LeetCode #2511 — EASY

Maximum Enemy Forts That Can Be Captured

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 0-indexed integer array forts of length n representing the positions of several forts. forts[i] can be -1, 0, or 1 where:

-1 represents there is no fort at the i^th position.
0 indicates there is an enemy fort at the i^th position.
1 indicates the fort at the i^th the position is under your command.

Now you have decided to move your army from one of your forts at position i to an empty position j such that:

0 <= i, j <= n - 1
The army travels over enemy forts only. Formally, for all k where min(i,j) < k < max(i,j), forts[k] == 0.

While moving the army, all the enemy forts that come in the way are captured.

Return the maximum number of enemy forts that can be captured. In case it is impossible to move your army, or you do not have any fort under your command, return 0.

Example 1:

Input: forts = [1,0,0,-1,0,0,0,0,1]
Output: 4
Explanation:
- Moving the army from position 0 to position 3 captures 2 enemy forts, at 1 and 2.
- Moving the army from position 8 to position 3 captures 4 enemy forts.
Since 4 is the maximum number of enemy forts that can be captured, we return 4.

Example 2:

Input: forts = [0,0,1,-1]
Output: 0
Explanation: Since no enemy fort can be captured, 0 is returned.

Constraints:

1 <= forts.length <= 1000
-1 <= forts[i] <= 1

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array forts of length n representing the positions of several forts. forts[i] can be -1, 0, or 1 where: -1 represents there is no fort at the ith position. 0 indicates there is an enemy fort at the ith position. 1 indicates the fort at the ith the position is under your command. Now you have decided to move your army from one of your forts at position i to an empty position j such that: 0 <= i, j <= n - 1 The army travels over enemy forts only. Formally, for all k where min(i,j) < k < max(i,j), forts[k] == 0. While moving the army, all the enemy forts that come in the way are captured. Return the maximum number of enemy forts that can be captured. In case it is impossible to move your army, or you do not have any fort under your command, return 0.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers

Example 1

[1,0,0,-1,0,0,0,0,1]

Example 2

[0,0,1,-1]

Core Insight

What unlocks the optimal approach

For each fort under your command, check if you can move the army from here.
If yes, find the closest empty positions satisfying all criteria.
How can two-pointers be used to solve this problem optimally?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2511: Maximum Enemy Forts That Can Be Captured
class Solution {
    public int captureForts(int[] forts) {
        int n = forts.length;
        int ans = 0, i = 0;
        while (i < n) {
            int j = i + 1;
            if (forts[i] != 0) {
                while (j < n && forts[j] == 0) {
                    ++j;
                }
                if (j < n && forts[i] + forts[j] == 0) {
                    ans = Math.max(ans, j - i - 1);
                }
            }
            i = j;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2511: Maximum Enemy Forts That Can Be Captured
func captureForts(forts []int) (ans int) {
	n := len(forts)
	i := 0
	for i < n {
		j := i + 1
		if forts[i] != 0 {
			for j < n && forts[j] == 0 {
				j++
			}
			if j < n && forts[i]+forts[j] == 0 {
				ans = max(ans, j-i-1)
			}
		}
		i = j
	}
	return
}

# Accepted solution for LeetCode #2511: Maximum Enemy Forts That Can Be Captured
class Solution:
    def captureForts(self, forts: List[int]) -> int:
        n = len(forts)
        i = ans = 0
        while i < n:
            j = i + 1
            if forts[i]:
                while j < n and forts[j] == 0:
                    j += 1
                if j < n and forts[i] + forts[j] == 0:
                    ans = max(ans, j - i - 1)
            i = j
        return ans

// Accepted solution for LeetCode #2511: Maximum Enemy Forts That Can Be Captured
impl Solution {
    pub fn capture_forts(forts: Vec<i32>) -> i32 {
        let n = forts.len();
        let mut ans = 0;
        let mut i = 0;
        while i < n {
            let mut j = i + 1;
            if forts[i] != 0 {
                while j < n && forts[j] == 0 {
                    j += 1;
                }
                if j < n && forts[i] + forts[j] == 0 {
                    ans = ans.max(j - i - 1);
                }
            }
            i = j;
        }
        ans as i32
    }
}

// Accepted solution for LeetCode #2511: Maximum Enemy Forts That Can Be Captured
function captureForts(forts: number[]): number {
    const n = forts.length;
    let ans = 0;
    let i = 0;
    while (i < n) {
        let j = i + 1;
        if (forts[i] !== 0) {
            while (j < n && forts[j] === 0) {
                j++;
            }
            if (j < n && forts[i] + forts[j] === 0) {
                ans = Math.max(ans, j - i - 1);
            }
        }
        i = j;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.