1855

1855. Maximum Distance Between a Pair of Values (Medium)

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.

Related Topics:
Two Pointers, Binary Search, Greedy

Solution 1. Binary Search

// OJ: https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int maxDistance(vector<int>& A, vector<int>& B) {
        int ans = 0;
        for (int i = 0; i < A.size(); ++i) {
            int j = upper_bound(begin(B), end(B), A[i], greater()) - begin(B);
            ans = max(ans, max(0, j - 1 - i));
        }
        return ans;
    }
};

Solution 2.

For each A[i]: while A[i] <= B[i + diff], we keep increasing diff. Eventually i + diff will become invalid so diff - 1 is the answer.

// OJ: https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int maxDistance(vector<int>& A, vector<int>& B) {
        int diff = 1, M = A.size(), N = B.size();
        for (int i = 0; i < M && i + diff < N; ++i) {
            while (i + diff < N && A[i] <= B[i + diff]) ++diff;
        }
        return diff - 1;
    }
};