Formatted question description: https://leetcode.ca/all/1855.html

1855. Maximum Distance Between a Pair of Values

Level

Medium

Description

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 <= 10^5
• 1 <= nums2.length <= 10^5
• 1 <= nums1[i], nums2[j] <= 10^5
• Both nums1 and nums2 are non-increasing.

Solution

Use two pointers. Loop over index2 from nums2.length - 1 to 0. For each index2, find the smallest index1 such that nums1[index1] <= nums2[index2] and calculate index2 - index1 as the distance. Since the distance is at least 0, if index1 > index2, the distance is negative and will never exceed 0. Only the non-negative instances are stored to maintain the maximum distance.

• Java
• C++
• Python

• class Solution {
      public int maxDistance(int[] nums1, int[] nums2) {
          int maxDistance = 0;
          int length1 = nums1.length, length2 = nums2.length;
          int index1 = length1 - 1, index2 = length2 - 1;
          while (index1 >= 0 && index2 >= 0) {
              while (index1 >= 0 && nums1[index1] <= nums2[index2]) {
                  maxDistance = Math.max(maxDistance, index2 - index1);
                  index1--;
              }
              index2--;
          }
          return maxDistance;
      }
  }
  

• // OJ: https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/
  // 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;
      }
  };
  

• # 1855. Maximum Distance Between a Pair of Values
  # https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/
  
  class Solution:
      def maxDistance(self, nums1: List[int], nums2: List[int]) -> int:
      
          n1, n2 = len(nums1), len(nums2)
          res = 0
          i = j = 0
          
          while i < n1 or j < n2:
              if nums1[i] <= nums2[j]:
                  res = max(res, j - i)
                  if j < n2 - 1:
                      j += 1
                  elif i < n1 - 1:
                      i += 1
                  else:
                      break
              else:
                  if i < n1 - 1:
                      i += 1
                  elif j < n2 - 1:
                      j += 1
                  else:
                      break
          
          return res
  
  

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