Asked in: AMAZON
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All Testcases Passed ✔
Solution
def getNumberOfOperations(packageSequence):
array = [(i, index) for index, i in enumerate(packageSequence)]
array.sort()
last = float('inf')
// ... rest of solution available after purchaseExplanation
```
To approach this problem, you need to simulate the sorting process described, but doing so naively by iterating over the entire sequence multiple times would be inefficient and impractical for large input sizes (up to 10^6 elements). Instead, you should think about the pattern and behavior of the sorting operations and how to deduce the minimum number of passes required without explicitly simulating each full scan.
Here is a step-by-step thought process to guide your approach:
1. **Understanding the process:**
- You have a permutation of package identifiers from 1 to n.
- Initially, none of the packages are sorted, so sortedCount = 0.
- In each operation (or pass), you scan the array left to right.
- Whenever you encounter the package with identifier `sortedCount + 1`, you increment sortedCount (meaning you have successfully sorted that package).
- You skip over any packages that are not the next expected one.
- You repeat these passes until all packages (from 1 to n) are sorted.
2. **Observing the pattern:**
- Sorting each package requires that you encounter them in the order 1, 2, 3, ..., n.
- However, the packages are placed in some arbitrary order.
- During each pass, you only can "catch" the next expected package (sortedCount + 1).
- If the next expected package appears before any previously unsorted expected package, you can pick it up immediately.
- If some packages appear out of order, you will have to wait until subsequent passes.
3. **Key insight - continuity of sorted package sequence:**
- Think about the sequence and the longest segment where packages are already in the expected order when scanning the array.
- For example, if the package with identifier 1 appears before package 2, which appears before package 3 in the sequence, these can be collected in one operation.
- But if the package with identifier 2 appears before package 1, you cannot pick 2 before 1, so you will need more passes.
4. **Relation to longest increasing subsequence (LIS):**
- The problem relates closely to the concept of subsequences, but with a twist: the order you can pick packages is fixed and must be contiguous and incremental by exactly 1 each time.
- However, you don’t need to find the LIS itself. Instead, you need to know how many "breaks" or "discontinuities" in the expected order exist in the permutation.
5. **Using positions to analyze the sequence:**
- Create an array or mapping `pos` where `pos[i]` is the index of package `i` in `packageSequence`.
- This helps because you can easily check the relative positions of packages in the input.
6. **Identifying the longest chain of consecutive packages placed in ascending order by their positions:**
- If for packages i and i+1, `pos[i+1] > pos[i]`, then package i+1 appears after package i in the sequence, which means you can pick them in one go in order.
- Otherwise, if `pos[i+1] < pos[i]`, the order breaks, indicating that the package i+1 appears before package i, forcing a new operation (pass) to collect it.
7. **Counting operations:**
- Initialize the count of operations to 1 because you need at least one operation.
- Iterate from i = 1 to n-1:
- If `pos[i+1] < pos[i]`, it means the ordering breaks.
- Every such break means you must start a new operation.
- So, increment the count of operations.
8. **Why this works:**
- Each operation collects a strictly increasing sequence of packages by their position in the input.
- When the position sequence stops increasing, a new operation is needed to collect the next set of packages.
- Hence, the minimum number of operations equals the number of these ascending segments in the position array.
9. **Complexity considerations:**
- You only need to do one pass over the array to build `pos`.
- Then another pass through `pos` to count breaks.
- Both steps are O(n), which is efficient for input size up to 10^6.
10. **Example:**
- For packageSequence = [5, 3, 4, 1, 2]
- pos[1] = 3, pos[2] = 4, pos[3] = 1, pos[4] = 2, pos[5] = 0 (assuming zero-based indexing)
- Now, check the sequence pos[1], pos[2], ..., pos[5]:
- pos[2] (4) > pos[1] (3) → OK
- pos[3] (1) < pos[2] (4) → break → operation++ (now 2)
- pos[4] (2) > pos[3] (1) → OK
- pos[5] (0) < pos[4] (2) → break → operation++ (now 3)
- Total operations = 3
11. **Summary:**
- The problem reduces to finding how many increasing sequences exist when mapping package identifiers to their positions.
- The count of operations is the number of these sequences.
- This insight allows you to solve the problem efficiently without explicit simulation of each operation.
By following this approach, you focus on the inherent order of packages and their positions to determine the minimum number of operations needed, optimizing the solution for large inputs and meeting the problem constraints.
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