How do you solve Find Indices With Index and Value Difference II on LeetCode?

Find Indices With Index and Value Difference II (LeetCode #2905) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a sliding window can significantly reduce the number of comparisons needed.

What is the time complexity of Find Indices With Index and Value Difference II?

The optimal solution for Find Indices With Index and Value Difference II runs in O(n) time and uses O(n) space. This complexity is achieved because we only traverse the array once, and the deque operations (adding/removing) are efficient.

What is the brute force solution for Find Indices With Index and Value Difference II?

The brute force approach for Find Indices With Index and Value Difference II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible pair of indices to see if they meet the conditions. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Indices With Index and Value Difference II?

Find Indices With Index and Value Difference II uses the following concepts: Array, Two Pointers. The recommended patterns to study are: Sliding Window, Deque.

What are the interview tips for Find Indices With Index and Value Difference II?

Always clarify the problem constraints and edge cases before diving into coding. Think about how to reduce the number of comparisons — this often leads to better solutions. Practice explaining your thought process as you code; this helps interviewers understand your reasoning.

What are common mistakes when solving Find Indices With Index and Value Difference II?

Not considering the case where i and j can be the same index. Failing to correctly manage the sliding window boundaries.

#2905

Find Indices With Index and Value Difference II

Medium

ArrayTwo PointersSliding WindowDeque

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

We can optimize the search by using a sliding window approach. This allows us to maintain a range of indices and check the minimum and maximum values efficiently.

⚙️

Algorithm

5 steps

1Step 1: Initialize a deque to keep track of indices for the sliding window.
2Step 2: Iterate through each index i from 0 to n-1.
3Step 3: Remove indices from the deque that are out of the valid range (i - indexDifference).
4Step 4: Check the minimum and maximum values in the deque against nums[i] to see if they satisfy the valueDifference condition.
5Step 5: If a valid pair is found, return the indices. If no valid pairs are found by the end, return [-1, -1].

solution.py15 lines

1from collections import deque
2
3def find_indices(nums, indexDifference, valueDifference):
4    n = len(nums)
5    window = deque()
6    for i in range(n):
7        while window and window[0] < i - indexDifference:
8            window.popleft()
9        if window:
10            if abs(nums[i] - nums[window[0]]) >= valueDifference:
11                return [i, window[0]]
12            if abs(nums[i] - nums[window[-1]]) >= valueDifference:
13                return [i, window[-1]]
14        window.append(i)
15    return [-1, -1]

ℹ

Complexity note: This complexity is achieved because we only traverse the array once, and the deque operations (adding/removing) are efficient.

1Using a sliding window can significantly reduce the number of comparisons needed.
2Maintaining a deque allows us to efficiently track the minimum and maximum values within a range.

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