Minimum Number of Pushes to Type Word II

Medium

Minimum Number of Pushes to Type Word II

Medium

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 other than lowercase English letters?
What is the maximum length of the input word? Is there a limit to the number of distinct characters within the word?
If the input word is empty, should I return 0?
Is there a standard phone keypad layout I should assume, or is it provided?
If multiple key assignments result in the same minimum number of pushes, is any valid assignment acceptable?

Brute Force Solution

Approach

The brute-force approach for this typing problem is like trying out every possible keyboard layout and sequence of pushes. We essentially simulate typing the given word on every possible arrangement of letters on the keypad. We then count the number of pushes needed for each arrangement and find the minimum.

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

Consider every possible way to assign letters to the keypad buttons.
For each possible letter assignment, simulate typing the given word.
While simulating, count how many pushes are needed for each letter based on the button it's assigned to.
Add up the total number of pushes for the entire word for the current letter assignment.
Compare the number of pushes needed for the current letter assignment with the minimum number of pushes found so far.
If the current assignment requires fewer pushes, update the minimum number of pushes.
Repeat the process for all possible letter assignments.
After checking all possibilities, report the lowest number of pushes found as the answer.

Code Implementation

import itertools

def minimum_number_of_pushes_brute_force(word):
    # Initialize minimum pushes to a large value
    minimum_pushes = float('inf')
    letters = sorted(list(set(word)))
    
    # Iterate through all possible letter assignments
    for permutation in itertools.permutations(letters):
        letter_assignment = dict(zip(range(26), permutation))
        
        total_pushes = 0
        # Calculate pushes for current assignment
        for char in word:
            for key_number in range(26):
                if letter_assignment[key_number] == char:
                    # Determine number of pushes based on the key
                    pushes = (key_number // 8) + 1
                    total_pushes += pushes
                    break
        
        # Update minimum pushes if needed
        minimum_pushes = min(minimum_pushes, total_pushes)
    
    return minimum_pushes

Big(O) Analysis

Time Complexity

O((26!)*(N)) – The algorithm considers every possible assignment of the 26 letters of the alphabet to the keypad buttons, which is 26! (26 factorial). For each of these assignments, it simulates typing a word of length N. The typing simulation involves iterating through each character of the input word to determine the number of pushes needed based on the current letter assignment. Therefore, for each permutation it does N operations. Multiplying these gives (26!)*N. Since we only care about the dominant term in Big O notation, the time complexity is O((26!)*(N)).

Space Complexity

O(1) – The brute-force approach, as described, primarily involves simulating letter assignments and counting pushes. While it iterates through possible letter arrangements, it doesn't inherently create large, dynamically-sized data structures proportional to the input word's length (N). The memory used is for storing the current minimum number of pushes and temporary variables to track letter assignments and push counts during simulation, remaining constant regardless of N. Thus, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The core idea is to figure out how frequently each letter appears and then strategically assign the most frequent letters to the easiest-to-reach keys on a phone's keypad. This minimizes the total number of key presses needed to type the word. We focus on the most commonly used letters first.

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

Count how many times each letter shows up in the given word.
Rank the letters based on how many times they appear, from most frequent to least frequent.
Imagine the phone keypad. Some keys are easier to press because you only need to press them once to get the letter. Other keys require pressing twice, three times, or even four times to get to a specific letter.
Assign the most frequent letters to the keys that require the fewest presses. For example, the three most frequent letters should go on the keys that require only one press. The next three go on the keys that need two presses, and so on.
Calculate the total number of presses needed based on this assignment. For each letter, multiply its frequency by the number of presses needed to type it according to your assignment. Sum these values to get the minimum number of pushes.

Code Implementation

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

    # Sort letters by frequency, most frequent first.
    sorted_letter_frequencies = sorted(letter_frequencies.items(), key=lambda item: item[1], reverse=True)

    keypad_presses = 0
    press_count = 1
    letter_index = 0

    #Assign higher frequencies to lower press counts
    for letter, frequency in sorted_letter_frequencies:
        keypad_presses += frequency * press_count
        letter_index += 1
        if letter_index % 9 == 0:
            press_count += 1

    return keypad_presses

Big(O) Analysis

Time Complexity

O(n) – Let n be the length of the input word. Counting letter frequencies involves iterating through the word once, taking O(n) time. Ranking the letters based on frequency can be done using sorting. Assuming the distinct number of characters will usually be less than or equal to 26, treating it like a constant, we can say sorting takes constant time, or is dominated by the earlier O(n) operation. Assigning the most frequent letters to keys and calculating presses also takes constant time, given the constraints. Thus, the dominant operation is counting letter frequencies, which is O(n).

Space Complexity

O(1) – The algorithm uses a constant amount of extra space. It counts the frequency of each letter, which, since we're dealing with a fixed alphabet (26 lowercase letters), requires space for at most 26 counters regardless of the input string's length (N). Furthermore, the space needed to store the ranked letters based on frequency and calculate the total presses remains constant and independent of N.

Edge Cases

Case	How to Handle
Empty word list	Return 0 as no pushes are needed for an empty word list.
Null or empty input word	Treat null/empty word as requiring 0 pushes or throw an IllegalArgumentException, depending on desired behavior.
Word contains non-alphabetic characters	Filter out or reject words with non-alphabetic characters, or map them to a default key press sequence.
All words are identical	The frequency counting should still accurately reflect the character distribution, so the algorithm works correctly.
Very long words	The algorithm's time complexity is determined by the word length, so very long words will increase execution time.
Maximum-sized input word list (e.g., many words of considerable length)	Ensure the data structures used (e.g., frequency map) are appropriately sized to prevent memory issues and optimize lookup.
Word contains repeated characters consecutively, like 'AAA'	Character frequency is calculated correctly, handling repetitions without issue.
Words with characters that need more than one push on a key (e.g., 's' requires 3 pushes)	Ensure the push count is correctly calculated based on the character's position on the phone keypad for each key.