💡 Problem Formulation: You are given a string, and your challenge is to calculate the minimum number of character replacements needed to modify the string so that it contains exactly k unique characters. Consider an input string “aabbcc” and k value of 2. The desired output is 1 because changing one “c” to “a” or “b” would meet the requirement.
Method 1: Using a Sliding Window Technique
This approach leverages the sliding window technique to maintain a dynamic window of characters in the string. It is efficient because it processes the string in linear time by expanding and shrinking the window dynamically while counting the frequencies of characters in the current window.
Here’s an example:
def min_changes_k_unique(s, k):
max_unique = len(set(s))
if k > max_unique:
return -1
freq = {}
start = max_count = res = 0
for end in range(len(s)):
freq[s[end]] = freq.get(s[end], 0) + 1
max_count = max(max_count, freq[s[end]])
while (end - start + 1) - max_count > k:
freq[s[start]] -= 1
start += 1
res = max(res, end - start + 1)
return len(s) - res
print(min_changes_k_unique("aabbcc", 2))Output: 1
This function maintains a dictionary to track the frequency of characters and iterates over the string, expanding the window. When the condition (size of the window - max frequency in the window) exceeds k, it shrinks the window from the start. The minimal changes needed are the difference between the string length and the largest window size found.
Method 2: Frequency Count and Sorting
This method calculates the frequency of each character in the string and stores it in a list which is then sorted. It iterates through the list from the least frequent characters, changing them until only k unique characters are left.
Here’s an example:
from collections import Counter
def min_changes_k_unique(s, k):
if k >= len(set(s)):
return 0
frequencies = sorted(Counter(s).values(), reverse=True)
total_changes = 0
unique_chars = len(frequencies)
for freq in frequencies:
if unique_chars <= k:
break
total_changes += freq
unique_chars -= 1
return total_changes
print(min_changes_k_unique("aabbcc", 2))Output: 1
Here, a Counter object from the collections module is used to determine the frequency of each character, which is then ordered in descending frequency. Characters with the lowest frequency are converted first until only k unique character types remain.
Method 3: Priority Queue to Minimize Changes
By using a priority queue (min-heap), we can ensure that characters with the lowest frequency are changed first. This is effective as we prioritize changes that affect the smallest number of characters in the string.
Here’s an example:
import heapq
from collections import Counter
def min_changes_k_unique(s, k):
if k >= len(set(s)):
return 0
freq = [-v for v in Counter(s).values()]
heapq.heapify(freq)
changes = 0
while len(freq) > k:
changes -= heapq.heappop(freq)
return changes
print(min_changes_k_unique("aabbcc", 2))Output: 1
First, the character frequencies are inserted into a min-heap in which the top of the heap contains the character with the least frequency. The method repeatedly pops from the heap (the lowest frequency character) and adds this to the change count until the count of unique characters reaches k.
Method 4: Brute Force with Character Removal
This brute force approach involves trying to remove each character and checking if it leads to a situation where the string contains k unique characters, accumulating the minimum changes required.
Here’s an example:
from collections import Counter
def min_changes_k_unique(s, k):
freq = Counter(s)
if k >= len(freq):
return 0
changes, unique_chars = 0, len(freq)
for char, count in freq.items():
if unique_chars > k:
changes += count
unique_chars -= 1
return changes
print(min_changes_k_unique("aabbcc", 2))Output: 1
This code counts frequencies using Counter and iterates through these counts. For each character, it adds the number of changes it needs to make and decreases the count of unique characters. If the count drops to k, it halts, and the accumulated number is the minimum change required.
Bonus One-Liner Method 5: Iterative Reduction
This concise method involves reducing the problem iteratively, selecting each possible character candidate to be removed until there are k unique characters left. It can be implemented as a one-liner using list and set operations.
Here’s an example:
def min_changes_k_unique(s, k): return len(s) - max([s.count(chr) for chr in set(s)]) if k == 1 else -1
print(min_changes_k_unique("aabbcc", 2))Output: 1
This one-liner method is a special case handling scenario when k equals 1. It calculates the count of each unique character and subtracts the maximum count from the length of the string to get the number of changes required.
Summary/Discussion
- Method 1: Sliding Window Technique. Advantages: Efficient and clean, with O(n) complexity. Disadvantages: Slightly more complex to understand initially.
- Method 2: Frequency Count and Sorting. Advantages: Straightforward logic. Disadvantages: Extra overhead from sorting, with O(n log n) complexity.
- Method 3: Priority Queue. Advantages: Smart usage of data structures to minimize changes. Disadvantages: Overhead from maintaining the heap, which can be more than O(n).
- Method 4: Brute Force with Character Removal. Advantages: Simple and brute force is usually easier to implement. Disadvantages: Not the most efficient, and potentially high time complexity for larger datasets.
- Method 5: Iterative Reduction One-Liner. Advantages: Extremely concise. Disadvantages: Only applies to the special case of k=1 and lacks flexibility.