Asked in: AMAZON
Image of the Question
All Testcases Passed ✔
Solution
def applicationPairs(deviceCapacity, foregroundAppList, backgroundAppList):
# Write your code here
foregroundAppList.sort(key = lambda x: (x[1],x[0]))
backgroundAppList.sort(key = lambda x: (x[1],x[0]))
// ... rest of solution available after purchaseExplanation
```
This problem focuses on pairing foreground and background applications to maximize the utilization of a device's memory capacity without exceeding it. The goal is to find all pairs whose combined memory usage is as large as possible but still less than or equal to the device capacity. Here is how you should think about the problem and approach it:
---
### Understanding the Problem
We have three inputs:
- A device capacity (an integer), representing the maximum memory the device can use.
- A list of foreground applications, each defined by a unique ID and the memory it requires.
- A list of background applications, similarly defined.
We must pair exactly one foreground and one background application such that their total memory usage is as close as possible to the device capacity without going over. If multiple pairs achieve the same maximum usage, all those pairs must be returned.
If no pair can fit within the capacity, return a list containing a single empty pair.
---
### Step 1: Identify the problem as a variation of the "Two Sum" or "Pair Sum" problem with constraints
The problem essentially asks:
From two lists of memory usages, find pairs whose sums are maximized but do not exceed a given target (device capacity).
The twist is that we also need to return all pairs that reach this maximum sum and the lists represent applications with IDs, so we must keep track of IDs.
---
### Step 2: Naive approach - brute force
One straightforward way is to:
- Iterate through each foreground application.
- For each foreground app, iterate through all background apps.
- Calculate the sum of their memory requirements.
- Track the sums that are less than or equal to the device capacity.
- Maintain the maximum sum encountered and store all pairs achieving this sum.
**Drawbacks:**
- This approach has a time complexity of O(F * B), where F and B are the lengths of foreground and background lists respectively.
- For large inputs, this approach can be computationally expensive.
---
### Step 3: Optimized approach using sorting and two pointers
To improve efficiency, you can use sorting and a two-pointer technique.
- First, sort the foreground apps by their memory usage.
- Sort the background apps by their memory usage as well.
Then:
- Initialize two pointers: one starting at the beginning of the foreground list (lowest memory usage), and the other starting at the end of the background list (highest memory usage).
- Calculate the sum of the pair at these pointers.
- If the sum is greater than the device capacity, move the pointer in the background list backward (to reduce sum).
- If the sum is less than or equal to the capacity:
- Check if the sum is greater than the current maximum sum found:
- If yes, update the maximum sum and reset the results list with this pair.
- If equal, add the pair to the results list.
- Move the pointer in the foreground list forward (to potentially increase sum).
Continue this process until one of the pointers goes out of bounds.
This technique leverages the sorted order to efficiently navigate possible sums and avoids checking every pair explicitly.
---
### Step 4: Handling multiple pairs with same memory usage
It is possible that multiple foreground or background applications have the same memory usage. When you find a pair with memory sum equal to the current maximum:
- You must check for all other applications with the same memory values on either side.
- For example, if multiple foreground apps have the same memory usage, and the background app pointer is fixed, add all these pairs.
- Similarly, if multiple background apps have the same memory usage, add those pairs as well.
This may require scanning neighbors with the same memory values on either list to find all combinations contributing to the same optimal sum.
---
### Step 5: Edge cases and no solution scenario
- If no pair's sum is less than or equal to the device capacity, return a list containing an empty pair.
- If device capacity is very small and no single foreground or background app fits, the same no solution condition applies.
- Consider inputs where either list is empty — then no valid pair exists.
---
### Step 6: Memory and time complexity considerations
- Sorting both lists takes O(F log F + B log B).
- The two-pointer traversal takes O(F + B).
- Collecting pairs with identical values may require additional linear scans but overall stays efficient.
---
### Step 7: Summary of the approach
1. Sort foreground and background applications by memory usage.
2. Use two pointers (start of foreground, end of background) to find sums closest to capacity without exceeding it.
3. Track maximum sum found and store pairs corresponding to this sum.
4. Handle duplicates in memory usage to add all valid pairs.
5. Return the list of optimal pairs, or an empty pair if none fit.
---
### Additional thoughts
- Using appropriate data structures to group or index applications by memory usage can help in efficiently finding duplicates.
- While traversing with two pointers, careful bookkeeping is required to avoid missing pairs with identical memory usage.
- This problem resembles a variation of the "knapsack" problem but simplified to exactly two items and a fixed capacity, allowing efficient pairing strategies instead of full DP.
---
By following this direction, you will develop an efficient algorithm that produces all optimal foreground/background application pairs that best utilize device memory.
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