How do you solve Minimum Absolute Difference Queries on LeetCode?

Minimum Absolute Difference Queries (LeetCode #1906) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using a set to track unique elements can significantly reduce the number of comparisons needed.

What is the time complexity of Minimum Absolute Difference Queries?

The optimal solution for Minimum Absolute Difference Queries runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing unique elements.

What is the brute force solution for Minimum Absolute Difference Queries?

The brute force approach for Minimum Absolute Difference Queries is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of elements in the subarray to find the minimum absolute difference. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Absolute Difference Queries?

Minimum Absolute Difference Queries uses the following concepts: Array, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Absolute Difference Queries?

Always start with a brute force solution to understand the problem better. Think about how you can reduce the problem size or number of comparisons. Be prepared to discuss the trade-offs between time and space complexity.

What are common mistakes when solving Minimum Absolute Difference Queries?

Not considering the case where all elements in the subarray are the same. Failing to optimize by using data structures like sets or maps.

#1906

Minimum Absolute Difference Queries

Medium

ArrayPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

By leveraging sorting and a set to track unique elements, we can efficiently find the minimum absolute difference without checking every pair. This approach reduces the number of comparisons needed.

⚙️

Algorithm

5 steps

1Step 1: For each query, extract the subarray defined by the indices l and r.
2Step 2: Use a set to store unique elements from the subarray.
3Step 3: Sort the unique elements to easily find the minimum difference between consecutive elements.
4Step 4: Iterate through the sorted unique elements to find the minimum absolute difference.
5Step 5: If there are less than two unique elements, return -1; otherwise, return the minimum difference found.

solution.py15 lines

1# Full working Python code
2
3def min_absolute_difference(nums, queries):
4    results = []
5    for l, r in queries:
6        subarray = nums[l:r+1]
7        unique_elements = sorted(set(subarray))
8        if len(unique_elements) < 2:
9            results.append(-1)
10            continue
11        min_diff = float('inf')
12        for i in range(1, len(unique_elements)):
13            min_diff = min(min_diff, unique_elements[i] - unique_elements[i - 1])
14        results.append(min_diff)
15    return results

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing unique elements.

1Using a set to track unique elements can significantly reduce the number of comparisons needed.
2Sorting the unique elements allows for efficient calculation of minimum differences.

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