How do you solve Maximum Enemy Forts That Can Be Captured on LeetCode?

Maximum Enemy Forts That Can Be Captured (LeetCode #2511) 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: Identify forts under your command first.

What is the brute force solution for Maximum Enemy Forts That Can Be Captured?

The brute force approach for Maximum Enemy Forts That Can Be Captured is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of forts under your command and counting how many enemy forts can be captured between them. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Enemy Forts That Can Be Captured?

Maximum Enemy Forts That Can Be Captured uses the following concepts: Array, Two Pointers. The recommended patterns to study are: Two Pointers, Sliding Window.

What are the interview tips for Maximum Enemy Forts That Can Be Captured?

Clearly explain your thought process as you code. Consider edge cases and test your solution with them. Optimize your solution after writing the brute force version.

What are common mistakes when solving Maximum Enemy Forts That Can Be Captured?

Not checking both directions from the fort. Forgetting to handle edge cases like no forts under command.

#2511

Maximum Enemy Forts That Can Be Captured

Easy

ArrayTwo PointersTwo PointersSliding Window

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 approach uses a single pass to identify all forts under your command and then uses a two-pointer technique to efficiently count enemy forts between them. This reduces unnecessary checks and improves performance.

⚙️

Algorithm

3 steps

1Step 1: Initialize two pointers to traverse the forts array.
2Step 2: For each fort under your command, move the left pointer to find the nearest empty position on the left and the right pointer to find the nearest empty position on the right.
3Step 3: Count the enemy forts between the two pointers and update the maximum captured.

solution.py19 lines

1# Full working Python code
2
3def maxEnemyForts(forts):
4    max_captured = 0
5    n = len(forts)
6    for i in range(n):
7        if forts[i] == 1:
8            left = i - 1
9            right = i + 1
10            captured = 0
11            while left >= 0 and forts[left] == 0:
12                captured += 1
13                left -= 1
14            while right < n and forts[right] == 0:
15                captured += 1
16                right += 1
17            max_captured = max(max_captured, captured)
18    return max_captured
19

ℹ

Complexity note: This complexity is achieved because we only traverse the array once, checking each fort under your command and counting enemy forts in a single pass.

1Identify forts under your command first.
2Use two pointers to efficiently count enemy forts.

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