Minimum Deletions to Make String K-Special #20 Most Asked
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Topics:
ArraysGreedy Algorithms
You are given a string word and an integer k.
We consider word to be k-special if |freq(word[i]) - freq(word[j])| <= k for all indices i and j in the string.
Here, freq(x) denotes the frequency of the character x in word, and |y| denotes the absolute value of y.
Return the minimum number of characters you need to delete to make word k-special.
Example 1:
Input: word = "aabcaba", k = 0
Output: 3
Explanation: We can make word 0-special by deleting 2 occurrences of "a" and 1 occurrence of "c". Therefore, word becomes equal to "baba" where freq('a') == freq('b') == 2.
Example 2:
Input: word = "dabdcbdcdcd", k = 2
Output: 2
Explanation: We can make word 2-special by deleting 1 occurrence of "a" and 1 occurrence of "d". Therefore, word becomes equal to "bdcbdcdcd" where freq('b') == 2, freq('c') == 3, and freq('d') == 4.
Example 3:
Input: word = "aaabaaa", k = 2
Output: 1
Explanation: We can make word 2-special by deleting 1 occurrence of "b". Therefore, word becomes equal to "aaaaaa" where each letter's frequency is now uniformly 6.
Constraints:
1 <= word.length <= 1050 <= k <= 105wordconsists only of lowercase English letters.
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 range of possible values for k, and how does it relate to the length of the string s?
- What characters can the string 's' contain? Are we limited to ASCII characters, or can it be Unicode?
- If the input string 's' is empty, what should be the return value?
- Is 'k' guaranteed to be a positive integer? What should happen if 'k' is zero?
- If it's impossible to make the string k-special (e.g., all characters appear a different number of times, and deleting all but one character doesn't solve it), is there a specific value I should return, or can I throw an exception?
Brute Force Solution
Approach
To solve this problem using brute force, we explore every possible combination of deletions. We want to find the smallest number of character removals needed to make the string satisfy the 'K-special' condition by exhaustively checking all possibilities.
Here's how the algorithm would work step-by-step:
- Start by considering every possible way to remove characters from the string.
- For each of these possibilities, divide the resulting string into groups (or substrings) of exactly size 'K'. If the final group is smaller than 'K', discard this option.
- Check if each of these groups are equal to each other. If they are, this is a valid solution. If not, discard it.
- Count the number of characters that were removed to achieve this arrangement.
- Repeat the above steps for all possible removal combinations.
- After exploring all possibilities, select the solution with the minimum number of removals. If no valid solutions exist, indicate that it's not possible.
Code Implementation
def min_deletions_k_special_brute_force(input_string, k_value):
minimum_deletions = float('inf')
for substring_start_index in range(len(input_string)):
for substring_end_index in range(substring_start_index, len(input_string)):
substring = input_string[substring_start_index:substring_end_index + 1]
# Check if the substring meets the k-special condition
if is_k_special(substring, k_value):
# Calculate the number of deletions required
deletions_needed = len(input_string) - len(substring)
# Update the minimum deletions if necessary
minimum_deletions = min(minimum_deletions, deletions_needed)
if minimum_deletions == float('inf'):
return -1 # No k-special substring found
else:
return minimum_deletions
def is_k_special(input_string, k_value):
if not input_string:
return False
frequency_map = {}
for char in input_string:
frequency_map[char] = frequency_map.get(char, 0) + 1
# Check if all frequencies are divisible by k
for frequency in frequency_map.values():
if frequency % k_value != 0:
return False
return True
Big(O) Analysis
Time Complexity
O(2^n) – The brute force approach considers all possible subsets of the input string of length n, which corresponds to 2^n possibilities (each character can either be kept or deleted). For each of these subsets, we perform additional checks: dividing the remaining string into groups of size K, and comparing these groups. While the grouping and comparison take O(n) time in the worst case, the dominant factor remains the generation of all possible subsets. Therefore, the overall time complexity is O(2^n), primarily driven by exploring the exponential number of deletion combinations.
Space Complexity
O(2^N) – The brute force approach explores all possible combinations of deletions. This can be visualized as a binary tree where each node represents a decision to either delete or keep a character. In the worst-case scenario, where N is the length of the string, this tree has a depth of N, and the total number of nodes (and thus possible combinations stored at some point implicitly or explicitly during recursion or iteration) can grow up to 2^N. Therefore, the auxiliary space complexity is O(2^N) due to the need to store or process all possible combinations of deleted characters during the exhaustive search.
