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A system logs the operations performed on a folder with a series of strings. Each string represents an operation that is one of the following:

• "../": Move to the parent folder. (If you are already in the main folder, you stay in the main folder.)
• "./": Stay in the current folder.
• "x/": Move to a child folder named x. This is guaranteed to be a valid folder name.

You are starting in the main folder, and your task is to determine the minimum number of operations required to go back to the main folder after performing all the operations in the log.

Given an array logs where logs[i] is a string representing the i-th operation, return the minimum number of operations needed to go back to the main folder.

Clarifying Questions

1. Are there any invalid or unexpected strings in the logs array?
   • No, all strings provided in the logs array are guaranteed to be valid operations as specified.
2. Is the main folder considered to be at depth 0?
   • Yes, starting from the main folder is considered to be at depth 0.
3. Can we assume the length of logs array will not exceed certain limits for computational purposes?
   • Yes, you can assume a reasonable upper bound for the length of the logs array to ensure computational feasibility.

Strategy

1. Use a counter depth to keep track of the current folder’s depth, initialized to 0.
2. Iterate over each operation in the logs array:
   • If the operation is "../", decrease the depth by 1 if it’s greater than 0 (because you can’t go above the main folder).
   • If the operation is "./", do nothing as it means staying in the current folder.
   • For any other string like "x/", increase the depth by 1 because it means moving to a child folder.
3. After processing all operations, the value of depth will represent the minimum operations required to return to the main folder (which is just the value of depth itself).

Code

def minOperations(logs):
    depth = 0
    for log in logs:
        if log == "../":
            if depth > 0:
                depth -= 1
        elif log == "./":
            continue
        else: # it means "x/"
            depth += 1
    return depth

Time Complexity

• Time Complexity: The time complexity of this solution is (O(n)), where (n) is the number of elements in the logs array. This is because we are iterating over each log once.

• Space Complexity: The space complexity is (O(1)) as we are using only a single integer variable (depth), whose space usage does not depend on the input size.

This efficient approach ensures that we handle the operations in linear time with constant extra space.

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