Problem Description

In this problem, we are given two non-increasing integer arrays, nums1 and nums2. We define a valid pair as a pair of indices (i, j) such that:

1. i is an index from nums1 and j is an index from nums2.
2. i must be less than or equal to j (i <= j).
3. The value at nums1[i] must be less than the value at nums2[j] (nums1[i] < nums2[j]).

The distance for a valid pair (i, j) is calculated by the difference j - i. The goal of the problem is to return the maximum possible distance for any valid pair of indices. If there are no valid pairs, we should return 0.

Note that an array is non-increasing if each element is less than or equal to the previous element. This implies that the numbers in both nums1 and nums2 are sorted in non-increasing order (from largest to smallest).

Intuition

The intuition behind the solution involves recognizing that if a pair (i, j) is valid, then any pair (i, k) where k < j is also valid due to the non-increasing order of the arrays. This property allows us to use a two-pointer or a binary search approach to efficiently find the maximum distance. We choose the two-pointer approach here as it simplifies the implementation.

We start by initializing two pointers, i and j, at the beginning of nums1 and nums2 respectively. We then iterate through nums1 with i, trying to find the furthest j in nums2 that will form a valid pair. As we go, we keep track of the maximum distance ans.

The crucial observation is that since both nums1 and nums2 are non-increasing, once we find a j for a particular i such that nums1[i] <= nums2[j], we don't need to reset j back to i for the next i. Instead, we can continue from the current j, because if nums1[i] <= nums2[j] and nums1 is non-increasing, then nums1[i + 1] will be less than or equal to nums1[i], and hence also less than or equal to nums2[j]. We keep incrementing j until the condition nums1[i] <= nums2[j] is no longer met.

However, we need to calculate the distance j - i and store the maximum. Since j is incremented in the inner loop until nums1[i] <= nums2[j] is false, at this point j is actually one index past the valid pair, so we compute j - i - 1 to get the correct distance for considering the current index i.

As a result, the algorithm efficiently calculates the maximum distance of any valid pair by scanning through both arrays only once.

Learn more about Greedy, Two Pointers and Binary Search patterns.

Solution Approach

The given reference solution approach indicates using a Binary Search algorithm. However, the provided solution code actually uses a two-pointer technique, which intuitively exploits the sorted nature of the arrays to find the maximum distance. Below, we'll walk through the implementation of the two-pointer solution given in the code.

The code defines a class Solution with a method maxDistance which accepts two arrays, nums1 and nums2.

1. Initialization: We begin by initializing a few variables:

   • m, the length of nums1
   • n, the length of nums2
   • ans, which will keep track of the maximum distance found among valid pairs. It is set to 0 initially.
   • Two pointers i and j, both set to 0, to iterate through nums1 and nums2 respectively.

2. Iterating with Two Pointers: We use a while loop to iterate through nums1 with i as long as i < m (ensuring we don't go out of bounds).

3. Finding the Furthest j: For every index i, we have an inner while loop that seeks the furthest j such that nums1[i] <= nums2[j]. This loop runs as long as j < n. If the condition is met, j is incremented by 1, moving the second pointer ahead in nums2.

4. Tracking the Maximum Distance: Once we are either out of bounds in nums2 or the condition is no longer satisfied, the inner loop breaks and we use the maximum function max(ans, j - i - 1) to update our answer with the largest distance found so far. The -1 is there to correct for the j index that is now at an invalid pair, having just moved past the last valid nums2[j].

5. Incrementing i: After checking and possibly updating the ans with the maximum distance for the current i, we increment i by 1 and repeat the above steps until all elements of nums1 have been considered.

6. Returning the Answer: When the loop terminates, ans will hold the maximum distance of any valid pair (i, j), or it will remain 0 if no valid pairs existed. This value is then returned as the result of the method.

The two-pointer approach is efficient because it leverages the sorted properties of nums1 and nums2 to avoid redundant checks. It doesn't require additional data structures and runs in linear time, as each pointer only passes through its respective array once.

Thus, the implementation uses a methodical two-pointer strategy to measure the maximum distance between valid pairs across two sorted arrays.

Example Walkthrough

Let's walk through an example to illustrate the solution approach in practice.

Suppose we have the following inputs: nums1 = [4, 2, 2] nums2 = [3, 2, 0]

We want to find the maximum possible distance for any valid pair of indices.

1. We start with i = 0 in nums1 and j = 0 in nums2, with ans = 0 as we haven't found any valid pairs yet.

2. In the outer loop, i is now at index 0 and nums1[0] is 4. The inner while loop starts iterating j through nums2. nums2[j] is 3, which is less than nums1[i], so we don't have a valid pair. We continue to the next j without changing ans.

3. Now, j = 1 and nums2[j] is 2. It's still not greater than nums1[i] = 4, so we move to the next j.

4. At j = 2, we have nums2[j] equal to 0, which is also not greater than 4. Having reached the end of nums2 without finding a valid pair for i = 0, we increment i to 1 and start over with j = 0.

5. Now i = 1 and nums1[i] is 2. Starting again from the beginning of nums2, we find that nums2[0] is 3, which is greater than nums1[1]. This is a valid pair (i=1, j=0). Hence, we update ans = max(ans, 0 - 1 - 1), but this gives us -2, which is not an increase over 0, so ans stays at 0.

6. We increment j to see if there is a pair with a larger distance. Now j is 1 and since nums1[i] = 2 is already less than nums2[j] = 2, we stop the inner while loop and update ans. Now ans = max(0, 1 - 1 - 1), which remains 0 as the indices are the same.

7. Finally, we increment i to 2. nums1[i] is still 2, and we try with nums2[j] starting from the beginning. At j = 0, we once again have a valid pair (i=2, j=0) and update ans to max(0, 0 - 2 - 1), giving us -3, which doesn't increase ans. Since all values are non-increasing and there's no greater element in nums2 than 3, we can conclude there's no need to iterate j further for i = 2.

After considering all elements of nums1 and finding no valid pair with a positive distance, the function returns the ans, which in this case remains 0 as there were no valid pairs that satisfied the condition for a positive maximum distance.

This example demonstrates how the two-pointer approach navigates through both arrays, cleverly using the condition of non-increasing order to find the valid pairs and calculating the maximum possible distance.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code is O(m + n). Here, m is the length of nums1 and n is the length of nums2. This is because the two pointers i and j only move forward through their respective arrays, with i incrementing in every outer loop, and j potentially moving forward during each inner loop until nums1[i] <= nums2[j] no longer holds. Since each pointer goes through its respective array at most once, the time complexity is linear with respect to the sizes of the input arrays.

Space Complexity

The space complexity of the code is O(1). No additional space is used that grows with the input size. The variables ans, i, j, m, and n each use a constant amount of space.

Learn more about how to find time and space complexity quickly using problem constraints.