Optimal Solution
Approach
The goal is to find the fewest letters to remove from a string so that every chunk of 'k' consecutive letters has the same number of occurrences of each letter of the alphabet. To do this efficiently, we calculate how often each letter appears in each 'k'-letter chunk and remove the least frequent letters to equalize the counts. We then compare how many letters we remove for each of the 26 possible characters to decide which character to optimize towards.
Here's how the algorithm would work step-by-step:
- Divide the string into overlapping chunks, each containing 'k' consecutive letters.
- For each chunk, count how many times each letter of the alphabet appears.
- For each letter of the alphabet, imagine making that letter the 'special' letter.
- Within each chunk, calculate how many letters you would need to delete to make the counts of other letters equal to the count of that special letter.
- Add up all the deletions needed across all chunks, for each potential special letter.
- Pick the special letter that results in the fewest total deletions.
- Return the minimum number of required deletions.
Code Implementation
def minimum_deletions_to_make_string_k_special(input_string, group_size):
string_length = len(input_string)
total_deletions = 0
# Iterate through the string in steps of group_size.
for i in range(0, string_length, group_size):
group = input_string[i:i + group_size]
group_length = len(group)
letter_counts = {}
# Count occurrences of each letter in the current group.
for letter in group:
letter_counts[letter] = letter_counts.get(letter, 0) + 1
most_frequent_letter_count = 0
# Find the frequency of the most frequent letter.
for letter in letter_counts:
most_frequent_letter_count = max(most_frequent_letter_count, letter_counts[letter])
# Add the deletions needed for the current group.
total_deletions += group_length - most_frequent_letter_count
return total_deletionsBig(O) Analysis
Time Complexity
O(26 * n * k) – The algorithm iterates through the string of length 'n' to create 'n-k+1' overlapping chunks of size 'k'. For each of these chunks, it counts the frequency of each of the 26 letters of the alphabet, which takes O(k) time per chunk. Then, for each of the 26 letters, it calculates the number of deletions needed across all chunks, which involves iterating through the 'n-k+1' chunks. Thus, the total complexity is O(26 * (n-k+1) * k). Assuming k is smaller than n, the overall time complexity can be approximated as O(26 * n * k).
Space Complexity
O(K) – The algorithm stores the character counts for each chunk of size K. In each chunk, we count the occurrences of each alphabet character, requiring a space proportional to the alphabet size, which is constant (26). While the algorithm iterates through all possible 'special' letters, this doesn't introduce additional space complexity as it reuses the existing count data structures. Therefore, the auxiliary space is dominated by the space to store character counts for each chunk, which depends only on the chunk size K and is independent of the string's length N.
Edge Cases
| Case | How to Handle |
|---|---|
| Empty string s | Return 0 since an empty string is trivially k-special for any k (although k=0 requires a check). |
| Null string s | Throw IllegalArgumentException or return -1 to indicate invalid input. |
| k = 0 | If k is 0, any non-empty string cannot be made k-special, so return the length of the string s; otherwise, an empty string is a valid k-special string, so return 0. |
| k is larger than the string length | Return the string length since all characters need to be deleted to fulfill the k-special condition. |
| String contains only one distinct character | If the string length is a multiple of k, return 0; otherwise, return length % k. |
| Maximum string length | Ensure the frequency counting mechanism (e.g., hash map) scales efficiently with large string sizes to avoid performance bottlenecks; consider using array-based frequency counting if character set is limited. |
| String with all unique characters and k > 1 | Return the string length since you need to delete all characters to satisfy the condition. |
| Integer overflow when calculating deletions | Use appropriate data types (e.g., long) for storing character counts and deletions, especially when dealing with very large string lengths to avoid integer overflow issues. |
Empty string s
How to Handle:
Return 0 since an empty string is trivially k-special for any k (although k=0 requires a check).
Null string s
How to Handle:
Throw IllegalArgumentException or return -1 to indicate invalid input.
k = 0
How to Handle:
If k is 0, any non-empty string cannot be made k-special, so return the length of the string s; otherwise, an empty string is a valid k-special string, so return 0.
k is larger than the string length
How to Handle:
Return the string length since all characters need to be deleted to fulfill the k-special condition.
String contains only one distinct character
How to Handle:
If the string length is a multiple of k, return 0; otherwise, return length % k.
Maximum string length
How to Handle:
Ensure the frequency counting mechanism (e.g., hash map) scales efficiently with large string sizes to avoid performance bottlenecks; consider using array-based frequency counting if character set is limited.
String with all unique characters and k > 1
How to Handle:
Return the string length since you need to delete all characters to satisfy the condition.
Integer overflow when calculating deletions
How to Handle:
Use appropriate data types (e.g., long) for storing character counts and deletions, especially when dealing with very large string lengths to avoid integer overflow issues.
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