How do you solve Maximum Distance Between a Pair of Values on LeetCode?

Maximum Distance Between a Pair of Values (LeetCode #1855) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Both arrays are sorted in non-increasing order, allowing for efficient searching.

What is the brute force solution for Maximum Distance Between a Pair of Values?

The brute force approach for Maximum Distance Between a Pair of Values is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible pair of indices (i, j) to find valid pairs and calculate their distances. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Distance Between a Pair of Values?

Maximum Distance Between a Pair of Values uses the following concepts: Array, Two Pointers, Binary Search. The recommended patterns to study are: Two Pointers, Sliding Window.

What are the interview tips for Maximum Distance Between a Pair of Values?

Always analyze the properties of the input data; they can lead to more efficient solutions. Practice explaining your thought process clearly, as communication is key in interviews. Consider edge cases and test your solution with them to ensure robustness.

What are common mistakes when solving Maximum Distance Between a Pair of Values?

Not recognizing the non-increasing property of the arrays, leading to inefficient solutions. Failing to handle edge cases where no valid pairs exist.

#1855

Maximum Distance Between a Pair of Values

Medium

ArrayTwo PointersBinary SearchTwo PointersSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

By leveraging the non-increasing property of the arrays, we can use a two-pointer technique to efficiently find the maximum distance without checking every pair.

⚙️

Algorithm

5 steps

1Step 1: Initialize two pointers, i for nums1 and j for nums2, both starting at 0.
2Step 2: While i is less than the length of nums1 and j is less than the length of nums2, check if nums1[i] <= nums2[j].
3Step 3: If valid, calculate the distance j - i and update max_distance. Increment j to check further elements in nums2.
4Step 4: If not valid, increment i to check the next element in nums1.
5Step 5: Return max_distance after the loop.

solution.py12 lines

1# Full working Python code
2
3def maxDistance(nums1, nums2):
4    max_distance = 0
5    i, j = 0, 0
6    while i < len(nums1) and j < len(nums2):
7        if nums1[i] <= nums2[j]:
8            max_distance = max(max_distance, j - i)
9            j += 1
10        else:
11            i += 1
12    return max_distance

ℹ

Complexity note: This complexity is linear because we are only traversing each array once with the two pointers, making it much more efficient than the brute force approach.

1Both arrays are sorted in non-increasing order, allowing for efficient searching.
2Using two pointers helps avoid unnecessary comparisons and reduces time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.