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

Problem Link

https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/

Problem Statement

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.

Approach 1: Two Pointers

As both arrays are sorted, we can use two pointers $i$ and $j$ to iterate $nums1$ and $nums2$ respectively. If $nums1[i] <= nums2[j]$, we can update the answer and increase pointer $j$ by $1$, else increase pointer $i$ by $1$.

Written by @wingkwong

class Solution {
public:
    int maxDistance(vector<int>& nums1, vector<int>& nums2) {
        int n = nums1.size(), m = nums2.size();
        int i = 0, j = 0, ans = 0;
        // nums1 = [55,30,5,4,2], nums2 = [100,20,10,10,5]
        while(i < n && j < m) {
            if(nums1[i] <= nums2[j]) {
                ans = max(ans, j - i); 
                j++;
            } else {
                i++;
            }
        }
        return ans;
    }
};

Approach 2: Lower Bound

Using the same idea as approach 1 but with lower bound.

Written by @wingkwong

class Solution {
public:
    int maxDistance(vector<int>& nums1, vector<int>& nums2) {
        int ans = 0, n = nums1.size(), m = nums2.size();
        for (int i = 0; i < n; i++) {
            auto it = lower_bound(nums2.rbegin(), nums2.rend(), nums1[i]) - nums2.rbegin();
            int j = m - 1 - it;
            ans = max(ans, j - i);
        }
        return ans;
    }
};