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 <= 1e51 <= nums1[i], nums2[j] <= 1e5- Both
nums1andnums2are non-increasing.
Approach 1: Two Pointers
As both arrays are sorted, we can use two pointers and to iterate and respectively. If , we can update the answer and increase pointer by , else increase pointer by .
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.
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;
}
};