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Minimum Number of Keypresses

Medium

Asked by:

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Greedy AlgorithmsArrays

You are given a string s consisting of lowercase English letters. A key press of length k is defined as pressing one key for k times consecutively.

For example, the string "aabbaa" can be typed by two key presses of length 2 ("aa" and "aa") and two key presses of length 1 ("b" and "b").

You have to rearrange the string s such that the total length of key presses is minimized. Return the minimum length of key presses after rearranging the string s.

Example 1:

Input: s = "aabbcc"
Output: 6
Explanation: The optimal arrangement is "aabbcc" which has 6 key presses (each letter is pressed once).

Example 2:

Input: s = "aabbaa"
Output: 4
Explanation: The optimal arrangement is "aaaaabb" which has 2 key presses ("aaaaaa" and "bb").

Example 3:

Input: s = "aaabbbbccccddeeeeefffffggggghhhhhhiiiiiijjjjjjkkkkkkllllllmmmmmnnnnnnooooooppppppqqqqqqrrrrrrssssssttttttuuuuuuvvvvvvwwwwwwxxxxxxzzzzzzz"
Output: 13

Constraints:

1 <= s.length <= 10⁵
s consists of lowercase English letters.

Try coding it in LeetCode

Minimum Number of Keypresses

Medium

Asked by:

33 views

Topics:

Greedy AlgorithmsArrays

You are given a string s consisting of lowercase English letters. A key press of length k is defined as pressing one key for k times consecutively.

For example, the string "aabbaa" can be typed by two key presses of length 2 ("aa" and "aa") and two key presses of length 1 ("b" and "b").

You have to rearrange the string s such that the total length of key presses is minimized. Return the minimum length of key presses after rearranging the string s.

Example 1:

Input: s = "aabbcc"
Output: 6
Explanation: The optimal arrangement is "aabbcc" which has 6 key presses (each letter is pressed once).

Example 2:

Input: s = "aabbaa"
Output: 4
Explanation: The optimal arrangement is "aaaaabb" which has 2 key presses ("aaaaaa" and "bb").

Example 3:

Input: s = "aaabbbbccccddeeeeefffffggggghhhhhhiiiiiijjjjjjkkkkkkllllllmmmmmnnnnnnooooooppppppqqqqqqrrrrrrssssssttttttuuuuuuvvvvvvwwwwwwxxxxxxzzzzzzz"
Output: 13

Constraints:

1 <= s.length <= 10⁵
s consists of lowercase 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:

What is the maximum length of the input string, and can it contain characters beyond lowercase English letters?
Are there any constraints on the frequency of each character in the input string?
If multiple key assignments result in the same minimum number of keypresses, is any valid assignment acceptable?
Is the empty string a valid input, and if so, what should the output be?
Can you provide a specific example of the mapping from characters to keypresses that the problem is asking for?

Brute Force Solution

Approach

The brute force strategy for finding the minimum number of keypresses involves trying every possible arrangement of assigning letters to keys. We explore all combinations and calculate the total keypresses for each combination. Finally, we pick the arrangement that results in the fewest keypresses.

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

Consider all possible ways to assign each unique letter to a number from 1 to 9, representing the keys on a phone.
For each possible assignment, figure out which press it is for each letter (first, second, or third) based on how many letters are assigned to that key.
Calculate the total number of key presses needed for the entire input string, given the current letter assignment.
Keep track of the minimum number of key presses seen so far, and update it whenever you find an assignment with fewer key presses.
After checking all possible assignments, return the smallest number of key presses you found.

Code Implementation

import itertools

def minimum_number_of_keypresses_brute_force(input_string):
    unique_characters = sorted(list(set(input_string)))
    minimum_keypresses = float('inf')

    # Generate all possible permutations of assigning unique characters to keys (1-9)
    for permutation in itertools.permutations(range(1, 10), len(unique_characters)):
        
        character_to_key_assignment = {}
        for index, character in enumerate(unique_characters):
            character_to_key_assignment[character] = permutation[index]

        # Calculate keypress counts for each key based on character assignment
        key_to_characters_assignment = {key: [] for key in range(1, 10)}
        for character, key in character_to_key_assignment.items():
            key_to_characters_assignment[key].append(character)

        current_keypresses = 0
        for character in input_string:
            key = character_to_key_assignment[character]
            characters_on_key = key_to_characters_assignment[key]
            
            # Determine the keypress count based on the position of the character on the key
            keypress_count = characters_on_key.index(character) + 1
            current_keypresses += keypress_count
        
        # Update the minimum if current is the lowest        minimum_keypresses = min(minimum_keypresses, current_keypresses)

    return minimum_keypresses

Big(O) Analysis

Time Complexity

O(A! * N) – The brute force solution considers all possible assignments of characters to keys. If there are A unique letters in the input string, there are A! (A factorial) ways to assign these letters to the 9 keys. For each possible assignment, we iterate through the input string of length N to calculate the total keypresses. Therefore, the overall time complexity is O(A! * N), where A is the number of unique characters and N is the length of the input string. Note that in the worst case A can be approximately equal to N (all characters unique).

