Freedom Trail

Hard

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StringsDynamic Programming

In the video game Fallout 4, the quest "Road to Freedom" requires players to reach a metal dial called the "Freedom Trail Ring" and use the dial to spell a specific keyword to open the door.

Given a string ring that represents the code engraved on the outer ring and another string key that represents the keyword that needs to be spelled, return the minimum number of steps to spell all the characters in the keyword.

Initially, the first character of the ring is aligned at the "12:00" direction. You should spell all the characters in key one by one by rotating ring clockwise or anticlockwise to make each character of the string key aligned at the "12:00" direction and then by pressing the center button.

At the stage of rotating the ring to spell the key character key[i]:

You can rotate the ring clockwise or anticlockwise by one place, which counts as one step. The final purpose of the rotation is to align one of ring's characters at the "12:00" direction, where this character must equal key[i].
If the character key[i] has been aligned at the "12:00" direction, press the center button to spell, which also counts as one step. After the pressing, you could begin to spell the next character in the key (next stage). Otherwise, you have finished all the spelling.

Example 1:

Input: ring = "godding", key = "gd"
Output: 4
Explanation:
For the first key character 'g', since it is already in place, we just need 1 step to spell this character. 
For the second key character 'd', we need to rotate the ring "godding" anticlockwise by two steps to make it become "ddinggo".
Also, we need 1 more step for spelling.
So the final output is 4.

Example 2:

Input: ring = "godding", key = "godding"
Output: 13

Constraints:

1 <= ring.length, key.length <= 100
ring and key consist of only lower case English letters.
It is guaranteed that key could always be spelled by rotating ring.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the ring and key contain duplicate characters, and if so, how should I handle them?
Are the ring and key guaranteed to contain only lowercase English letters, or can they contain other characters?
If the key cannot be formed from the ring, what should I return?
What is the maximum length of the ring and key?
Is the 'ring' circular in that rotating past the end brings you back to the beginning?

Brute Force Solution

Approach

The brute force approach to the Freedom Trail problem explores every possible sequence of button presses to spell out the target phrase. It essentially tries every combination, regardless of efficiency, to find the shortest path. It's like trying every single door in a maze until you find the exit.

Here's how the algorithm would work step-by-step:

Start at the beginning of the target phrase.
Imagine the dial on the lock. For each letter in the phrase, check every single letter on the dial.
For each letter on the dial, calculate how many clicks it would take to get from the current position on the dial to that letter.
Consider each of those clicks as a potential 'next step'.
Continue doing this for the rest of the phrase. So if the phrase has ten letters, each dial-letter combination is being considered for each of those letters.
After reaching the end of the phrase, count the total number of clicks each sequence took to complete the entire phrase.
Compare all the different click-sequences and choose the sequence with the fewest total clicks. That's the answer.

Code Implementation

def freedom_trail_brute_force(ring, key):
    def calculate_distance(current_index, target_character_index, ring_length):
        distance = abs(current_index - target_character_index)
        return min(distance, ring_length - distance)

    def find_shortest_path(current_index, key_index):
        # Base case: If we've reached the end of the key, return 0 clicks.
        if key_index == len(key):
            return 0

        minimum_clicks = float('inf')

        # Iterate through all possible characters on the ring.
        for ring_index, ring_character in enumerate(ring):
            if ring_character == key[key_index]:
                # Recursively explore paths from this ring index.

                distance = calculate_distance(current_index, ring_index, len(ring))

                total_clicks = distance + find_shortest_path(ring_index, key_index + 1) + 1
                minimum_clicks = min(minimum_clicks, total_clicks)

        return minimum_clicks

    # Start at index 0 of the ring.
    starting_index = 0
    return find_shortest_path(starting_index, 0)

Big(O) Analysis

Time Complexity

O(m^n) – The brute force approach explores all possible combinations of dial positions for each character in the target phrase. Let m be the length of the ring string and n be the length of the key string. For each of the n characters in the key, we iterate through all m positions on the ring. This creates a branching factor of m for each character. Consequently, we explore m possibilities for the first character, m*m for the first two characters and so on, resulting in m multiplied by itself n times. Therefore, the time complexity is O(m^n).

Space Complexity

O(1) – The brute force approach, as described, doesn't explicitly create any auxiliary data structures that scale with the input size. The described process calculates clicks and compares them; it does not maintain any significant temporary lists, hash maps, or other data structures related to the length of the ring or the key. Therefore, the space used is constant, regardless of the input size.

Optimal Solution

Approach

The most efficient way to solve this puzzle is to use dynamic programming. We figure out the shortest path to spell each letter in the key sequence given the circular nature of the ring. We build up from the beginning, always choosing the quickest way to the next letter.

