How do you solve Maximum Distance in Arrays on LeetCode?

Maximum Distance in Arrays (LeetCode #624) 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: The maximum distance can be achieved by comparing the smallest element of one array with the largest element of another.

What is the time complexity of Maximum Distance in Arrays?

The optimal solution for Maximum Distance in Arrays runs in O(n) time and uses O(1) space. This complexity is optimal because we only make a single pass through the arrays, updating our min and max values as we go.

What is the brute force solution for Maximum Distance in Arrays?

The brute force approach for Maximum Distance in Arrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible pair of integers from different arrays to find the maximum distance. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Distance in Arrays?

Maximum Distance in Arrays uses the following concepts: Array, Greedy. The recommended patterns to study are: Array, Greedy.

What are the interview tips for Maximum Distance in Arrays?

Always check if the input data has any constraints that can simplify your approach. Think about how you can reduce the number of comparisons by leveraging properties of the data (like sorting). Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Maximum Distance in Arrays?

Not considering that the arrays are sorted, leading to unnecessary comparisons. Failing to update the min and max values correctly during iterations.

#624

Maximum Distance in Arrays

Medium

ArrayGreedyArrayGreedy

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)

The optimal solution leverages the fact that the arrays are sorted. By only considering the minimum and maximum values of the first and last arrays, we can efficiently compute the maximum distance without needing to check every pair.

⚙️

Algorithm

3 steps

1Step 1: Initialize `min_val` as the first element of the first array and `max_val` as the last element of the first array.
2Step 2: Loop through each subsequent array, updating `min_val` and `max_val` based on the first and last elements of the current array.
3Step 3: Calculate the maximum distance using the absolute differences between the current `min_val` and `max_val`.

solution.py11 lines

1# Full working Python code
2
3def maxDistance(arrays):
4    min_val = arrays[0][0]
5    max_val = arrays[0][-1]
6    max_distance = 0
7    for i in range(1, len(arrays)):
8        max_distance = max(max_distance, abs(arrays[i][-1] - min_val), abs(max_val - arrays[i][0]))
9        min_val = min(min_val, arrays[i][0])
10        max_val = max(max_val, arrays[i][-1])
11    return max_distance

ℹ

Complexity note: This complexity is optimal because we only make a single pass through the arrays, updating our min and max values as we go.

1The maximum distance can be achieved by comparing the smallest element of one array with the largest element of another.
2Sorting of arrays allows us to efficiently access the minimum and maximum values.

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