2903. Find Indices With Index and Value Difference I

Problem Description

In this problem, you're given an integer array called nums, with a length of n. You're also given two integers: indexDifference and valueDifference. Your task is to find two indices i and j such that both i and j are within the range from 0 to n - 1 and they meet the following criteria:

1. The absolute difference between i and j must be at least indexDifference,
2. The absolute difference between the values at nums[i] and nums[j] must be at least valueDifference.

In terms of the outcome, you need to return an array called answer. This array should consist of the indices [i, j] if such a pair of indices exists. If there are multiple valid pairs, you can return any one of them. If no valid pairs are found, then return [-1, -1]. An interesting point to note is that according to the problem statement, i and j can be the same index, which implies that indexDifference could potentially be 0.

Intuition

The primary intuition behind the solution is the usage of a sliding window technique, combined with the maintenance of the minimum and maximum values within the window. The sliding window is defined by two pointers, i and j, that maintain a distance apart specified by indexDifference. The pointers are used as markers to capture a subarray within nums to check against our two conditions.

At the outset, i starts at the position indexDifference, and j starts at 0. By doing this, the gap between i and j reflects the indexDifference requirement of our problem.

We maintain two variables, mi and mx, to keep track of the indices where the minimum and maximum values are found within our sliding window that ends at the current j index. While sliding i further along the array, we update mi and mx to account for the entry of new values into the window and the exit of old values.

When updating mi and mx, if nums[j] is less than nums[mi], we reassign mi to j, because we have found a new minimum. Conversely, if nums[j] is greater than nums[mx], we reassign mx to j due to identifying a new maximum.

After every movement of i and update of mi and mx, we check our two conditions against nums[i] (the current value at i). If the difference between nums[i] and the value at nums[mi] is greater than or equal to valueDifference, we have found a valid pair [mi, i]. Alternatively, if the difference between the maximum value (nums[mx]) and nums[i] is also greater than or equal to valueDifference, then [mx, i] is a valid pair. In this situation, we immediately return the pair as it meets our requirements.

If we reach the end of the array without finding a pair that satisfies both conditions, we conclude that no such pair exists, and we return the default output [-1, -1].

The approach's efficiency comes from the fact that it avoids checking every possible pair of indices, which would otherwise lead to a less efficient solution with a higher time complexity.

Solution Approach

The solution uses a sliding window approach, which involves moving a fixed-size window across the array to examine sub-sections one at a time. This allows for checking the conditions over smaller segments in a single pass through the array, making the solution more efficient than a brute force approach that would involve examining all possible index pairs.

To implement this technique, the algorithm maintains two pointers: i and j. These pointers define the bounds of the sliding window. The pointer i starts at the index equal to indexDifference while j starts at 0, thus immediately satisfying the condition abs(i - j) >= indexDifference because i - j is initialized to indexDifference.

As the algorithm iterates over the array, starting from i = indexDifference, it keeps track of the indices of the minimum and the maximum values found so far to the left of j. These indices are stored in variables mi and mx, respectively.

1for i in range(indexDifference, len(nums)):
2    j = i - indexDifference

Within the loop, we first check whether the current element at index j changes the minimum or maximum:

1if nums[j] < nums[mi]:
2    mi = j
3if nums[j] > nums[mx]:
4    mx = j

After updating mi and mx, we check if nums[i] differs enough from the minimum or maximum value to satisfy the valueDifference condition:

1if nums[i] - nums[mi] >= valueDifference:
2    return [mi, i]
3if nums[mx] - nums[i] >= valueDifference:
4    return [mx, i]

If either of these checks succeeds, the function immediately returns the corresponding pair of indices, as they meet both prescribed conditions. If the function reaches the end of the array without returning, this means no valid pairs were found, and thus it returns [-1, -1].

In terms of data structures, no additional structures are needed beyond the use of a few variables to keep track of the indices and values encountered. This algorithm is space-efficient because it operates directly on the input array without requiring extra space proportional to the input size.

The choice of a sliding window and keeping track of minimum and maximum values eliminates the need to compare every element with every other element, thereby reducing the time complexity from O(n^2) to O(n), where n is the length of the input array.

Example Walkthrough

Let's walk through an example to illustrate the solution approach described above. Consider the integer array nums = [1, 2, 3, 4, 5], with indexDifference = 3 and valueDifference = 3.

Our task is to find indices i and j such that abs(i - j) >= indexDifference and abs(nums[i] - nums[j]) >= valueDifference.

According to the given solution approach, we initialize the sliding window by setting i to indexDifference and j to 0. This immediately satisfies the first condition as the difference between the indices i = 3 and j = 0 is 3, which is equal to indexDifference.

We then maintain mi and mx to keep track of the minimum and maximum values within the window.

Here's how we proceed step by step:

1. On the first iteration where i = 3:
   • We have j = 0
   • We initiate mi to j since there are no previous values to compare with, and similarly, mx is also initiated to j
   • The elements under consideration are [nums[0], nums[3]] i.e., [1, 4]
   • We see that nums[3] - nums[0] = 4 - 1 = 3, which is equal to valueDifference
   • Since nums[3] - nums[0] fulfills the valueDifference condition, we return [0, 3] as the indices that satisfy both conditions.

In this example, we directly found a pair that met both conditions, and thus we would return [0, 3]. However, if we needed to continue the iteration, we would:

2. Increment i to the next position and decrement j to keep the window size constant while satisfying the indexDifference. If nums[j] changes the minimum or maximum within the new window, update mi or mx accordingly.
3. Check if nums[i] differs enough from nums[mi] or nums[mx] as explained previously.
4. If we find a pair, we return it. If not, we continue iterating until i reaches the end of the array.

If no valid pairs are found by the end of the array, we return [-1, -1] as specified.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code is O(n). This is achieved by iterating over the array once from indexDifference to the length of the array len(nums). Only constant time checks and updates are performed within the loop, leading to a linear time complexity relative to the array's size.

Space Complexity

The space complexity of the code is O(1). No additional space proportional to the input size is used. Only a fixed number of variables mi, mx, and j are used, which occupy constant space regardless of the input array size.

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