#2903

Easy

auto_awesomeTwo-pointer scanning with invariant tracking

LeetCode Problem Workspace

Find Indices With Index and Value Difference I

Find two indices in an array where their difference in both index and value meets specified thresholds.

Problem Statement

You are given a 0-indexed integer array nums of length n, an integer indexDifference, and an integer valueDifference.

Your task is to find two indices i and j, both in the range [0, n - 1], such that the absolute difference in indices satisfies abs(i - j) >= indexDifference and the absolute difference in values satisfies abs(nums[i] - nums[j]) >= valueDifference. Return the indices as an array [i, j] if found, or [-1, -1] if no valid pair exists.

Examples

Example 1

Input: nums = [5,1,4,1], indexDifference = 2, valueDifference = 4

Output: [0,3]

In this example, i = 0 and j = 3 can be selected. abs(0 - 3) >= 2 and abs(nums[0] - nums[3]) >= 4. Hence, a valid answer is [0,3]. [3,0] is also a valid answer.

Example 2

Input: nums = [2,1], indexDifference = 0, valueDifference = 0

Output: [0,0]

In this example, i = 0 and j = 0 can be selected. abs(0 - 0) >= 0 and abs(nums[0] - nums[0]) >= 0. Hence, a valid answer is [0,0]. Other valid answers are [0,1], [1,0], and [1,1].

Example 3

Input: nums = [1,2,3], indexDifference = 2, valueDifference = 4

Output: [-1,-1]

In this example, it can be shown that it is impossible to find two indices that satisfy both conditions. Hence, [-1,-1] is returned.

Constraints

1 <= n == nums.length <= 100
0 <= nums[i] <= 50
0 <= indexDifference <= 100
0 <= valueDifference <= 50

Solution Approach

Brute-force Approach

One simple solution is to use a brute-force approach where we compare every pair of indices to check if they satisfy the given conditions. This will involve two nested loops to check all possible pairs. While this is a valid solution, its time complexity can be high, especially for large arrays.

Two-pointer Approach

The optimal approach uses a two-pointer scanning technique. One pointer traverses the array while maintaining a valid window of indices that satisfy the conditions. This reduces unnecessary comparisons and allows for a more efficient solution compared to brute-force.

Invariant Tracking

By tracking the invariants of both index and value differences as we move through the array, we can quickly identify pairs that meet the requirements. This allows for skipping over many invalid pairs and further optimizes the two-pointer approach.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The brute-force solution has a time complexity of O(n^2), where n is the length of the array, since we need to check all pairs. The two-pointer approach can reduce this to O(n), making it more efficient for larger inputs. Both approaches require O(1) additional space, assuming the input array is not modified.

What Interviewers Usually Probe

The candidate demonstrates familiarity with both brute-force and optimized solutions.
The candidate considers edge cases such as arrays with small lengths or when no solution exists.
The candidate may discuss trade-offs between clarity and efficiency in choosing the solution.

Common Pitfalls or Variants

Common pitfalls

Relying too heavily on the brute-force approach without considering optimization strategies like two-pointer scanning.
Forgetting to handle edge cases like when the array is very small or when no valid pair can be found.
Incorrectly assuming that the two-pointer approach always works for every problem; ensuring correct invariant tracking is crucial.

Follow-up variants

What if there is a constraint that both indexDifference and valueDifference are always 0?
How would the problem change if the array is sorted?
What happens if nums contains duplicate values?

FAQ

What is the time complexity of the brute-force solution for this problem?

The brute-force solution has a time complexity of O(n^2), as it involves checking all pairs of indices in the array.

How can I optimize the brute-force approach for this problem?

You can optimize the brute-force approach by using a two-pointer technique to reduce unnecessary comparisons and improve efficiency.

How does two-pointer scanning apply to this problem?

Two-pointer scanning allows you to reduce the search space by maintaining a valid range of indices that satisfy the difference conditions.

What happens if no valid indices are found in the array?

If no valid indices are found, return [-1, -1] as the output, indicating no solution exists.

What is the pattern behind this problem's solution?

This problem relies on two-pointer scanning with invariant tracking to efficiently find a pair of indices that satisfy the required differences.

terminal

Solution

Solution 1: Two Pointers + Maintaining Maximum and Minimum Values

We use two pointers $i$ and $j$ to maintain a sliding window with a gap of $indexDifference$, where $j$ and $i$ point to the left and right boundaries of the window, respectively. Initially, $i$ points to $indexDifference$, and $j` points to $0$.

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class Solution:
    def findIndices(
        self, nums: List[int], indexDifference: int, valueDifference: int
    ) -> List[int]:
        mi = mx = 0
        for i in range(indexDifference, len(nums)):
            j = i - indexDifference
            if nums[j] < nums[mi]:
                mi = j
            if nums[j] > nums[mx]:
                mx = j
            if nums[i] - nums[mi] >= valueDifference:
                return [mi, i]
            if nums[mx] - nums[i] >= valueDifference:
                return [mx, i]
        return [-1, -1]

Continue Practicing

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