Space Complexity

O(1) – The provided brute force algorithm primarily uses a fixed number of variables to store the minimum keypresses found so far and to iterate through the possible letter assignments. Although conceptually it explores all possible assignments of letters to phone keys and calculates the keypresses for each, it does not explicitly store all these assignments or intermediate keypress counts in large data structures. Therefore, the auxiliary space used remains constant, regardless of the length of the input string, N, making the space complexity O(1).

Optimal Solution

Approach

The most efficient approach is to assign the most frequent letters to the easiest key presses, meaning those requiring the fewest number of presses. After counting the occurrences of each letter, we'll prioritize the most frequently used ones.

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

First, count how many times each letter appears in all the words.
Then, sort the letters based on how frequently they appear, putting the most frequent letters first.
Assign the most frequent letters to the keys that need one keypress. Then, assign the next most frequent letters to the keys that need two keypresses, and so on.
Keep doing this until all letters have been assigned a number of keypresses.
For each letter, multiply its frequency by the number of keypresses it requires.
Add up all these products to get the minimum number of keypresses required.

Code Implementation

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

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

    total_key_presses = 0
    key_press_count = 1
    keys_assigned = 0

    # Assign letters to key presses based on frequency.
    for letter, count in sorted_letters:
        total_key_presses += count * key_press_count
        keys_assigned += 1
        if keys_assigned % 9 == 0:
            key_press_count += 1

    return total_key_presses

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm first counts the frequency of each letter, which takes O(n) time where n is the total number of characters in the input words. Then, it sorts the letters based on their frequency. Sorting typically takes O(k log k) time, where k is the number of unique letters (at most 26). Since k is constant and relatively small, we can approximate the sorting time as O(log k) or consider the number of input word n. This sorting step dominates the overall time complexity. Therefore, the total time complexity is O(n + k log k), which simplifies to O(n log n) when the number of words affects the letter occurrences.

Space Complexity

O(1) – The space complexity is dominated by the storage needed to count letter frequencies and sort them. While counting frequencies could use a hash map of size 26 (constant), the sorting process, although conceptually in-place, might use a stack of logarithmic depth during the recursive calls (e.g., quicksort or mergesort). Given that there are at most 26 letters, the sort operations occupy constant space. Therefore, we only need to store a fixed number of variables, like the frequencies and intermediate variables during sorting, leading to constant space usage regardless of the input string's length. Thus the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input string	Return 0 if the input string is null or empty as no keypresses are needed.
Input string with a single character	Return 1, as a single character needs only one keypress.
Input string with all identical characters	The solution should correctly calculate keypresses by assigning higher keypress counts to these frequent characters.
Input string with a very long length	Ensure the solution scales efficiently, considering potential memory or time complexity issues with very large inputs; time complexity should ideally be O(n log n) from sorting, and can be optimized further by avoiding full sorting with use of data structures.
Input string containing non-alphabetic characters (e.g., numbers, symbols)	Filter out or ignore any non-alphabetic characters or handle them according to the problem specification (e.g., treat them as having the lowest frequency).
Input string with a highly skewed distribution of character frequencies	The solution should correctly assign keypresses even with uneven distributions, by giving the most frequent characters the lowest number of presses.
String with exactly 26 distinct characters	The first occurrence of each character will have a keypress of 1, and subsequent occurrences depend on frequency, requiring more keypresses than characters appearing before.
Very long string with repeating patterns, but non-uniform distribution	The character frequency count must be accurate to prevent errors in assigning keypresses based on actual occurrences and sorting these occurrences to correctly assign keypress value.

Null or empty input string

How to Handle:

Return 0 if the input string is null or empty as no keypresses are needed.

Input string with a single character

How to Handle:

Return 1, as a single character needs only one keypress.

Input string with all identical characters

How to Handle:

The solution should correctly calculate keypresses by assigning higher keypress counts to these frequent characters.

Input string with a very long length

How to Handle:

Ensure the solution scales efficiently, considering potential memory or time complexity issues with very large inputs; time complexity should ideally be O(n log n) from sorting, and can be optimized further by avoiding full sorting with use of data structures.

Input string containing non-alphabetic characters (e.g., numbers, symbols)

How to Handle:

Filter out or ignore any non-alphabetic characters or handle them according to the problem specification (e.g., treat them as having the lowest frequency).

Input string with a highly skewed distribution of character frequencies

How to Handle:

The solution should correctly assign keypresses even with uneven distributions, by giving the most frequent characters the lowest number of presses.

String with exactly 26 distinct characters

How to Handle:

The first occurrence of each character will have a keypress of 1, and subsequent occurrences depend on frequency, requiring more keypresses than characters appearing before.

Very long string with repeating patterns, but non-uniform distribution

How to Handle:

The character frequency count must be accurate to prevent errors in assigning keypresses based on actual occurrences and sorting these occurrences to correctly assign keypress value.

