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Formatted question description: https://leetcode.ca/all/2511.html
2511. Maximum Enemy Forts That Can Be Captured
Description
You are given a 0indexed 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 thei^{th}
position.0
indicates there is an enemy fort at thei^{th}
position.1
indicates the fort at thei^{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
wheremin(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
Solutions

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; } }

class Solution { public: int captureForts(vector<int>& forts) { int n = forts.size(); 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 = max(ans, j  i  1); } } i = j; } return ans; } };

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

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, ji1) } } i = j } return } func max(a, b int) int { if a > b { return a } return b }

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; }

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 } }