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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.
1var maxDistance = function(nums1, nums2) {
2 let i = 0, j = 0, maxDist = 0;
3 while (i < nums1.length && j < nums2.length) {
4 if (nums1[i] <= nums2[j]) {
5 maxDist = Math.max(maxDist, j - i);
6 j++;
7 } else {
8 i++;
9 }
10 }
11 return maxDist;
12};The JavaScript solution incorporates the two-pointer approach with similar mechanics, using `Math.max` to store the maximal distance.
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).
The Python solution incorporates binary search to quickly locate the farthest valid `j` for each `i`, enhancing performance while maintaining simplicity.