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:

What is the maximum length of the input string, and can it contain characters beyond lowercase English letters?
Are there any constraints on the frequency of each character in the input string?
If multiple key assignments result in the same minimum number of keypresses, is any valid assignment acceptable?
Is the empty string a valid input, and if so, what should the output be?
Can you provide a specific example of the mapping from characters to keypresses that the problem is asking for?

Brute Force Solution

Approach

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

Consider all possible ways to assign each unique letter to a number from 1 to 9, representing the keys on a phone.
For each possible assignment, figure out which press it is for each letter (first, second, or third) based on how many letters are assigned to that key.
Calculate the total number of key presses needed for the entire input string, given the current letter assignment.
Keep track of the minimum number of key presses seen so far, and update it whenever you find an assignment with fewer key presses.
After checking all possible assignments, return the smallest number of key presses you found.

Code Implementation

import itertools

def minimum_number_of_keypresses_brute_force(input_string):
    unique_characters = sorted(list(set(input_string)))
    minimum_keypresses = float('inf')

    # Generate all possible permutations of assigning unique characters to keys (1-9)
    for permutation in itertools.permutations(range(1, 10), len(unique_characters)):
        
        character_to_key_assignment = {}
        for index, character in enumerate(unique_characters):
            character_to_key_assignment[character] = permutation[index]

        # Calculate keypress counts for each key based on character assignment
        key_to_characters_assignment = {key: [] for key in range(1, 10)}
        for character, key in character_to_key_assignment.items():
            key_to_characters_assignment[key].append(character)

        current_keypresses = 0
        for character in input_string:
            key = character_to_key_assignment[character]
            characters_on_key = key_to_characters_assignment[key]
            
            # Determine the keypress count based on the position of the character on the key
            keypress_count = characters_on_key.index(character) + 1
            current_keypresses += keypress_count
        
        # Update the minimum if current is the lowest        minimum_keypresses = min(minimum_keypresses, current_keypresses)

    return minimum_keypresses

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, count how many times each letter appears in all the words.
Then, sort the letters based on how frequently they appear, putting the most frequent letters first.
Assign the most frequent letters to the keys that need one keypress. Then, assign the next most frequent letters to the keys that need two keypresses, and so on.
Keep doing this until all letters have been assigned a number of keypresses.
For each letter, multiply its frequency by the number of keypresses it requires.
Add up all these products to get the minimum number of keypresses required.

Code Implementation

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

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

    total_key_presses = 0
    key_press_count = 1
    keys_assigned = 0

    # Assign letters to key presses based on frequency.
    for letter, count in sorted_letters:
        total_key_presses += count * key_press_count
        keys_assigned += 1
        if keys_assigned % 9 == 0:
            key_press_count += 1

    return total_key_presses

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input string	Return 0 if the input string is null or empty as no keypresses are needed.
Input string with a single character	Return 1, as a single character needs only one keypress.
Input string with all identical characters	The solution should correctly calculate keypresses by assigning higher keypress counts to these frequent characters.
Input string with a very long length	Ensure the solution scales efficiently, considering potential memory or time complexity issues with very large inputs; time complexity should ideally be O(n log n) from sorting, and can be optimized further by avoiding full sorting with use of data structures.
Input string containing non-alphabetic characters (e.g., numbers, symbols)	Filter out or ignore any non-alphabetic characters or handle them according to the problem specification (e.g., treat them as having the lowest frequency).
Input string with a highly skewed distribution of character frequencies	The solution should correctly assign keypresses even with uneven distributions, by giving the most frequent characters the lowest number of presses.
String with exactly 26 distinct characters	The first occurrence of each character will have a keypress of 1, and subsequent occurrences depend on frequency, requiring more keypresses than characters appearing before.
Very long string with repeating patterns, but non-uniform distribution	The character frequency count must be accurate to prevent errors in assigning keypresses based on actual occurrences and sorting these occurrences to correctly assign keypress value.

Null or empty input string

How to Handle:

Return 0 if the input string is null or empty as no keypresses are needed.

Input string with a single character

How to Handle:

Return 1, as a single character needs only one keypress.

Input string with all identical characters

How to Handle:

The solution should correctly calculate keypresses by assigning higher keypress counts to these frequent characters.

Input string with a very long length

How to Handle:

Input string containing non-alphabetic characters (e.g., numbers, symbols)

How to Handle:

Filter out or ignore any non-alphabetic characters or handle them according to the problem specification (e.g., treat them as having the lowest frequency).

Input string with a highly skewed distribution of character frequencies

How to Handle:

The solution should correctly assign keypresses even with uneven distributions, by giving the most frequent characters the lowest number of presses.

String with exactly 26 distinct characters

How to Handle:

The first occurrence of each character will have a keypress of 1, and subsequent occurrences depend on frequency, requiring more keypresses than characters appearing before.

Very long string with repeating patterns, but non-uniform distribution

How to Handle:

The character frequency count must be accurate to prevent errors in assigning keypresses based on actual occurrences and sorting these occurrences to correctly assign keypress value.