Asked in: Amazon
Image of the Question
All Testcases Passed ✔
Solution
def sortOrders(sorterOrder):
# Write your code here
arr = [(i, idx) for idx,i in enumerate(sorterOrder)]
arr.sort()
// ... rest of solution available after purchaseExplanation
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To approach this problem, it is crucial to understand the process described and how the sorting progresses through each operation.
Step 1: Understand the problem setup
- You have n packages, uniquely identified from 1 to n.
- The packages are arranged in a specific order given by the permutation sorterOrder.
- The goal is to sort packages in the order 1, 2, 3, ..., n (the natural ascending order).
- Initially, sortedCount = 0, meaning no package has been sorted yet.
- Each operation consists of scanning through all packages from left to right.
- During this scan, if a package matches sortedCount + 1 (i.e., the next expected package in the sorted sequence), it gets sorted, and sortedCount increments by 1.
- Packages that don't match the next expected package are ignored.
- The process repeats operations until all packages are sorted.
- The question is: how many full operations (scans through the array) are needed to sort all packages?
Step 2: Analyze what happens in one operation
- Each operation involves scanning the entire list from the beginning to the end.
- During this scan, the manager only looks to sort the next expected package in the natural order.
- If a package matches sortedCount + 1, it is "sorted," so sortedCount increases by one.
- Potentially, multiple packages can be sorted in a single operation if they appear in increasing order corresponding to the next expected package.
- If the packages needed for sorting are scattered, many operations may be needed.
Step 3: Key insight - longest consecutive subsequence
- The process resembles sorting by incrementally finding the next expected package.
- If packages are arranged in a way that some packages appear in ascending order according to their position in the sorting sequence, they can be sorted consecutively in one operation.
- If the next expected package appears far to the right, it requires multiple operations before it is reached again.
- Thus, the problem boils down to how the packages are arranged relative to the natural order and how "long" the longest consecutive increasing subsequence in terms of expected sorting order is.
Step 4: Mapping package identifiers to their positions
- To understand how many operations are needed, consider mapping each package's identifier to its position in sorterOrder.
- For example, position[package_id] = index where package_id occurs in sorterOrder.
- Now, consider the natural order sequence 1, 2, 3, ..., n.
- For the packages to be sorted with minimal operations, the positions of consecutive packages in sorterOrder should ideally be increasing.
- This means that if position[1] < position[2] < position[3] < ... < position[k], these k packages can be sorted consecutively in one operation without having to restart.
Step 5: Finding the longest consecutive ascending subsequence in positions
- The key is to find the longest subsequence where positions of consecutive package IDs increase by exactly 1 or simply increase in order.
- More specifically, the problem requires identifying the longest consecutive segment of the natural sequence where the positions are strictly increasing.
- This longest consecutive run tells us how many packages can be sorted in one operation without interruption.
Step 6: Calculating the minimum number of operations
- Once we find the length of the longest consecutive ascending subsequence (let’s call it longestRun), it indicates the maximum number of packages sorted consecutively in one operation.
- Since total packages are n, the number of operations required would be roughly:
total_operations = n - longestRun + 1
- This is because in the best case, sorting the longest run can be done in one operation, and the rest of the packages need their operations as well.
Step 7: Algorithm outline
- Step 1: Create an array position of size n, where position[i] represents the index of package i+1 in sorterOrder.
- Step 2: Iterate through packages 1 to n in natural order.
- Step 3: Track the length of consecutive positions where position[i+1] > position[i].
- Step 4: Keep updating longestRun to the maximum length of these consecutive increasing runs.
- Step 5: Compute the final answer as n - longestRun + 1.
Step 8: Validate with the example
- Using the example sorterOrder = [5, 3, 4, 1, 2]:
- Map package IDs to positions:
1 → position 3
2 → position 4
3 → position 1
4 → position 2
5 → position 0
- Check consecutive positions for packages in natural order:
From 1 to 2: positions 3 → 4 (increasing by 1) → good consecutive run length 2
From 2 to 3: positions 4 → 1 (not increasing) → run breaks
From 3 to 4: positions 1 → 2 (increasing by 1) → run length 2
From 4 to 5: positions 2 → 0 (not increasing)
- The longest run is 2 (either for packages 1-2 or 3-4).
- Total operations = n - longestRun + 1 = 5 - 2 + 1 = 4.
Step 9: Edge cases and complexity
- For sorted sequences (sorterOrder already sorted), the longestRun = n, so only 1 operation is needed.
- For reverse order, longestRun is 1, so n operations are needed.
- The algorithm runs in O(n) since it requires just one pass to map positions and one pass to find the longest consecutive increasing run.
Step 10: Summary
- The problem reduces to identifying the longest consecutive ascending subsequence of positions in sorterOrder mapped by natural package IDs.
- The minimum number of operations is linked to how many packages can be sorted in one continuous pass.
- Efficient mapping and tracking consecutive runs enable a linear time solution, suitable for large inputs.
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