Here's how the algorithm would work step-by-step:

First, create a memory table to store the minimum number of steps needed to spell each character of the key, given the circular ring's layout.
Begin with the first character of the key sequence. We're initially positioned at the first character of the ring.
For each character in the ring that matches the current key character, calculate the number of steps it takes to reach that matching character by going both clockwise and counter-clockwise.
Choose the direction (clockwise or counter-clockwise) that results in the fewest steps to reach that character. Add one step for actually pressing the button to select the right letter.
Store the total number of steps required to reach the current key character from the starting position (including the button press) into the memory table.
Repeat the previous steps for each subsequent character in the key sequence. The starting position for finding the next key character is now the location of the previously spelled character.
When you reach the last character of the key sequence, the value stored in the memory table will be the minimum number of steps required to spell the entire sequence.

Code Implementation

def freedom_trail(ring, key):    ring_length = len(ring)
    key_length = len(key)

    # dp_table[i] stores min steps to spell key[0...i]
    dp_table = [float('inf')] * key_length

    # Initialize the first character of the key
    for i in range(ring_length):
        if ring[i] == key[0]:
            clock_wise_distance = abs(i - 0)
            counter_clock_wise_distance = ring_length - clock_wise_distance
            dp_table[0] = min(clock_wise_distance, counter_clock_wise_distance) + 1

    # Iterate through the key to populate the dp table
    for i in range(1, key_length):
        for j in range(ring_length):
            if ring[j] == key[i]:
                min_distance = float('inf')
                for k in range(ring_length):
                    if ring[k] == key[i-1]:
                        clock_wise_distance = abs(j - k)
                        counter_clock_wise_distance = ring_length - clock_wise_distance
                        min_distance = min(min_distance, min(clock_wise_distance, counter_clock_wise_distance) + dp_table[i-1])

                # Add one step for pressing the button
                dp_table[i] = min(dp_table[i], min_distance + 1)

    # Find the minimum steps to spell the entire key
    return dp_table[key_length - 1]

Big(O) Analysis

Time Complexity

O(m*n) – The algorithm iterates through each character 'm' of the key sequence. For each key character, it iterates through the entire ring of 'n' characters to find matching characters. For each matching character, the shortest distance (clockwise or counter-clockwise) is computed, which takes constant time. Therefore, the dominant operation is iterating through the ring 'n' for each character 'm' in the key, leading to a time complexity of O(m*n).

Space Complexity

O(M*N) – The solution uses a memory table (dynamic programming table) to store the minimum number of steps needed to spell each character of the key. Let N be the length of the key sequence, and M be the length of the ring string. The memory table has dimensions related to the length of the key and possible positions on the ring. In this case, the memory table is of size M*N, therefore, the space complexity is O(M*N).

Edge Cases

Case	How to Handle
Empty ring or key	Return 0 immediately as no rotations are needed for empty inputs.
Ring or key is null	Throw an IllegalArgumentException to indicate invalid input.
Ring and key contain the same single character	Return the length of the key since we must dial each character.
Ring contains duplicate characters; key contains duplicate characters	The algorithm should correctly handle duplicates when searching for optimal paths.
Key contains characters not present in the ring	Return Integer.MAX_VALUE (or a similarly large value) since the key can't be formed.
Very large ring/key sizes causing stack overflow with recursion	Consider using iterative dynamic programming to avoid stack overflow.
Ring and key are very long strings of the same character	The algorithm should find the shortest path by choosing the direction to rotate based on distance.
Integer overflow when accumulating rotations	Use long to store the minimum rotations at each step to avoid overflow.

Empty ring or key

How to Handle:

Return 0 immediately as no rotations are needed for empty inputs.

Ring or key is null

How to Handle:

Throw an IllegalArgumentException to indicate invalid input.

Ring and key contain the same single character

How to Handle:

Return the length of the key since we must dial each character.

Ring contains duplicate characters; key contains duplicate characters

How to Handle:

The algorithm should correctly handle duplicates when searching for optimal paths.

Key contains characters not present in the ring

How to Handle:

Return Integer.MAX_VALUE (or a similarly large value) since the key can't be formed.

Very large ring/key sizes causing stack overflow with recursion

How to Handle:

Consider using iterative dynamic programming to avoid stack overflow.

Ring and key are very long strings of the same character

How to Handle:

The algorithm should find the shortest path by choosing the direction to rotate based on distance.

Integer overflow when accumulating rotations

How to Handle:

Use long to store the minimum rotations at each step to avoid overflow.