Maximum Distance Between a Pair of Values

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).

Constraints:

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

Approaches

Checkout 3 different approaches to solve Maximum Distance Between a Pair of Values. Click on different approaches to view the approach and algorithm in detail.

Iteration with Binary Search

This approach improves upon the brute-force method by optimizing the search for the index j. For each index i in nums1, instead of linearly scanning nums2, we use binary search to find the largest valid index j. This is possible because nums2 is a non-increasing (sorted) array.

Algorithm

Initialize maxDistance = 0.
Iterate i from 0 to nums1.length - 1.
For each i, perform a binary search on nums2 in the range [i, nums2.length - 1] for the value nums1[i].
The goal of the binary search is to find the largest index j such that nums2[j] >= nums1[i].
Let this index be found_j.
If a valid found_j exists (i.e., it's not -1), update maxDistance = max(maxDistance, found_j - i).
Return maxDistance.

We can optimize the inner loop of the brute-force approach. For a fixed i, we want to find the largest j such that j >= i and nums2[j] >= nums1[i]. Since nums2 is non-increasing, the values nums2[j] that are greater than or equal to nums1[i] will appear at the beginning of the array. We can use binary search to efficiently find the rightmost index j that satisfies this condition.

For each i from 0 to nums1.length - 1, we perform a binary search on the subarray nums2 starting from index i. The search aims to find the largest j where nums2[j] >= nums1[i]. If such a j is found, we calculate j - i and update our maxDistance.

class Solution {
    public int maxDistance(int[] nums1, int[] nums2) {
        int maxDistance = 0;
        int n = nums1.length;
        int m = nums2.length;
        for (int i = 0; i < n; i++) {
            // Binary search for the largest j >= i with nums2[j] >= nums1[i]
            int low = i;
            int high = m - 1;
            int best_j = -1;
            
            while (low <= high) {
                int mid = low + (high - low) / 2;
                if (nums2[mid] >= nums1[i]) {
                    // This is a potential answer, try to find a larger j
                    best_j = mid;
                    low = mid + 1;
                } else {
                    // nums2[mid] is too small, need to search in the left part
                    high = mid - 1;
                }
            }
            
            if (best_j != -1) {
                maxDistance = Math.max(maxDistance, best_j - i);
            }
        }
        return maxDistance;
    }
}

Complexity Analysis

Time Complexity: O(N * log M), where N is the length of `nums1` and M is the length of `nums2`. The outer loop runs N times, and each binary search takes O(log M) time.Space Complexity: O(1), as we only use a few variables for the loops and binary search.

Pros and Cons

Pros:

Significantly faster than the brute-force approach.
Passes the time limits for the given constraints.

Cons:

More complex to implement correctly than the brute-force approach.
Not as efficient as the optimal two-pointer solution.

Code Solutions

Checking out 4 solutions in different languages for Maximum Distance Between a Pair of Values. Click on different languages to view the code.

/** * @param {number[]} nums1 * @param {number[]} nums2 * @return {number} */ var maxDistance =
  function (nums1, nums2) {
    let ans = 0;
    let m = nums1.length;
    let n = nums2.length;
    for (let i = 0; i < m; ++i) {
      let left = i;
      let right = n - 1;
      while (left < right) {
        const mid = (left + right + 1) >> 1;
        if (nums2[mid] >= nums1[i]) {
          left = mid;
        } else {
          right = mid - 1;
        }
      }
      ans = Math.max(ans, left - i);
    }
    return ans;
  };

Video Solution

Watch the video walkthrough for Maximum Distance Between a Pair of Values

Algorithms:

Binary Search

Patterns:

Two Pointers

Data Structures:

Array