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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.
1def maxDistance(nums1, nums2):
2 i, j, max_dist = 0, 0, 0
3 while i < len(nums1) and j < len(nums2):
4 if nums1[i] <= nums2[j]:
5 max_dist = max(max_dist, j - i)
6 j += 1
7 else:
8 i += 1
9 return max_distIn Python, the two-pointer technique is slipped into a function, looping through with pointer `i` on `nums1` and `j` on `nums2` to find valid pairs, updating `max_dist` accordingly.
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).
#include <algorithm>
using namespace std;
int binarySearch(vector<int>& arr, int left, int right, int target) {
while (left <= right) {
int mid = left + (right - left) / 2;
if (arr[mid] < target) {
right = mid - 1;
} else {
left = mid + 1;
}
}
return right;
}
int maxDistance(vector<int>& nums1, vector<int>& nums2) {
int maxDist = 0;
for (int i = 0; i < nums1.size(); ++i) {
int j = binarySearch(nums2, i, nums2.size() - 1, nums1[i]);
if (j >= i) {
maxDist = max(maxDist, j - i);
}
}
return maxDist;
}The C++ version uses an inline `binarySearch` function to determine the maximum valid `j` position for each element in `nums1` by using binary search on `nums2`.