Minimum Distance to Type a Word Using Two Fingers

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Dynamic Programming

You have a keyboard layout as shown above in the X-Y plane, where each English uppercase letter is located at some coordinate.

For example, the letter 'A' is located at coordinate (0, 0), the letter 'B' is located at coordinate (0, 1), the letter 'P' is located at coordinate (2, 3) and the letter 'Z' is located at coordinate (4, 1).

Given the string word, return the minimum total distance to type such string using only two fingers.

The distance between coordinates (x₁, y₁) and (x₂, y₂) is |x₁ - x₂| + |y₁ - y₂|.

Note that the initial positions of your two fingers are considered free so do not count towards your total distance, also your two fingers do not have to start at the first letter or the first two letters.

Example 1:

Input: word = "CAKE"
Output: 3
Explanation: Using two fingers, one optimal way to type "CAKE" is: 
Finger 1 on letter 'C' -> cost = 0 
Finger 1 on letter 'A' -> cost = Distance from letter 'C' to letter 'A' = 2 
Finger 2 on letter 'K' -> cost = 0 
Finger 2 on letter 'E' -> cost = Distance from letter 'K' to letter 'E' = 1 
Total distance = 3

Example 2:

Input: word = "HAPPY"
Output: 6
Explanation: Using two fingers, one optimal way to type "HAPPY" is:
Finger 1 on letter 'H' -> cost = 0
Finger 1 on letter 'A' -> cost = Distance from letter 'H' to letter 'A' = 2
Finger 2 on letter 'P' -> cost = 0
Finger 2 on letter 'P' -> cost = Distance from letter 'P' to letter 'P' = 0
Finger 1 on letter 'Y' -> cost = Distance from letter 'A' to letter 'Y' = 4
Total distance = 6

Constraints:

2 <= word.length <= 300
word consists of uppercase English letters.

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:

Is the input word guaranteed to contain only uppercase English letters?
Can the input word be empty or null? If so, what should I return?
If multiple finger placements result in the same minimum distance, is any one of them acceptable?
Are the coordinates of the letters (A-Z) fixed as described (A at (0,0), B at (0,1), etc.) or could they be different?
What is the maximum length of the input word?

Brute Force Solution

Approach

The brute force method for this typing problem involves trying every possible finger placement combination. We will explore all the ways to type the word, calculating the distance for each combination. Finally, we will find the combination with the smallest total distance.

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

Consider each letter in the word, one at a time.
For the first letter, imagine placing either your left or right finger on that letter. These are your two starting points.
For the second letter, for each of the previous finger placements (left or right on the first letter), try placing either the left or right finger on the second letter. Now you have four possibilities.
Calculate the distance between the finger positions for each possibility. For instance, if the left finger was on the first letter and the right finger is now on the second letter, calculate the distance between those letters on the keyboard.
Continue this process for each subsequent letter in the word. For each letter, explore placing either the left or right finger on it, considering all the finger positions from the previous letter.
Keep track of the total distance for each of these possibilities as you type the whole word.
After exploring every possible finger placement combination for the entire word, compare the total distances calculated for each combination.
The combination with the smallest total distance is the optimal solution.

Code Implementation

def minimum_distance_two_fingers_brute_force(word):
    keyboard = ["ABCDEF", "GHIJKL", "MNOPQR", "STUVWX", "YZ"]

    def get_coordinates(character):
        for row_index, row in enumerate(keyboard):
            if character in row:
                col_index = row.index(character)
                return (row_index, col_index)
        return None

    def calculate_distance(coord1, coord2):
        if coord1 is None or coord2 is None:
            return 0
        return abs(coord1[0] - coord2[0]) + abs(coord1[1] - coord2[1])

    def brute_force(index, left_finger_position, right_finger_position):
        if index == len(word):
            return 0

        letter = word[index]
        letter_coordinates = get_coordinates(letter)

        # Explore placing the left finger on the current letter
        left_distance = calculate_distance(left_finger_position, letter_coordinates) \
            + brute_force(index + 1, letter_coordinates, right_finger_position)

        # Explore placing the right finger on the current letter
        right_distance = calculate_distance(right_finger_position, letter_coordinates) \
            + brute_force(index + 1, left_finger_position, letter_coordinates)

        return min(left_distance, right_distance)

    # Initiate the recursive calls.  The starting positions are 'empty' represented by None.
    return brute_force(0, None, None)

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores every possible combination of left and right finger placements for each letter in the word. For a word of length n, each letter has two choices (left or right finger), leading to 2 * 2 * ... * 2 (n times) = 2^n possible combinations. The algorithm calculates the distance for each of these 2^n combinations. Thus, the time complexity is O(2^n).

Space Complexity

O(2^N) – The brute force approach explores all possible finger placements. For each letter in the word, it considers two options: placing the left or right finger. This leads to a binary branching tree where each level corresponds to a letter in the word. Since each level potentially doubles the number of paths explored, the number of paths grows exponentially with the word length N. Each path needs to keep track of the total distance, which could implicitly involve storing intermediate distances for each letter position explored. Therefore, the space complexity is O(2^N) as it needs to explore all combinations, and potentially store the cumulative distances for them.

Optimal Solution

Approach

The best way to solve this problem is to use dynamic programming, which means breaking it down into smaller, overlapping subproblems and solving each subproblem only once. We will store the solutions to the subproblems to avoid recalculating them, making the overall solution efficient. The key is to consider the current letter and the positions of both fingers, and calculate the minimum cost to type the rest of the word from that point.

