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

Find Indices With Index and Value Difference I (LeetCode #2903) 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 HashMap allows for efficient lookups of indices.

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

The optimal solution for Find Indices With Index and Value Difference I runs in O(n) time and uses O(n) space. This complexity is achieved by using a HashMap to store indices, allowing us to check conditions in constant time as we iterate through the array.

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

The brute force approach for Find Indices With Index and Value Difference I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of indices in the array to see if they meet the given conditions. This is straightforward but can be slow for larger 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 I?

Find Indices With Index and Value Difference I uses the following concepts: Array, Two Pointers. The recommended patterns to study are: Hash Map, Array.

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

Clarify the constraints and edge cases with the interviewer. Discuss the brute force approach before jumping to optimizations. Practice explaining your thought process while coding.

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

Not considering edge cases where indexDifference or valueDifference is zero. Failing to check both conditions before returning indices.

#2903

Find Indices With Index and Value Difference I

Easy

ArrayTwo PointersHash MapArray

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)

Using a sliding window approach allows us to efficiently find indices that meet the conditions without checking every possible pair. We maintain a window of valid indices and check conditions in linear time.

⚙️

Algorithm

5 steps

1Step 1: Initialize a HashMap to store the indices of the values as we iterate through the array.
2Step 2: For each index 'i', check if there are any indices 'j' in the HashMap that satisfy the conditions.
3Step 3: If a valid pair is found, return [i, j].
4Step 4: Update the HashMap with the current index for the current value.
5Step 5: If no valid pair is found after all iterations, return [-1, -1].

solution.py9 lines

1def find_indices(nums, indexDifference, valueDifference):
2    n = len(nums)
3    value_map = {}
4    for i in range(n):
5        for value in value_map:
6            if abs(i - value_map[value]) >= indexDifference and abs(nums[i] - value) >= valueDifference:
7                return [value_map[value], i]
8        value_map[nums[i]] = i
9    return [-1, -1]

ℹ

Complexity note: This complexity is achieved by using a HashMap to store indices, allowing us to check conditions in constant time as we iterate through the array.

1Using a HashMap allows for efficient lookups of indices.
2Understanding the conditions helps in optimizing the search.

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