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One simple yet effective way to tackle this problem is to track the minimum and maximum values as we iterate over the arrays, since each array is sorted in ascending order. We initialize the maximum distance as zero. For each array except the first one, calculate the possible distance using the current array's first element with the global maximum, and the current array's last element with the global minimum. Update the global minimum and maximum as you traverse each array. This approach ensures all distance calculations only consider meaningful pairs from different arrays and efficiently computes the result in a single pass.
Time Complexity: O(m) where m is the number of arrays. Space Complexity: O(1) as we are only using constant extra space.
1using System;
2using System.Collections.Generic;
3
4public class Solution {
5 public int MaxDistance(IList<IList<int>> arrays) {
6 int maxDist = 0;
7 int minVal = arrays[0][0];
8 int maxVal = arrays[0][arrays[0].Count - 1];
9
10 for (int i = 1; i < arrays.Count; i++) {
11 int first = arrays[i][0];
12 int last = arrays[i][arrays[i].Count - 1];
13 maxDist = Math.Max(maxDist, Math.Abs(last - minVal));
maxDist = Math.Max(maxDist, Math.Abs(maxVal - first));
minVal = Math.Min(minVal, first);
maxVal = Math.Max(maxVal, last);
}
return maxDist;
}
}This C# solution follows similar logic to other implementations, employing List<> and index operations to compute required global boundaries as computations ensue through the List collection.
A naive brute force approach would involve iterating over each pair of arrays and comparing all possible combinations of elements from both arrays. While this method ensures completeness of coverage for all potential maximum distance computations, it results in a less efficient O(m2 * n) complexity where n is the average number of elements in one array. It serves as a solid baseline comparison for the more efficient approach.
Time Complexity: O(m2 * n) where m is the number of arrays and n is the average number of elements in each array. Space Complexity: O(1), constant space aside from an output variable.
#include <cmath>
using namespace std;
int maxDistance(vector<vector<int>>& arrays) {
int maxDist = 0;
for (int i = 0; i < arrays.size(); i++) {
for (int j = i + 1; j < arrays.size(); j++) {
int dist1 = abs(arrays[i][0] - arrays[j].back());
int dist2 = abs(arrays[j][0] - arrays[i].back());
maxDist = max(maxDist, max(dist1, dist2));
}
}
return maxDist;
}The brute force approach checks each viable pair of arrays, calculating and updating the potential maximum distance as each computation occurs, reinforcing exhaustive checking at the expense of greater computation volume.