doyedele1

'''
    Explanation:
        - history:   [leetcode, youtube, facebook, google]
        - urlPosition:               i
        visit: pop out all urls after the urlPosition, append the url to visit and move the urlPosition
        
        - history:   [leetcode, youtube, facebook, google], steps = 2
        - urlPosition:               i
        back and forward --> you don't want to go out of bound because of the steps
        

doyedele1

'''
    Explanation:
        DFS
            [
                [1,1,1,1],
                [1,1,1,1],
                [1,1,1,1],
                [1,1,1,1]
            ]
            

doyedele1

'''
    Explanation:
        cache --> using memory to speed up the computer algorithms
        * all the key values are positive so we can return something that won't clash with -1 when the key is not found in the cache
        - We could use OrderedDict library in Python which remembers the order in which elements are inserted
        
        - Get or put requires us to delete or insert at different positions. We can use a doubly linked list because deletion and insertions are constant time, but we also need the linked list for proper ordering.
        - When we delete things, we pop it off the right (most recent used)
        left <-> [1,1] <-> [2,2] <-> right
        lru                          mru

doyedele1

'''
    Explanation:
        3[a2[c]]
        - Number in front of a character shows how many times that character will be multiplied
        - 3 applies to everything in the bracket
        - acc * 3 ==> accaccacc
        
        
        - Nested brackets will make the solution tricky/complicated
        - We will have to solve the inner brackets before the outer brackets --> recursive solution

doyedele1

'''
    Explanation I: Brute-force solution
        - Convert the transactions to individual item of the valid data types and store it in an array. i.e. alice = string, 20 = integer, 800 = integer and mtv = string
        - Loop through the new array and find where amount > 1000. Append that to result array and convert the transaction to a string splitting each item by comma
            - Nested loop to check for same name, and if time difference is less than or equal to 60 and if different city,
                - Append that to the res array and convert the transaction to a string splitting each item by comma
        
        - TC - O(n-squared)
        - SC - O(n)
'''

doyedele1

'''
    Explanation:
    - deeedbbcccbdaa
    [[d,1], [e,3]]
    [[d,2]]

    TC - O(2n) = O(n) where n is the size of the input string
    SC - O(n)
'''

doyedele1

'''
    Explanation:
        - Major steps to determine which candies to crush,
            1. Check three or more candies to be crushed in rows
            2. Check three or more candies to be crushed in columns
            3. Dropping the candies
            
            Steps 1 & 2: * This is an issue when considering columns --> [3,1,1,1,1,5] = [3,0,0,0,0,5]
            Can we check for rows and columns simultaneously? Yes. We can use a slider of three at a time and check for the absolute values of the three integers. If the absolute values are the same, then we can change the integers to negative.
            

doyedele1

 '''
        Explanation:
            - We have to pick equal number of A and B
            Initially,
            - Can we see if we can send all candidates to city A?
            - Now, which of the candidates can we send to B? We will pick the candidate for which we gained the maximum (the most negative number)

            10      +  30   +   400   +  30
            +10     +170        -350     -10

doyedele1

'''
    Explanation:
        - We will have a recursive helper (flatten_helper) function that takes in the head pointer (curr = head, tail = head)
        - While curr is not null,
            - Here we can store curr, next and tail on the first node (head)
            - If no child,
                current = next and next will point to next of current
            - If it has child, 
                We flatten the child. i.e. child_tail will be flatten_helper function called on child. Recursive function here
                curr --> next = child

doyedele1

'''
    Explanation:
        - id passed in checkIn and checkOut: unique identifier to our customer
        - Each customer can only be checked into a station one place at a time
        
        1. Create a map called arrivals. key = id, value = tuple containing stationName and time --> (stationName, t)
        2. To keep track of all previous travels between any two stations, we create another map called travels. key = stationNames separated by a delimiter, value = count, total --> [count, total]
        
        Example:
        Calls                   Arrivals                Travels