💡 Problem Formulation: The challenge is to locate a position within an array such that the product of elements in the subarray to the left is equal to the product of elements in the subarray to the right. Given an input array like [1, 2, 3, 6, 5, 6], the desired output is the index 3, since the product of the subarray [1, 2, 3] is equal to the product of the subarray [5, 6], both yielding 6.
Method 1: Brute Force Approach
The brute force approach involves iteratively calculating the product of subarrays on either side of each array element and comparing them until a balance point is found. This method is straightforward but may not be efficient for large arrays as its time complexity is O(n^2), where n is the length of the array.
Here’s an example:
def find_balanced_index(arr):
for i in range(1, len(arr)):
if prod(arr[:i]) == prod(arr[i+1:]):
return i
return -1
def prod(subarr):
result = 1
for num in subarr:
result *= num
return result
# Testing the function
print(find_balanced_index([1, 2, 3, 6, 5, 6]))Output: 3
In this code snippet, we define a function find_balanced_index that looks for the index dividing the array into two subarrays with equal products. The helper function prod calculates the product of the elements in a subarray. The function is used to compare the products iteratively in a loop.
Method 2: Prefix and Suffix Product
This approach optimizes the brute force method by precomputing the prefix and suffix products for the array. This way, we only need to calculate the products once, making the comparison at each index O(1), and reduces the overall time complexity to O(n).
Here’s an example:
def find_balanced_index(arr):
prefix_prod = [1] * (len(arr) + 1)
suffix_prod = [1] * (len(arr) + 1)
for i in range(len(arr)):
prefix_prod[i+1] = prefix_prod[i] * arr[i]
for i in reversed(range(len(arr))):
suffix_prod[i] = suffix_prod[i+1] * arr[i]
for i in range(1, len(arr)):
if prefix_prod[i] == suffix_prod[i+1]:
return i
return -1
print(find_balanced_index([1, 2, 3, 6, 5, 6]))Output: 3
This code snippet demonstrates how to use prefix and suffix product arrays to find the balanced index efficiently. It first calculates the products of elements from the start (prefix) and from the end (suffix) of the array. It then iterates through the array to find a matching product.
Method 3: Division Approach
The division approach involves computing the total product of the entire array and then iterating through the array while keeping track of the running prefix product. At each step, we can determine the suffix product by dividing the total product by the current prefix product. This avoids using additional space for the suffix product array.
Here’s an example:
from functools import reduce
from operator import mul
def find_balanced_index(arr):
total_prod = reduce(mul, arr, 1)
prefix_prod = 1
for i, num in enumerate(arr):
if prefix_prod == total_prod // prefix_prod // num:
return i
prefix_prod *= num
return -1
print(find_balanced_index([1, 2, 3, 6, 5, 6]))Output: 3
The given code uses the reduce function to calculate the total product of the array and divides it by the running prefix product to determine if the current index divides the array into two subarrays with equal products.
Method 4: Logarithm Based Approach
Since multiplication can cause overflow for large numbers, a logarithm based approach can be utilized. By summing the logarithms of the array elements, we can perform addition instead of multiplication, preventing overflow and following a similar logic for comparison.
Here’s an example:
import math
def find_balanced_index(arr):
log_sum = sum(math.log(num) for num in arr)
prefix_log_sum = 0
for i, num in enumerate(arr):
prefix_log_sum += math.log(num)
if math.isclose(prefix_log_sum, log_sum - prefix_log_sum):
return i
return -1
print(find_balanced_index([1, 2, 3, 6, 5, 6]))Output: 3
By adding logs instead of multiplying numbers, we avoid potentially large numbers and overflow issues. The function checks for a balanced index by comparing the sum of log values before and after each index.
Bonus One-Liner Method 5: Pythonic Approach with itertools
Python’s itertools module can greatly simplify the process. Utilizing accumulate() to compute the prefix products and a reversed list comprehension, this one-liner demonstrates a neat, pythonic way to find the balanced index.
Here’s an example:
from itertools import accumulate from operator import mul def find_balanced_index(arr): preprod = list(accumulate(arr, mul))[:-1] sufprod = list(accumulate(reversed(arr), mul))[:-1] return next((i for i, pp in enumerate(preprod) if pp == sufprod[~i]), -1) print(find_balanced_index([1, 2, 3, 6, 5, 6]))
Output: 3
This concise one-liner leverages accumulate() from itertools to generate prefix and suffix product lists and finds the balanced index using a generator expression.
Summary/Discussion
- Method 1: Brute Force Approach. Simple and easy to implement. Not suitable for large arrays due to its O(n^2) time complexity.
- Method 2: Prefix and Suffix Product. More efficient than brute force with an O(n) time complexity. Requires additional space to store product arrays.
- Method 3: Division Approach. Time-efficient with a constant space complexity. Needs handling for zero values to avoid division by zero errors.
- Method 4: Logarithm Based Approach. Prevents overflow by working with sums of logs. Requires careful handling to avoid floating-point comparison issues.
- Method 5: Pythonic Approach with itertools. Concise and idiomatic, but relies on Python-specific libraries and may be less intuitive for beginners.