Asked in: Nokia
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All Testcases Passed ✔
Solution
def homework(N, Points, M):
distances = []
for i in range(N):
for j in range(i + 1, N):
// ... rest of solution available after purchaseExplanation
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To solve this problem, first understand the modified definition of distance between two points. Instead of the usual Euclidean or Manhattan distance, the distance here is defined as the minimum of the absolute differences between their x-coordinates and y-coordinates: min(|x1 - x2|, |y1 - y2|). This means the distance between two points depends only on the smaller difference between their horizontal or vertical distances.
Given N unique points, we need to find the M-th smallest distance among all possible pairs of points according to this new distance metric. There are a total of N*(N-1)/2 pairs, so directly computing and sorting all distances is expensive for large N. Hence, the solution must be more efficient.
Start by considering the properties of the distance metric. Since the distance depends on the minimum of the horizontal or vertical difference, focus on how the points align when sorted by their x-coordinates and separately when sorted by their y-coordinates. Points close in either x or y dimension can generate small distances.
To proceed, think about the following key observations:
1. **Pairwise Distances From Sorted Coordinates**: If you sort the points by their x-coordinates, the closest x differences will be among adjacent points in this sorted list. Similarly, sorting by y-coordinates will reveal the closest y differences. Thus, the minimum distances will come from pairs close in either dimension.
2. **Distance Candidates**: The set of candidate distances that can possibly be the M-th smallest will come from differences of x-values or y-values among pairs. Specifically, since the distance is the minimum of |x1 - x2| and |y1 - y2|, each candidate must be one of these differences.
3. **Binary Search on Distance**: Since we want the M-th smallest distance without enumerating all distances, consider applying a binary search on the range of possible distances. The smallest possible distance is 0 (if points share an x or y coordinate) or at least 1 if coordinates differ, and the largest possible distance is limited by the maximum coordinate difference.
4. **Counting Pairs Within Distance Threshold**: For a given candidate distance `d` during binary search, determine how many pairs have distance ≤ `d`. The crux is to efficiently count pairs where min(|x1 - x2|, |y1 - y2|) ≤ d.
To count these:
- First, consider all pairs where |x1 - x2| ≤ d. Since points are sorted by x, this can be done using a two-pointer technique, moving a sliding window over points whose x difference is within `d`.
- Within that window, count pairs whose |y1 - y2| ≤ d. This sub-condition is more challenging because the points within the window may have widely varying y-coordinates.
- To handle this efficiently, consider storing the y-values of points in the current window in a data structure supporting quick range queries (like a balanced BST or segment tree), allowing counting of points whose y-coordinate is within [y_i - d, y_i + d].
5. **Combining Counts**: The count obtained above gives the number of pairs with |x1 - x2| ≤ d and |y1 - y2| ≤ d. However, since the distance is the minimum of these two differences, pairs where |y1 - y2| ≤ d but |x1 - x2| > d should also be counted. So you also need to count pairs where |y1 - y2| ≤ d and min(|x1 - x2|, |y1 - y2|) ≤ d, which simplifies to pairs where either coordinate difference is ≤ d.
This suggests counting pairs in both sorted-by-x and sorted-by-y order and combining results carefully to avoid double counting.
6. **Adjusting the Binary Search**: With a method to count how many pairs have distance ≤ d, perform a binary search over d. For each mid value, compute the count. If count is at least M, you know the M-th smallest distance is at most d, so you move left. Otherwise, move right.
7. **Edge Cases**: Consider cases with points sharing the same x or y coordinate, which result in zero distance. Also consider when M is very small (like 1) or very large (close to total number of pairs). Ensure that indexing and counts are handled carefully to match 1-based indexing of M.
In summary, the approach involves:
- Sorting points by x and by y.
- Using a binary search over possible distances.
- For each candidate distance, efficiently counting pairs whose distance according to the custom metric is ≤ candidate.
- Using data structures or sliding window techniques to achieve counting efficiently.
- Returning the distance where the count reaches the M-th smallest.
This approach avoids enumerating all pairs and uses problem structure for efficient computation.
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