💡 Problem Formulation: Imagine needing to find the shortest distance of a specified character ‘c’ from every index in a given string. For example, given the input string “algorithm” and the character ‘a’, the desired output would be a list [0,1,2,3,4,5,6,7,8], representing the distance from ‘a’ at each index.
Method 1: Brute Force Search
This method involves checking the distance from the character ‘c’ to every other character in the string, updating the minimum distance whenever a closer one is found. Function specification: Given a string and a character, it returns an array with the minimum distances to the specified character at each index.
Here’s an example:
def closest_distance_brute(s, c):
result = []
for i in range(len(s)):
min_dist = len(s)
for j in range(len(s)):
if s[j] == c:
min_dist = min(min_dist, abs(i - j))
result.append(min_dist)
return result
print(closest_distance_brute("loveleetcode", 'e'))
Output: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]
This code defines a function closest_distance_brute which takes a string and a character as arguments and returns a list of minimum distances. For each index, it calculates the distance to every occurrence of the character and keeps the smallest one.
Method 2: Two-Pass Method
This method improves on brute force by scanning the string twice. The first pass finds distances for all indices from left to right, and the second pass updates them from right to left. Function specification: Scans string in two passes to efficiently find the closest distance of the character ‘c’ at each index.
Here’s an example:
def closest_distance_twopass(s, c):
result = [len(s)] * len(s)
prev = float('inf')
for i in range(len(s)):
if s[i] == c:
prev = i
result[i] = min(result[i], abs(i - prev))
for i in range(len(s) - 1, -1, -1):
if s[i] == c:
prev = i
result[i] = min(result[i], abs(i - prev))
return result
print(closest_distance_twopass("loveleetcode", 'e'))
Output: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]
This code snippet defines the closest_distance_twopass function, which initializes a result list with maximum initial distances. It reduces distances on two passes using a previously encountered instance of the character ‘c’.
Method 3: Dynamic Programming
Utilizing dynamic programming, this method calculates the minimum distances with optimal substructure and overlapping subproblems properties. Function Specification: Dynamically computes the closest distances to character ‘c’ to reduce redundant computations.
Here’s an example:
def closest_distance_dp(s, c):
n = len(s)
result = [0 if char == c else n for char in s]
for i in range(1, n):
result[i] = min(result[i], result[i - 1] + 1)
for i in range(n - 2, -1, -1):
result[i] = min(result[i], result[i + 1] + 1)
return result
print(closest_distance_dp("loveleetcode", 'e'))
Output: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]
This snippet implements closest_distance_dp, which first initializes the list with the distance as 0 for indices where ‘c’ is present, otherwise a large number. It then iteratively updates the list in forwards and backwards directions, each time using the result of the previous index to get the current minimum distance.
Method 4: Using Built-in Functions
The built-in index() function can be exploited to find the distances by catching exceptions when the character ‘c’ is not found. Function Specification: Leverages Python’s built-in functions to track the closest ‘c’ on both sides of each index.
Here’s an example:
def closest_distance_builtin(s, c):
result = []
last_c = -1
for i, char in enumerate(s):
try:
next_c = s.index(c, i)
except ValueError:
next_c = float('inf')
if char == c:
last_c = i
result.append(min(i - last_c, next_c - i) if last_c != -1 else next_c - i)
return result
print(closest_distance_builtin("loveleetcode", 'e'))
Output: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]
This code defines closest_distance_builtin which uses index() to find the closest ‘c’ after the current index and keeps track of the last occurrence of ‘c’ for calculating minimum distance from each position.
Bonus One-Liner Method 5: Using List Comprehension
This concise method uses list comprehension with built-in functions to compute the distances in a single line. Function specification: Provides a compact one-liner solution using list comprehension and min/max functions.
Here’s an example:
s = "loveleetcode"
c = 'e'
print([min(abs(i - j) if s[j] == c else float('inf') for j in range(len(s))) for i in range(len(s))])
Output: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]
This one-liner calculates the minimum distance from ‘c’ to each index ‘i’ by traversing the entire string for each ‘i’ within a list comprehension, therefore combining the concept of brute force within a condensed expression.
Summary/Discussion
- Method 1: Brute Force Search. Simple to understand. High time complexity, especially for long strings.
- Method 2: Two-Pass Method. More efficient than brute force. Linear time complexity, but still requires two passes over the data.
- Method 3: Dynamic Programming. Makes clever use of previous computations. Efficient, with two passes over the data, similar to Method 2 but more elegant.
- Method 4: Using Built-in Functions. Simplifies code using Python’s powerful built-ins. It can be less transparent about how the distances are computed due to hidden complexity within the built-in functions.
- Bonus Method 5: Using List Comprehension. Offers the most concise code. While elegant, it can be less efficient and harder to debug due to its condensed nature.