💡 Problem Formulation: Consider an array of integers and a positive integer k. The challenge is to find the length of the smallest contiguous sublist that can be removed so that the sum of the remaining elements in the array is divisible by k. For example, given an array [1, 2, 3, 4, 5] and k=5, removing the sublist [1, 4] leaves us with [2, 3, 5], which sums up to 10, a multiple of 5. The desired output for this example would be 2, the length of the removed sublist.
Method 1: Brute Force
This method involves checking all possible sublists to find the smallest one that meets the condition. For each starting index, sublists of increasing sizes are generated and checked for divisibility by k after removing them from the array. It’s a straightforward but inefficient method, with a time complexity of O(n^3).
Here’s an example:
def find_smallest_sublist(arr, k):
n = len(arr)
for length in range(1, n+1):
for start in range(n-length+1):
temp_arr = arr[:start] + arr[start+length:]
if sum(temp_arr) % k == 0:
return length
return n
# Example array and k
print(find_smallest_sublist([1, 2, 3, 4, 5], 5))The output of this code will be:
2
The function find_smallest_sublist takes an array and a divisor k, iterates over all possible sublists, testing the condition for each, and returns the length of the smallest sublist that works. Although simple, its poor time complexity makes it unsuitable for large arrays.
Method 2: Prefix Sums and Hashing
This method uses prefix sums and hashing to achieve a more efficient solution. By keeping track of the sum of elements up to each index, we can quickly compute the sum of any sublist. A hash map is used to store the sum modulo k. The first repeat of a particular mod value indicates that a subsequence can be removed to form a sum divisible by k.
Here’s an example:
def smallest_sublist_div_by_k(arr, k):
mod_map = {0: -1}
curr_sum = 0
result = len(arr)
for i in range(len(arr)):
curr_sum += arr[i]
mod = curr_sum % k
if mod in mod_map:
result = min(result, i - mod_map[mod])
else:
mod_map[mod] = i
return result if result < len(arr) else -1
# Example array and k
print(smallest_sublist_div_by_k([1, 2, 3, 4, 5], 5))The output of this code will be:
2
The smallest_sublist_div_by_k function leverages prefix sums and a hash map to efficiently compute the result. This reduces the time complexity significantly compared to the brute force approach, though it requires understanding more complex concepts like modulo operations and hash tables.
Method 3: Sliding Window
A sliding window approach can help in reducing the time complexity by dynamically adjusting the size of the window to keep track of the sublist sums. When a sum that satisfies the divisibility after removal is found, the window is resized to continue searching for potentially smaller sublists.
Here’s an example:
def find_sublist_by_sliding_window(arr, k):
total_sum = sum(arr)
window_sum = 0
min_length = float('inf')
left = 0
for right in range(len(arr)):
window_sum += arr[right]
while (total_sum - window_sum) % k == 0 and right - left + 1 < min_length:
min_length = right - left + 1
window_sum -= arr[left]
left += 1
return min_length if min_length != float('inf') else -1
# Example array and k
print(find_sublist_by_sliding_window([1, 2, 3, 4, 5], 5))The output of this code will be:
2
The function find_sublist_by_sliding_window maintains a running sum within a window of the array. By adjusting the window borders, it finds the smallest sublist which, if removed, leaves a sum divisible by k. This method finds the solution much faster than a brute force, especially on longer lists.
Method 4: Cumulative Sum with Look-ahead
This method is a variant of prefix sums but utilizes a look-ahead technique. Instead of searching for an existing prefix sum lookup, the algorithms calculate the needed sum directly and search for it ahead in the prefix sums array. This approach harnesses the fact that the cumulative sum of the sublist being looked at is the amount required to make the entire array’s sum divisible by k.
Here’s an example:
def find_sublist_with_lookahead(arr, k):
prefix_sums = [0]
for num in arr:
prefix_sums.append((prefix_sums[-1] + num) % k)
need = sum(arr) % k
min_len = len(arr)
sum_dict = {}
for i, p_sum in enumerate(prefix_sums):
if p_sum in sum_dict:
min_len = min(min_len, i - sum_dict[p_sum])
sum_dict[p_sum] = i
return min_len
# Example array and k
print(find_sublist_with_lookahead([1, 2, 3, 4, 5], 5))The output of this code will be:
2
The function find_sublist_with_lookahead calculates prefix sums modulo k and uses a hash table to find the smallest subarray sum needed to make the remainder 0 when divided by k. It is similar to Method 2 but with a subtle difference in approach that might offer efficiency improvements in certain cases.
Bonus One-Liner Method 5: List Comprehension with Min Function
For those who love Python’s one-liners and comprehensions, a very concise solution using a combination of list comprehension and the built-in min function can be used. However, this method may suffer from the same inefficiency as the brute force approach, but it showcases Python’s expressive power.
Here’s an example:
arr = [1, 2, 3, 4, 5] k = 5 min_len = min([i - j for i in range(len(arr) + 1) for j in range(i) if sum(arr[j:i]) % k == 0], default=-1) print(min_len)
The output of this code will be:
2
This one-liner computes the length of every possible sublist whose removal would result in an array sum divisible by k, then finds the smallest such length. It’s a pythonic but computationally heavy approach.
Summary/Discussion
- Method 1: Brute Force. Simple to understand and implement. However, highly inefficient and impractical for large datasets.
- Method 2: Prefix Sums and Hashing. Significantly more efficient than the brute force approach. Requires a good grasp of hashing and prefix sums.
- Method 3: Sliding Window. Dynamic in nature and often more efficient than brute force. Can provide better performance especially on larger datasets.
- Method 4: Cumulative Sum with Look-ahead. Offers potential efficiency improvements. It’s a less common but interesting approach to the problem.
- Bonus One-Liner Method 5: List Comprehension with Min Function. Very concise and idiomatic Python. Not recommended for large datasets due to inefficiency similar to the brute force method.