How do you solve Minimum Absolute Difference Between Two Values on LeetCode?

Minimum Absolute Difference Between Two Values (LeetCode #3880) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Valid pairs require specific values (1 and 2).

What is the time complexity of Minimum Absolute Difference Between Two Values?

The optimal solution for Minimum Absolute Difference Between Two Values runs in O(n) time and uses O(1) space. The complexity is O(n) because we traverse the array only once, maintaining the last seen indices.

What is the brute force solution for Minimum Absolute Difference Between Two Values?

The brute force approach for Minimum Absolute Difference Between Two Values is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible pair of indices to find valid pairs where nums[i] is 1 and nums[j] is 2. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Absolute Difference Between Two Values?

Minimum Absolute Difference Between Two Values uses the following concepts: Array, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Absolute Difference Between Two Values?

Explain your thought process clearly. Consider edge cases, like arrays without valid pairs. Optimize your solution after brute force.

What are common mistakes when solving Minimum Absolute Difference Between Two Values?

Forgetting to check both indices before calculating the difference. Not initializing variables properly.

#3880

Minimum Absolute Difference Between Two Values

Easy

ArrayEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Instead of checking all pairs, we can keep track of the last seen indices of 1 and 2. This allows us to calculate the minimum difference in a single pass through the array.

⚙️

Algorithm

3 steps

1Step 1: Initialize variables to store the last seen index of 1 and 2, and a variable for the minimum difference.
2Step 2: Iterate through the array; update the last seen index for 1 or 2 when found.
3Step 3: Whenever both indices are updated, calculate the absolute difference and update the minimum difference.

solution.py12 lines

1def minAbsDifference(nums):
2    last1, last2, min_diff = -1, -1, float('inf')
3    for i, num in enumerate(nums):
4        if num == 1:
5            last1 = i
6            if last2 != -1:
7                min_diff = min(min_diff, abs(last1 - last2))
8        elif num == 2:
9            last2 = i
10            if last1 != -1:
11                min_diff = min(min_diff, abs(last1 - last2))
12    return min_diff if min_diff != float('inf') else -1

ℹ

Complexity note: The complexity is O(n) because we traverse the array only once, maintaining the last seen indices.

1Valid pairs require specific values (1 and 2).
2Tracking indices reduces unnecessary comparisons.

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