Minimum Number of Pushes to Type Word II
Medium
You are given a string word containing 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.
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 = "xyzxyzxyzxyz" Output: 12 Explanation: The remapped keypad given in the image provides the minimum cost. "x" -> one push on key 2 "y" -> one push on key 3 "z" -> one push on key 4 Total cost is 1 * 4 + 1 * 4 + 1 * 4 = 12 It can be shown that no other mapping can provide a lower cost. Note that the key 9 is not mapped to any letter: it is not necessary to map letters to every key, but to map all the letters.
Example 3:
Input: word = "aabbccddeeffgghhiiiiii" Output: 24 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 "f" -> one push on key 7 "g" -> one push on key 8 "h" -> two pushes on key 9 "i" -> one push on key 9 Total cost is 1 * 2 + 1 * 2 + 1 * 2 + 1 * 2 + 1 * 2 + 1 * 2 + 1 * 2 + 2 * 2 + 6 * 1 = 24. It can be shown that no other mapping can provide a lower cost.
Can you write an algorithm for the problem?
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_pushesBig(O) Analysis
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_pressesBig(O) Analysis
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. |
