leetcode_5751

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). Example 4:

Input: nums1 = [5,4], nums2 = [3,2] Output: 0 Explanation: There are no valid pairs, so return 0.

Constraints:

1 <= nums1.length <= 105 1 <= nums2.length <= 105 1 <= nums1[i], nums2[j] <= 105 Both nums1 and nums2 are non-increasing.

Solutions

1. two pointers O(n)

class Solution {
public:
    int maxDistance(vector<int>& nums1, vector<int>& nums2) {
        int i = 0, j = 0, res = 0;
        while (j < nums2.size()) {
            int n2 = nums2[j];
            while (i < nums1.size() && nums1[i] > n2) {
                i++;
            }
            if (i <= j && i < nums1.size())
                res = max(res, j - i);
            j++;
        }
        return res;
    }
};

1. binary search O(nlog(n))