💡 Problem Formulation: In the context of array processing, a common task involves identifying subarrays whose elements sum to an odd number. Given an input array like [1, 2, 3, 4], we aim to find the count of all possible contiguous subarrays where the sum is odd. For this input, the desired output is 4, corresponding to the subarrays [1], [1, 2, 3], [2, 3], and [3].
Method 1: Brute Force Approach
Using the brute force approach, one simply iterates through all possible contiguous subarrays, calculates their sums, and counts the ones with an odd sum. It requires iterating with two nested loops and can be quite computationally expensive, especially for large arrays.
Here’s an example:
def count_odd_sum_subarrays(arr):
odd_count = 0
for start in range(len(arr)):
for end in range(start, len(arr)):
if sum(arr[start:end + 1]) % 2 != 0:
odd_count += 1
return odd_count
print(count_odd_sum_subarrays([1, 2, 3, 4]))Output: 4
This code snippet defines a function count_odd_sum_subarrays that iterates over all subarrays using two nested loops and checks if the sum is odd. If it is, the count is incremented.
Method 2: Prefix Sum Optimization
The prefix sum optimization method involves creating an auxiliary array that holds the sum of elements from the start up to each index. This simplifies the computation of subarray sums and can help reduce the time complexity.
Here’s an example:
def count_odd_sum_subarrays(arr):
odd_count = 0
prefix_sums = [0]
for num in arr:
prefix_sums.append(prefix_sums[-1] + num)
for start in range(len(arr)):
for end in range(start, len(arr)):
if (prefix_sums[end + 1] - prefix_sums[start]) % 2 != 0:
odd_count += 1
return odd_count
print(count_odd_sum_subarrays([1, 2, 3, 4]))Output: 4
This method uses a prefix sum array to find the sum of subarrays in constant time. However, it still employs two nested loops to explore all subarray possibilities.
Method 3: Using Cumulative Frequency
This clever method implies maintaining a running total of odd and even cumulative sums, and using these counts to determine the possible subarrays yielding an odd sum based on previous sums.
Here’s an example:
def count_odd_sum_subarrays(arr):
result = odd_sum = even_sum = 0
for num in arr:
if (odd_sum + num) % 2 != 0:
result += even_sum + 1
odd_sum, even_sum = even_sum + 1, odd_sum
else:
result += odd_sum
odd_sum, even_sum = even_sum, odd_sum + 1
return result
print(count_odd_sum_subarrays([1, 2, 3, 4]))Output: 4
The code snippet maintains running totals of even and odd sums and calculates the number of odd sum subarrays based on these running totals, thus optimizing the computation.
Method 4: Hash Map with Running Sum
The hash map with running sum approach uses a dictionary to store the frequencies of the sums (modulo 2) encountered so far, and compute the desired count based on the properties of odd and even numbers.
Here’s an example:
def count_odd_sum_subarrays(arr):
sum_freq = {0: 1}
running_sum = odd_count = 0
for num in arr:
running_sum += num
if running_sum % 2 != 0:
odd_count += sum_freq.get(0, 0)
else:
odd_count += sum_freq.get(1, 0)
sum_freq[running_sum % 2] = sum_freq.get(running_sum % 2, 0) + 1
return odd_count
print(count_odd_sum_subarrays([1, 2, 3, 4]))Output: 4
This code utilizes a hash map to keep track of the parity of the running sums, enabling the efficient calculation of the number of odd sum subarrays without inspecting every subarray individually.
Bonus One-Liner Method 5: List Comprehension with itertools
The itertools library in Python contains a function accumulate, which can be used to create a list of running sums. Coupled with list comprehension, one can extract the relevant count in a compact form.
Here’s an example:
from itertools import accumulate
def count_odd_sum_subarrays(arr):
return sum((s % 2 != 0) for s in accumulate(arr))
print(count_odd_sum_subarrays([1, 2, 3, 4]))Output: 4
This elegant one-liner uses accumulate to create an iterator of running sums, from which we use a generator expression to count the odd sums in a single line of code.
Summary/Discussion
- Method 1: Brute Force Approach. It’s straightforward but inefficient for large arrays due to its O(n^2) complexity.
- Method 2: Prefix Sum Optimization. Slightly more efficient than brute force, but still suffers from high time complexity due to nested loops.
- Method 3: Using Cumulative Frequency. Highly efficient with O(n) complexity. Relies on the insight that summing subarrays can be tracked with just two running totals.
- Method 4: Hash Map with Running Sum. Like Method 3, it offers O(n) complexity. Utilizes hash map for efficient tracking of sum frequencies.
- Bonus Method 5: List Comprehension with itertools. Elegant and concise, but might be confusing for beginners and does not necessarily offer the same performance benefits as Methods 3 or 4.