Crawler Log Folder

summary

• The problem is to find the minimum number of operations needed to go back to the main folder after a series of change folder operations.

approach

• The solution uses a stack data structure to keep track of the current folder path
• When a folder operation is encountered, the corresponding action is performed on the stack
• If the operation is '../', the top element is removed from the stack
• If the operation is './', the top element is left unchanged
• If the operation is 'x/', the top element is replaced with 'x'
• The minimum number of operations needed to go back to the main folder is the number of times '../' is performed minus the number of times './' is performed.

complexity

• O(n), where n is the number of operations in the logs array

explain

• The solution iterates over the logs array and performs the corresponding action on the stack for each operation
• The stack is used to keep track of the current folder path
• When the stack is empty and an '../' operation is encountered, the solution returns 0
• Otherwise, the solution returns the difference between the number of '../' operations and the number of './' operations.

Solution Code:

class Solution {
public:
    int minOperations(vector& logs) {
        int cnt = 0;
        for(int i = 0; i < logs.size(); i++){
            if(!logs[i].compare("../")){
                cnt = cnt - 1 < 0 ? 0 : cnt - 1;
            } else if (!logs[i].compare("./")){

            } else {
                cnt++;
            }
        }
        return cnt;
    }
};