Asked in: UBER
Image of the Question
All Testcases Passed ✔
Solution
def solvecount(left, right):
if not left or not right:
return 0, []
n1 = len(left)
// ... rest of solution available after purchaseExplanation
```
To analyze the efficiency and behavior of the given sorting algorithm and determine the number of swaps it performs, start by understanding exactly what the algorithm does in each step:
1. The algorithm repeatedly searches for the "smallest" pair of indices (i, j) such that i < j and arr[i] > arr[j].
- "Smallest" here means the first such pair encountered when scanning from left to right and, for each i, scanning j from i+1 onward. This lexicographical ordering means that among all pairs where arr[i] > arr[j], the algorithm always chooses the pair with the smallest i, and among those with the same i, the smallest j.
2. After finding this pair, it swaps arr[i] and arr[j].
3. It continues this process until no such pair remains, meaning the array is sorted.
---
### Step 1: Recognize What the Algorithm Is Doing Conceptually
This algorithm is a form of sorting that at each step identifies the first inversion pair in the array — that is, the earliest pair (in lexicographical order of indices) where the order is incorrect — and fixes it by swapping the two elements involved. Because it always picks the first inversion to fix, it "bubbles" smaller elements leftwards in the array one swap at a time.
This contrasts with standard sorting methods that might look globally at all pairs, or bubble sort which swaps adjacent elements. Here, the swap may be between any two elements, but the first inversion found is always resolved before proceeding.
---
### Step 2: Understanding the Inversions and Their Resolution
- An inversion is a pair (i, j) such that i < j and arr[i] > arr[j].
- The total number of inversions in the array represents how far the array is from being sorted.
- Each swap removes at least one inversion but can create or move others.
- Since the algorithm fixes inversions in lex order, it consistently moves the smaller elements earlier in the array.
---
### Step 3: Link the Number of Swaps to Inversions
- It might seem intuitive that the total number of swaps equals the number of inversions initially present.
- But because the swaps can be between non-adjacent elements, and because each swap changes the array configuration, counting inversions at each step is not trivial.
- However, a crucial observation is that the algorithm always fixes the leftmost inversion first. So each swap places a smaller element closer to its correct position.
- This effectively means each swap decreases the total number of inversions by at least one.
- Because elements are distinct and the process is deterministic, the number of swaps is exactly equal to the number of inversions in the initial array.
---
### Step 4: Counting Inversions Efficiently
- Directly counting inversions by checking all pairs would be O(n^2), which is too slow for n up to 10^5.
- Use a well-known technique like a Fenwick tree (Binary Indexed Tree) or a Segment Tree to count the number of inversions efficiently in O(n log n).
- Alternatively, merge sort can be adapted to count inversions during the merge step.
- This approach counts how many elements to the right of each element are smaller than it.
---
### Step 5: Translate the Problem to Counting Inversions
- Since each swap resolves the leftmost inversion, and the process continues until no inversions remain, the total number of swaps performed is equal to the initial number of inversions.
- This means the algorithm’s swap count is exactly the inversion count of the original array.
---
### Step 6: Implications and Algorithm Outline
- To solve the problem efficiently:
- Compute the number of inversions in the input array.
- Return this count as the number of swaps performed by the given sorting algorithm.
---
### Step 7: Why Does This Work?
- Each swap eliminates at least one inversion — the inversion between the chosen pair (i, j).
- Because the algorithm picks the smallest i first, it ensures the inversion that is fixed is the leftmost inversion.
- After swapping, the leftmost inversion involving that i disappears.
- The process cannot create inversions involving a smaller index than i, thus ensuring progress.
- The overall effect is that each swap corresponds exactly to removing one inversion.
- Eventually, when all inversions are removed, the array is sorted.
---
### Step 8: Example Walkthrough
- For example, with arr = [5, 1, 4, 2], count the inversions:
- (5, 1), (5, 4), (5, 2), (4, 2) → 4 inversions.
- The algorithm does 4 swaps, consistent with the inversion count.
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### Step 9: Summary
- The key to solving this problem efficiently lies in recognizing the connection between the algorithm’s swaps and the concept of inversions.
- Counting inversions is a classical problem with well-known O(n log n) solutions.
- Once the inversion count is known, that number is exactly the number of swaps the algorithm performs.
- This insight avoids simulating the entire algorithm, which would be too slow for large n.
- Thus, the approach is to:
1. Count the initial number of inversions in the array.
2. Return that count as the number of swaps.
This method provides an efficient and elegant solution to the problem.
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