Minimum Number of Pushes to Type Word I

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Greedy Algorithms

You are given a string word containing distinct lowercase English letters.

Telephone keypads have keys mapped with distinct collections of lowercase English letters, which can be used to form words by pushing them. For example, the key 2 is mapped with ["a","b","c"], we need to push the key one time to type "a", two times to type "b", and three times to type "c" .

It is allowed to remap the keys numbered 2 to 9 to distinct collections of letters. The keys can be remapped to any amount of letters, but each letter must be mapped to exactly one key. You need to find the minimum number of times the keys will be pushed to type the string word.

Return the minimum number of pushes needed to type word after remapping the keys.

An example mapping of letters to keys on a telephone keypad is given below. Note that 1, *, #, and 0 do not map to any letters.

Example 1:

Input: word = "abcde"
Output: 5
Explanation: The remapped keypad given in the image provides the minimum cost.
"a" -> one push on key 2
"b" -> one push on key 3
"c" -> one push on key 4
"d" -> one push on key 5
"e" -> one push on key 6
Total cost is 1 + 1 + 1 + 1 + 1 = 5.
It can be shown that no other mapping can provide a lower cost.

Example 2:

Input: word = "xycdefghij"
Output: 12
Explanation: The remapped keypad given in the image provides the minimum cost.
"x" -> one push on key 2
"y" -> two pushes on key 2
"c" -> one push on key 3
"d" -> two pushes on key 3
"e" -> one push on key 4
"f" -> one push on key 5
"g" -> one push on key 6
"h" -> one push on key 7
"i" -> one push on key 8
"j" -> one push on key 9
Total cost is 1 + 2 + 1 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 12.
It can be shown that no other mapping can provide a lower cost.

Constraints:

1 <= word.length <= 26
word consists of lowercase English letters.
All letters in word are distinct.

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 input word contain characters beyond lowercase English letters (a-z)?
Is the input word guaranteed to be non-empty?
Is there a limit to the length of the input word?
Are we assuming a standard phone keypad layout?
If multiple sequences of pushes result in the minimum number, is any one acceptable?

Brute Force Solution

Approach

The brute force strategy for this problem involves trying every single combination of button pushes to create the target word. We'll methodically go through all possible sequences, no matter how inefficient, until we find the shortest one. Think of it as trying every possible password combination until you get the right one.

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

Consider all possible sequences of button pushes, including ones that are far too long.
For each sequence, check if pushing those buttons produces the target word.
If a sequence creates the word, figure out how many button pushes were needed.
Keep track of the shortest sequence that works.
After checking every single possible sequence of button pushes, pick the shortest one we found.

Code Implementation

def min_pushes_brute_force(word):
    shortest_sequence_length = float('inf')
    keyboard = [["a", "b", "c"], ["d", "e", "f"], ["g", "h", "i"],
                ["j", "k", "l"], ["m", "n", "o"], ["p", "q", "r", "s"],
                ["t", "u", "v"], ["w", "x", "y", "z"]]

    def generate_sequences(current_sequence, current_word):
        nonlocal shortest_sequence_length

        if len(current_word) > len(word):
            return

        if current_word == word:
            shortest_sequence_length = min(shortest_sequence_length, len(current_sequence))
            return

        # Try adding each possible button push to the current sequence
        for button_index in range(len(keyboard)):
            for letter_index, letter in enumerate(keyboard[button_index]):

                # We try all combinations, inefficiently
                generate_sequences(current_sequence + [(button_index + 2, letter_index + 1)], current_word + letter)

    generate_sequences([], "")

    if shortest_sequence_length == float('inf'):
        return -1

    # Calculate total number of pushes based on the shortest sequence.
    total_pushes = 0
    for _, press_count in get_button_press_counts(find_shortest_sequence(word, keyboard)):
        total_pushes += press_count * keyboard_press_count_multiplier(press_count)

    return total_pushes

def get_button_press_counts(button_press_sequence):
    button_counts = {}
    for button in button_press_sequence:
        if button in button_counts:
            button_counts[button] += 1
        else:
            button_counts[button] = 1
    return button_counts.items()

def keyboard_press_count_multiplier(keyboard_press_count):
    return keyboard_press_count

def find_shortest_sequence(word, keyboard):
    button_press_sequence = []
    for char in word:
        for button_index in range(len(keyboard)):
            if char in keyboard[button_index]:
                button_press_sequence.append(button_index + 2)
                break
    return button_press_sequence

Big(O) Analysis

Time Complexity

O(k^n) – The brute force approach considers all possible sequences of button pushes. Let n be the length of the target word and k be the number of possible button presses (effectively the size of the keypad, approximately 10 if considering digits only). For each character in the target word, we are exploring 'k' possibilities. Therefore, in the worst case, we're exploring all sequences of length n, where each position has k choices. This results in k^n possible sequences. The complexity is thus exponential in the length of the word.

