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2511. Maximum Enemy Forts That Can Be Captured

Description

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.

 

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

Solution 1: Two Pointers

Thinking

A move starts at a friendly fort, crosses a run of empty cells, and stops at an enemy fort; the captured count is the zeros in between. \(n\le 1000\) allows enumerating endpoints, but each \(i\) need not rescan to the right.

Park \(i\) on a nonzero cell and let \(j\) skip the following zeros to the next nonzero. Opposite signs mean \(j-i-1\) zeros can update the answer; then set \(i\) to \(j\) so the array is walked once.

We use a pointer \(i\) to traverse the array \(forts\), and a pointer \(j\) to start traversing from the next position of \(i\) until it encounters the first non-zero position, i.e., \(forts[j] \neq 0\). If \(forts[i] + forts[j] = 0\), then we can move the army between \(i\) and \(j\), destroying \(j - i - 1\) enemy forts. We use the variable \(ans\) to record the maximum number of enemy forts that can be destroyed.

The time complexity is \(O(n)\), and the space complexity is \(O(1)\). Where \(n\) is the length of the array forts.

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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
    }
}

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