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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.
1#include <stdio.h>
2#include <math.h>
3
4int maxDistance(int** arrays, int arraysSize, int* arraysColSize) {
5 int maxDist = 0;
6 int minVal = arrays[0][0];
7 int maxVal = arrays[0][arraysColSize[0] - 1];
8
9 for (int i = 1; i < arraysSize; i++) {
10 int first = arrays[i][0];
11 int last = arrays[i][arraysColSize[i] - 1];
12
13 // Calculate distances and update maxDist
14 maxDist = fmax(maxDist, fabs(last - minVal));
15 maxDist = fmax(maxDist, fabs(maxVal - first));
16
17 // Update global min and max
18 minVal = fmin(minVal, first);
19 maxVal = fmax(maxVal, last);
20 }
21 return maxDist;
22}
23This solution iterates over the given arrays while keeping track of the global minimum and maximum values seen so far. It utilizes these for calculating potential maximum distances between the current array's element boundaries and previous minima/maxima without redundant calculations.
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.
The brute force solution systematically checks distance calculations across every possible pair of separate arrays, albeit with lesser theoretical efficiency due to the intrinsic nested loop structure causing significant execution time overhead.