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This method employs two pointers to traverse both arrays simultaneously. Start with `i` at the beginning of `nums1` and `j` at the beginning of `nums2`. Try to find the largest `j` for each `i` while maintaining the condition `nums1[i] <= nums2[j]` and `i <= j`. The distance `j - i` is calculated and the maximum one is stored. Iterate through both arrays by incrementing `j` while ensuring the condition holds true, otherwise increment `i`.
Time Complexity: O(n + m), where n and m are lengths of nums1 and nums2 respectively.
Space Complexity: O(1), only a few variables are used.
1#include <stdio.h>
2int maxDistance(int* nums1, int nums1Size, int* nums2, int nums2Size) {
3 int i = 0, j = 0, maxDist = 0;
4 while (i < nums1Size && j < nums2Size) {
5 if (nums1[i] <= nums2[j]) {
6 if (j - i > maxDist) {
7 maxDist = j - i;
8 }
9 j++;
10 } else {
11 i++;
12 }
13 }
14 return maxDist;
15}The solution uses two pointers initialized at the start of `nums1` and `nums2`. It checks if `nums1[i] <= nums2[j]` and calculates the distance `j-i`. If valid, move `j` to explore a potential longer distance. If invalid, increment `i` to find a valid pair. The largest distance found is stored and returned.
Utilizing binary search on `nums2` for each element in `nums1` can optimize the search for each valid `j` index. This approach leverages the sorted nature of the arrays. For each element in `nums1`, perform a binary search in `nums2` to find the farthest possible valid `j`.
Time Complexity: O(n * log(m)), where n is the length of nums1 and m is the length of nums2.
Space Complexity: O(1).
1
The JavaScript solution uses an embedded `binarySearch` function to optimize the search for a valid `j` index, iterating efficiently over `nums1`.