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

Understand that at any point, one finger is on the keyboard, and the other is either on another key or hasn't been used yet. We want to minimize the effort needed to type the word.
Define a function that calculates the distance between any two letters on the keyboard. Treat the keyboard as a grid, and calculate the Manhattan distance (sum of the absolute differences in x and y coordinates).
Create a table to store the minimum cost to type the remainder of the word, given the positions of the two fingers. The table will have dimensions related to the word length and possible finger positions.
Start filling the table from the end of the word towards the beginning. This 'bottom-up' approach ensures that when we calculate the cost for a particular letter, we already know the optimal costs for the letters that follow.
For each letter in the word and each possible combination of finger positions, consider two options: move the left finger to the current letter or move the right finger to the current letter.
Calculate the cost of each option by adding the distance the finger needs to move to the cost of typing the rest of the word (which we've already calculated and stored in our table).
Choose the option with the minimum cost and store it in the table. This represents the optimal way to type the rest of the word from that state.
The final result is the minimum cost to type the entire word, starting with both fingers unassigned (or assigned to some starting position). This value will be stored in our table and can be retrieved easily.

Code Implementation

def minimum_distance_to_type_a_word(word):
    def calculate_distance(letter1, letter2):
        if letter1 == -1 or letter2 == -1:
            return 0
        letter1_row, letter1_col = divmod(ord(letter1) - ord('A'), 6)
        letter2_row, letter2_col = divmod(ord(letter2) - ord('A'), 6)
        return abs(letter1_row - letter2_row) + abs(letter1_col - letter2_col)

    word_length = len(word)

    # Initialize the DP table to store min costs.
    dp_table = [[([0] * 27) for _ in range(27)] for _ in range(word_length + 1)]

    # Iterate from the end of the word to the beginning
    for index in range(word_length - 1, -1, -1):
        for left_finger_position in range(27):
            for right_finger_position in range(27):
                current_letter_code = ord(word[index]) - ord('A')

                # Calculate cost of moving the left finger.
                move_left_finger = calculate_distance(
                    chr(left_finger_position + ord('A')) if left_finger_position < 26 else -1,
                    word[index]
                ) + dp_table[index + 1][current_letter_code][right_finger_position]

                # Calculate cost of moving the right finger.
                move_right_finger = calculate_distance(
                    chr(right_finger_position + ord('A')) if right_finger_position < 26 else -1,
                    word[index]
                ) + dp_table[index + 1][left_finger_position][current_letter_code]

                # Store the minimum cost.
                dp_table[index][left_finger_position][right_finger_position] = min(
                    move_left_finger, move_right_finger
                )

    #The result is the minimum cost starting with no fingers used.
    return dp_table[0][26][26]

Big(O) Analysis

Time Complexity

O(n*300*300) – The dynamic programming solution iterates through each letter in the input word, which has length n. For each letter, it considers all possible positions for the two fingers on the keyboard. Since there are 26 letters (plus the initial unassigned state), each finger can be in one of approximately 300 positions (treating the keyboard as a grid). Thus, for each letter in the word, the algorithm examines on the order of 300*300 possible finger combinations. Calculating the minimum cost for each state takes constant time. The time complexity is therefore O(n * 300 * 300), which because the 300*300 represents a constant value can also be represented as O(n).

Space Complexity

O(N) – The primary space complexity comes from the dynamic programming table used to store the minimum cost to type the remainder of the word. This table has dimensions related to the word length and possible finger positions, essentially storing costs for each letter and each possible finger combination. Specifically, if we consider the fingers can be in any of the 26 keyboard positions or unassigned (represented as -1), the table would have dimensions dependent on N (the length of the word) and two finger positions (27 possibilities each), leading to a space requirement proportional to N. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty word input	Return 0 as no typing is needed, or throw an IllegalArgumentException if null input is disallowed.
Word with only one character	Return 0 as only one finger needs to move.
Word with two identical characters	The solution should correctly calculate the distance from the start position to the first character and from there to the second.
Word with only two characters, far apart on the keyboard	Ensures maximum distance calculation is correct, covering the largest possible key displacement.
Very long word to test DP table size limits	Verify that the DP table size does not cause memory issues or integer overflow during indexing; consider using memoization if recursion depth becomes a problem.
Word with all the same character repeated many times	The algorithm should only need to move one finger once, and the rest of the time, keep the other finger at its initial position.
Word requiring alternating finger movements between distant keys	This case tests the effectiveness of the dynamic programming approach in selecting optimal finger assignments.
Word contains non-alphabetic characters	Handle the case by either ignoring the non-alphabetic characters, throwing an exception, or mapping them to appropriate positions (e.g., based on ASCII value).

Null or empty word input

How to Handle:

Return 0 as no typing is needed, or throw an IllegalArgumentException if null input is disallowed.

Word with only one character

How to Handle:

Return 0 as only one finger needs to move.

Word with two identical characters

How to Handle:

The solution should correctly calculate the distance from the start position to the first character and from there to the second.

Word with only two characters, far apart on the keyboard

How to Handle:

Ensures maximum distance calculation is correct, covering the largest possible key displacement.

Very long word to test DP table size limits

How to Handle:

Verify that the DP table size does not cause memory issues or integer overflow during indexing; consider using memoization if recursion depth becomes a problem.

Word with all the same character repeated many times

How to Handle:

The algorithm should only need to move one finger once, and the rest of the time, keep the other finger at its initial position.

Word requiring alternating finger movements between distant keys

How to Handle:

This case tests the effectiveness of the dynamic programming approach in selecting optimal finger assignments.

Word contains non-alphabetic characters

How to Handle:

Handle the case by either ignoring the non-alphabetic characters, throwing an exception, or mapping them to appropriate positions (e.g., based on ASCII value).