Error in Understanding Activation Conditions

The activation condition states that element i can be activated when the number of active elements is strictly less than limit[i]. A common mistake is using "less than or equal to" instead.

Wrong: "Can activate when active count ≤ limit[i]"
Correct: "Can activate when active count < limit[i]"

This means if limit[i] = 3, you can activate when there are 0, 1, or 2 active elements, but NOT when there are already 3 active elements.

4. Not Handling Edge Cases with Python Slicing

While Python's negative slicing [-lim:] handles cases where the list has fewer than lim elements gracefully (it returns all elements), developers coming from other languages might worry about this and add unnecessary checks:

Overcomplicated (unnecessary):

if len(value_list) >= limit_value:
    total_sum += sum(value_list[-limit_value:])
else:
    total_sum += sum(value_list)

Simple and correct:

total_sum += sum(value_list[-limit_value:])

Python's slicing automatically handles the case where the slice extends beyond the list boundaries.