Introduction

In this blog post, we will explore how to solve the LeetCode problem "2511" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 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

Explanation

• Iterate through the forts array, keeping track of the last seen position of a friendly fort (1).
  • When encountering an empty cell (-1), continue iterating. When encountering another fort (either friendly or enemy), check if the current fort and the last seen friendly fort are such that the path between them only contains enemy forts (0s). If so, update the maximum captured forts count.
  • Return the maximum captured forts count.

• Runtime complexity: O(n), where n is the length of the forts array. Storage complexity: O(1).

Code

    def captureForts(forts):
    n = len(forts)
    max_captured = 0
    last_fort = -1

    for i in range(n):
        if forts[i] == 1:
            if last_fort != -1:
                captured = 0
                valid = True
                start = min(i, last_fort) + 1
                end = max(i, last_fort)
                for j in range(start, end):
                    if forts[j] != 0:
                        valid = False
                        break
                    captured += 1
                if valid:
                    max_captured = max(max_captured, captured)
            last_fort = i
        elif forts[i] == 0:
            pass
        else:  # forts[i] == -1
            pass

        if forts[i] == 0:
            if last_fort != -1:
                captured = 0
                valid = True
                start = min(i, last_fort) + 1
                end = max(i, last_fort)

                if forts[last_fort] == 1:
                  for j in range(start, end):
                      if forts[j] != 0:
                          valid = False
                          break
                      captured += 1
                  if valid:
                      max_captured = max(max_captured, captured)

    last_fort = -1
    for i in range(n-1, -1, -1):
        if forts[i] == 1:
            if last_fort != -1:
                captured = 0
                valid = True
                start = min(i, last_fort) + 1
                end = max(i, last_fort)
                for j in range(start, end):
                    if forts[j] != 0:
                        valid = False
                        break
                    captured += 1
                if valid:
                    max_captured = max(max_captured, captured)
            last_fort = i
        elif forts[i] == 0:
            pass
        else:  # forts[i] == -1
            pass

        if forts[i] == 0:
            if last_fort != -1:
                captured = 0
                valid = True
                start = min(i, last_fort) + 1
                end = max(i, last_fort)

                if forts[last_fort] == 1:
                  for j in range(start, end):
                      if forts[j] != 0:
                          valid = False
                          break
                      captured += 1
                  if valid:
                      max_captured = max(max_captured, captured)

    return max_captured