1855 Maximum Distance Between a Pair of Values

You are given two non-increasing 0-indexed integer arrays nums1​​​​​​ and nums2​​​​​​.

A pair of indices (i, j), where 0 <= i < nums1.length and 0 <= j < nums2.length, is valid if both i <= j and nums1[i] <= nums2[j]. The distance of the pair is j - i​​​​.

Return the maximum distance of any valid pair (i, j). If there are no valid pairs, return 0.

An array arr is non-increasing if arr[i-1] >= arr[i] for every 1 <= i < arr.length.

Example 1:

Input: nums1 = [55,30,5,4,2], nums2 = [100,20,10,10,5]
Output: 2
Explanation: The valid pairs are (0,0), (2,2), (2,3), (2,4), (3,3), (3,4), and (4,4).
The maximum distance is 2 with pair (2,4).

Example 2:

Input: nums1 = [2,2,2], nums2 = [10,10,1]
Output: 1
Explanation: The valid pairs are (0,0), (0,1), and (1,1).
The maximum distance is 1 with pair (0,1).

Example 3:

Input: nums1 = [30,29,19,5], nums2 = [25,25,25,25,25]
Output: 2
Explanation: The valid pairs are (2,2), (2,3), (2,4), (3,3), and (3,4).
The maximum distance is 2 with pair (2,4).

Constraints:

• 1 <= nums1.length, nums2.length <= 105

• 1 <= nums1[i], nums2[j] <= 105

• Both nums1 and nums2 are non-increasing.

这个题，很经典的双指针，而且有点2sum的感觉，只是这里的判断条件不是sum == target，而是i <= j && nums1[i] <= nums2[j]。如果符合条件，我们算max，如果不符合的时候，我们移动指针。因为条件限制了i <= j，然后max是算j - i，所以j一定要比i大，不符合条件，同时向右。然后还有一种情况是，i <= j, j到了结尾了。这种情况也不需要特判，因为j - i求的是max，就算继续移动i向右也不可能得到更大的值。所以两个条件都归纳到else里了。T:O(n + m)，S: O(1)

public int maxDistance(int[] nums1, int[] nums2) {
    if (nums1 == null || nums2 == null || 
        nums1.length == 0 || nums2.length == 0) {
        return -1;
    }

    int max = 0;
    int i = 0;
    int j = 0;
    while (i < nums1.length && j < nums2.length) {
        if (i <= j && nums1[i] <= nums2[j]) {
            max = Math.max(j - i, max);
            j++;
        } else {
            i++;
            j++;
        }
    }        

    return max;
}