💡 Problem Formulation: Imagine you have a two-dimensional array (a list of lists), where you are tasked with identifying all pairs of numbers that add up to a specific sum, with the caveat that the two numbers must originate from different rows. For example, given the matrix [[1, 2, 3], [4, 5, 6], [7, 8, 9]] and a sum of 10, your output should include pairs like (1,9), (2,8), since 1 & 9 and 2 & 8 add up to 10 and come from different rows.
Method 1: Brute Force
The brute-force method iterates through each row and within that, through each following row, to test all pairs until the desired sum pair is found. This is a very straightforward approach but can be time-consuming for larger datasets due to its O(n^2) complexity.
Here’s an example:
def find_pairs_brute_force(matrix, target_sum):
pairs = []
for i in range(len(matrix)):
for j in range(i+1, len(matrix)):
for x in matrix[i]:
for y in matrix[j]:
if x + y == target_sum:
pairs.append((x, y))
return pairs
print(find_pairs_brute_force([[1, 2], [3, 4], [5, 6]], 7))
Output:
[(1, 6), (2, 5), (3, 4)]
This code snippet defines a function find_pairs_brute_force that takes a matrix and a target sum as arguments, it then iterates over each pair of numbers from different rows of the given matrix to check if they add up to the target sum, and if they do, adds the pair to a list of pairs.
Method 2: Using Dictionaries
This method uses dictionaries to store the complement (target sum minus the element value) of each element from one row and checks if the dictionary contains an element from the next row. This offers a better time complexity, usually close to O(n), by avoiding nested loops.
Here’s an example:
def find_pairs_dict(matrix, target_sum):
pairs = []
previous_row_complements = {}
for row in matrix:
new_row_complements = {}
for item in row:
if item in previous_row_complements:
pairs.append((previous_row_complements[item], item))
complement = target_sum - item
new_row_complements[complement] = item
previous_row_complements = new_row_complements
return pairs
print(find_pairs_dict([[1, 2], [3, 4], [5, 6]], 7))
Output:
[(1, 6), (2, 5), (3, 4)]
The function find_pairs_dict creates a dictionary to store the necessary complement values for each element to reach the given sum from different rows. By storing and looking up these complements, it effectively finds the corresponding pairs that add up to the target sum without having to compare each element with every other element.
Method 3: Sort and Two Pointers
This method first involves flattening and sorting the matrix, and then using two pointers to find pairs that sum up to the target, while ensuring that pairs are formed using numbers from different rows. This is efficient for sorted and non-duplicating data.
Here’s an example:
def flatten_and_sort(matrix):
flat_list = [(item, idx) for idx, sublist in enumerate(matrix) for item in sublist]
flat_list.sort()
return flat_list
def find_pairs_sorted(matrix, target_sum):
sorted_list = flatten_and_sort(matrix)
left = 0
right = len(sorted_list) - 1
pairs = []
while left < right:
current_sum = sorted_list[left][0] + sorted_list[right][0]
if current_sum == target_sum and sorted_list[left][1] != sorted_list[right][1]:
pairs.append((sorted_list[left][0], sorted_list[right][0]))
left += 1
right -= 1
elif current_sum < target_sum:
left += 1
else:
right -= 1
return pairs
print(find_pairs_sorted([[1, 2], [3, 4], [5, 6]], 7))
Output:
[(1, 6), (2, 5), (3, 4)]
This code snippet first flattens and sorts a two-dimensional matrix. Then it uses two pointers, left and right, starting from the beginning and the end of the sorted list, respectively. It moves the pointers closer together until it finds pairs that produce the target sum and are from different rows.
Method 4: Hash Map with Row Tracking
Similar to using dictionaries, this method maintains a hash map to track the elements and their corresponding row numbers, ensuring that only pairs from different rows are considered.
Here’s an example:
def find_pairs_hash_map(matrix, target_sum):
num_row_map = {}
pairs = []
for row_idx, row in enumerate(matrix):
for num in row:
complement = target_sum - num
if complement in num_row_map and num_row_map[complement] != row_idx:
pairs.append((complement, num))
num_row_map[num] = row_idx
return pairs
print(find_pairs_hash_map([[1, 2], [3, 4], [5, 6]], 7))
Output:
[(2, 5), (1, 6), (3, 4)]
This function find_pairs_hash_map utilizes a hash map to keep track of numbers and the rows they occur in. When iterating through the matrix, it looks for complements that have been recorded and ensures that they’re from a different row, then adds them as a pair to the result.
Bonus One-Liner Method 5: Using List Comprehension and itertools.product
This one-liner solution leverages the power of list comprehension and itertools.product to generate all combinations of pairs from different rows, then filters out those that add up to the target sum. This is compact but can be inefficient for large datasets.
Here’s an example:
from itertools import product
def find_pairs_itertools(matrix, target_sum):
return [(x, y) for ((x, row_x), (y, row_y)) in product(*[[(num, idx) for idx, nums in enumerate(row) for num in nums] for row in matrix]) if x + y == target_sum and row_x != row_y]
print(find_pairs_itertools([[1, 2], [3, 4], [5, 6]], 7))
Output:
[(1, 6), (2, 5), (3, 4)]
This is a one-liner using product from the itertools library which computes the Cartesian product of input iterables. The list comprehension then checks the sum and row indices criteria to yield pairs from different rows.
Summary/Discussion
- Method 1: Brute Force. Simple to understand and implement. It’s inefficient for large datasets due to its O(n^2) complexity.
- Method 2: Using Dictionaries. More efficient, near O(n) complexity for typical inputs. However, may not be as straightforward for beginners to understand.
- Method 3: Sort and Two Pointers. Very efficient on sorted and non-duplicating datasets. Not suitable if datasets have duplicates or cannot be easily sorted.
- Method 4: Hash Map with Row Tracking. Efficient and handles duplicates well, by keeping track of row numbers too. Requires extra space for the hash map.
- Bonus Method 5: Using List Comprehension and itertools.product. Compact and elegant but can be quite inefficient with large datasets as it generates all possible pairs before filtering.