💡 Problem Formulation: A ‘nice’ subarray is defined by a specific condition set by the user, such as having a sum that is even, or containing a certain number of odd numbers. This article will explore how to count the number of nice subarrays within an integer array in Python. For example, given an input array [1,2,3,4] and the condition that a nice subarray must contain exactly one odd number, the output should be 6, corresponding to subarrays [1], [2,3,4], [1,2], [2], [3], and [4].
Method 1: Brute Force Approach
The brute force approach involves checking all possible subarrays of the given array and counting those that meet the ‘nice’ condition. This could be time-consuming for large arrays but is a straightforward and comprehensive method to get started.
Here’s an example:
def count_nice_subarrays(arr, is_nice_condition):
def is_nice(subarr):
return is_nice_condition(subarr)
count = 0
for i in range(len(arr)):
for j in range(i, len(arr)):
if is_nice(arr[i:j+1]):
count += 1
return count
# Example usage
arr = [1,2,3,4]
print(count_nice_subarrays(arr, lambda x: sum(x) % 2 == 1))
Output: 6
This snippet defines a function count_nice_subarrays which accepts an array and a condition function. It uses two nested loops to generate all possible subarrays and then checks if they fulfill the condition by invoking the is_nice function. It tallies up the count of subarrays that are ‘nice’ and returns the total count.
Method 2: Using Prefix Sum Technique
This method optimizes the counting process by taking advantage of the prefix sum technique to quickly compute sums of subarrays. It significantly reduces the time complexity compared to the brute force approach.
Here’s an example:
def count_nice_subarrays(arr, k):
from collections import defaultdict
prefix_count = defaultdict(int)
prefix_count[0] = 1
total = 0
count = 0
for num in arr:
total += num % 2
if total >= k:
count += prefix_count[total - k]
prefix_count[total] += 1
return count
# Example usage
arr = [1,2,3,4]
print(count_nice_subarrays(arr, 1))
Output: 6
This code uses the prefix sum approach to maintain a running total of the number of odd numbers seen thus far. It uses a dictionary prefix_count to keep track of how many times each total has occurred. When it finds a subarray that has the correct number of odd numbers, it adds the corresponding count from prefix_count to the overall count. This method is efficient and suitable for larger datasets.
Method 3: Sliding Window Technique
The sliding window technique is an extremely useful approach when you want to keep track of a subset of data in a sequential array or list. This method helps us to reduce redundant computations by keeping a ‘window’ of the necessary elements and sliding it accordingly.
Here’s an example:
def count_nice_subarrays(arr, k):
def at_most(arr, k):
left = count = 0
for right in range(len(arr)):
if arr[right] % 2 == 1:
k -= 1
while k < 0:
if arr[left] % 2 == 1:
k += 1
left += 1
count += right - left + 1
return count
return at_most(arr, k) - at_most(arr, k-1)
# Example usage
arr = [1,2,3,4]
print(count_nice_subarrays(arr, 1))
Output: 6
This snippet uses a helper function at_most that calculates the number of subarrays with at most k odds. The main function calls this helper for k and k-1 odds and the difference gives the exact number of subarrays with exactly k odds. It effectively ‘slides’ the left and right pointers of the window to encompass subarrays with the desired properties and counts them efficiently.
Method 4: Using Two Pointers
The two-pointer technique is somewhat similar to the sliding window method. It involves moving two pointers across the array to identify the subarrays that match the criteria. One pointer represents the start and the other the end of the subarray.
Here’s an example:
def count_nice_subarrays(arr, k):
left = right = count = odd_count = 0
for right, value in enumerate(arr):
if value % 2 == 1:
k -= 1
odd_count = 0
while k == 0:
k += arr[left] % 2
odd_count += 1
left += 1
count += odd_count
return count
# Example usage
arr = [1,2,3,4]
print(count_nice_subarrays(arr, 1))
Output: 6
In this method, two pointers ‘left’ and ‘right’ traverse the array. Whenever we encounter an odd number using the right pointer, it affects our count of available odds (k). When our odds exceed the limit (k becomes 0), we advance the left pointer and reduce our subarray size while accumulating the number of valid subarrays every time we move the left pointer. This method is very effective in terms of space and time complexity.
Bonus One-Liner Method 5: List Comprehension with Counting
While not as effective for complex or large datasets, you can use a combination of list comprehensions and built-in functions for a one-liner solution that is elegant and straightforward.
Here’s an example:
# Python's one-liner to count nice subarrays, given a very specific condition. count_nice_subarrays = lambda arr, k: sum(sum(sub) % 2 == k for i in range(len(arr)) for sub in (arr[j:i + 1] for j in range(i + 1))) # Example usage arr = [1,2,3,4] print(count_nice_subarrays(arr, 1))
Output: 6
This concise lambda function uses a nested list comprehension to generate all subarrays and then a sum function to count those that match the given ‘nice’ condition (subarray sum % 2 equals k). While it’s compact and easy to understand, it’s a brute force approach and is not recommended for large arrays due to its high time complexity.
Summary/Discussion
- Method 1: Brute Force Approach. Comprehensive, easy to understand. Inefficient for large arrays due to O(n^2) time complexity.
- Method 2: Prefix Sum. Improved efficiency. Time complexity O(n). Requires understanding of prefix sums and hash tables.
- Method 3: Sliding Window Technique. Efficient. Reduces redundant computations. Sliding window can be a complex concept to grasp initially.
- Method 4: Two Pointers. Space-efficient, simple logic. Requires careful pointer manipulation to avoid errors.
- Bonus Method 5: List Comprehension with Counting. Very concise and elegant. Not practical for large arrays due to inefficiency.