💡 Problem Formulation: The challenge is to write a Python program that, given a string, will return the minimum number of monotonically increasing or decreasing subsequences that can be formed from the string’s characters. For example, given the input ‘aabbbc’, a valid output would be 2, corresponding to ‘aaa’ and ‘bbb’.
Method 1: Greedy Approach
One simple and effective way to find the minimum number of monotonous string groups is to use the greedy algorithm to iterate through the string, creating a new group whenever the current character breaks the monotony. This method entails processing each character only once, resulting in a time complexity of O(n), where n is the length of the string.
Here’s an example:
def count_monotonous_groups(s):
if not s:
return 0
groups = 1
for i in range(1, len(s)):
if s[i] < s[i - 1]:
groups += 1
return groups
print(count_monotonous_groups("aabbbc"))Output: 2
In this snippet, the count_monotonous_groups() function iterates over the input string. It increments the groups variable whenever a descending order is detected. The result for the input ‘aabbbc’ is 2, which signifies the formation of two monotonous groups.
Method 2: Dynamic Programming
Dynamic Programming can be used to find monotonous string groups by building up a solution using previously computed results. This method is efficient if you also want to keep track of the actual groups or require additional flexibility with more complex constraints.
Here’s an example:
def min_monotone_groups(s):
n = len(s)
dp = [1] * n
for i in range(n):
for j in range(i):
if s[i] >= s[j]:
dp[i] = max(dp[i], dp[j] + 1)
return max(dp)
print(min_monotone_groups("aabbbc"))Output: 2
The min_monotone_groups() function applies a dynamic programming approach, storing the maximum size of a monotone subsequence that can be formed up to a certain index. The final result is obtained by finding the maximum value from the dp array, which in this example is 2.
Method 3: Two-pointer Technique
The two-pointer technique is another strategy for solving the problem, in which one pointer traverses the string to identify a monotonous block and the other pointer marks the start of the next potential block. This is also efficient as it processes the string in a single pass.
Here’s an example:
def min_monotone_groups_tp(s):
left, right = 0, 0
groups = 0
while right < len(s):
if right == len(s) - 1 or s[right] > s[right + 1]:
groups += 1
left = right + 1
right += 1
return groups
print(min_monotone_groups_tp("aabbbc"))Output: 2
The min_monotone_groups_tp() function applies the two-pointer technique. It uses pointers to identify when a monotone group ends and increments the groups counter. It continues this until it has processed the entire input yielding the number of monotonous groups.
Method 4: Recursion with Memoization
For a more complex version of the problem that might include additional conditions, a recursive solution with memoization might be suitable. This approach involves breaking down the problem into smaller subproblems and caching the results.
Here’s an example:
def count_groups_memo(s, prev, index, memo):
if index == len(s):
return 0
if (prev, index) in memo:
return memo[(prev, index)]
included = 1 + count_groups_memo(s, s[index], index + 1, memo) if prev <= s[index] else float('inf')
excluded = count_groups_memo(s, prev, index + 1, memo)
memo[(prev, index)] = min(included, excluded)
return memo[(prev, index)]
def min_monotone_groups_rec(s):
return count_groups_memo(s, '', 0, {})
print(min_monotone_groups_rec("aabbbc"))Output: 2
The min_monotone_groups_rec() function utilizes recursion with a memoization dictionary to store previous computations. It makes two recursive calls, one including the current character in the subsequence and one excluding it, and then stores the minimum of the two outcomes.
Bonus One-Liner Method 5: Using itertools.groupby
Python’s itertools.groupby() function can be smartly used to solve the problem in a single line of code. This method is concise and relies on Python’s powerful standard library.
Here’s an example:
import itertools
def min_monotone_groups_itertools(s):
return sum(1 for _, g in itertools.groupby(s))
print(min_monotone_groups_itertools("aabbbc"))Output: 2
The min_monotone_groups_itertools() one-liner takes advantage of itertools.groupby() to group consecutive identical characters and count the number of such groups.
Summary/Discussion
- Method 1: Greedy Approach. Straightforward and efficient for monotonous strings. May not be suitable for complex variations of the problem.
- Method 2: Dynamic Programming. Provides a robust solution, usable in more complicated contexts. It can be overkill for simpler cases and is less efficient than the greedy method.
- Method 3: Two-pointer Technique. Simple and efficient. Similar in performance to the greedy approach with added clarity.
- Method 4: Recursion with Memoization. Scales well for added problem complexity. Smaller subproblems are computed once, potentially less efficient due to recursive overhead.
- Bonus One-Liner Method 5: Using
itertools.groupby(). Extremely concise, but readability might be impacted for those unfamiliar with the library. High elegance and Pythonic.