Space Complexity

O(1) – The brute force approach, as described, primarily involves generating and checking sequences of button pushes. While generating each sequence, temporary storage might be used to construct the sequence, but this memory is released after checking if the sequence produces the target word. The key aspect is that at any given time, the algorithm only needs to store the 'shortest sequence found so far' and a counter for button pushes, irrespective of the input string length N. Therefore, the auxiliary space used is constant, resulting in O(1) space complexity.

Optimal Solution

Approach

The key is to realize that some letters on the phone keypad are typed with one push, some with two, and some with three or four. We want to assign the most frequent letters to the keys that require fewer pushes to minimize the overall count. The general idea is to sort the letters based on how often they appear and then assign them to the keys that require the fewest presses.

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

First, count how many times each letter appears in the input word.
Next, rank the letters from most frequent to least frequent.
Imagine the phone keypad. The letters that need only one press are the most valuable, then the letters that need two, and so on.
Assign the most frequent letters to the keys needing only one press. Then assign the next most frequent to keys needing two presses, and so on.
Keep assigning letters based on their frequency until all letters in the word have been assigned to a key.
Finally, for each letter in the original word, determine the number of presses needed based on its assigned key and sum these values up. This sum represents the minimum number of pushes to type the word.

Code Implementation

def minimum_number_of_pushes(word):
    letter_counts = {}
    for letter in word:
        letter_counts[letter] = letter_counts.get(letter, 0) + 1

    # Sort letters by frequency in descending order.
    sorted_letter_counts = sorted(letter_counts.items(), key=lambda item: item[1], reverse=True)

    presses_needed = {}
    total_pushes = 0
    index = 0
    for i in range(1, 5):
        for j in range(9):

            # Ensure index is within the bounds of the sorted letters.
            if index < len(sorted_letter_counts):
                letter = sorted_letter_counts[index][0]
                presses_needed[letter] = i
                index += 1
            else:
                break
        else:
            continue
        break

    # Calculate total pushes based on letter frequencies.
    for letter in word:
        total_pushes += presses_needed[letter]

    return total_pushes

Big(O) Analysis

Time Complexity

O(n log n) – Counting the frequency of each character in the input word takes O(n) time, where n is the length of the word. Sorting the character frequencies dominates the time complexity. Sorting n character counts using an efficient algorithm like merge sort or quicksort takes O(n log n) time. The remaining operations, like assigning letters to keys and calculating the total presses, take O(n) time. Therefore, the overall time complexity is dominated by the sorting step, resulting in O(n log n).

Space Complexity

O(1) – The algorithm uses a frequency counter, which stores counts for each letter of the alphabet. Since the alphabet size is fixed (26 in this case), the space used by the counter is constant. The remaining operations involve assigning letters to keys and summing up pushes, which are performed using a fixed number of variables, independent of the length of the input word, N. Therefore, the auxiliary space used is constant, regardless of the input word's length.

Edge Cases

Case	How to Handle
Null or empty input string	Return 0, as no pushes are required for an empty word.
Input string with a single character	Return 1, as a single character always requires one push.
Input string containing characters outside the 'a' to 'z' range	Handle as an invalid input and either return an error code or throw an exception.
Input string with all the same characters (e.g., 'aaaa')	The frequency map ensures efficient counting even when all characters are identical, properly accumulating pushes.
Input string with all unique characters (26 characters long)	The solution should correctly iterate through all characters and calculate the pushes needed based on their frequency/order.
Very long input string (potentially exceeding memory constraints)	Ensure the frequency map is implemented using space efficient data structures, and avoid unnecessary memory allocation.
String with a skewed distribution where some letters are very frequent while others are rare	The solution will prioritize less frequent letters to use with single pushes, leading to an optimized count.
Integer overflow when calculating total presses if input string is extremely long	Use a data type with a large range (e.g., long) to store the total number of presses to prevent overflow.

Null or empty input string

How to Handle:

Return 0, as no pushes are required for an empty word.

Input string with a single character

How to Handle:

Return 1, as a single character always requires one push.

Input string containing characters outside the 'a' to 'z' range

How to Handle:

Handle as an invalid input and either return an error code or throw an exception.

Input string with all the same characters (e.g., 'aaaa')

How to Handle:

The frequency map ensures efficient counting even when all characters are identical, properly accumulating pushes.

Input string with all unique characters (26 characters long)

How to Handle:

The solution should correctly iterate through all characters and calculate the pushes needed based on their frequency/order.

Very long input string (potentially exceeding memory constraints)

How to Handle:

Ensure the frequency map is implemented using space efficient data structures, and avoid unnecessary memory allocation.

String with a skewed distribution where some letters are very frequent while others are rare

How to Handle:

The solution will prioritize less frequent letters to use with single pushes, leading to an optimized count.

Integer overflow when calculating total presses if input string is extremely long

How to Handle:

Use a data type with a large range (e.g., long) to store the total number of presses to prevent overflow.