You are given two integer arrays, source and target, both of length n. You are also given an array allowedSwaps where each allowedSwaps[i] = [ai, bi] indicates that you are allowed to swap the elements at index ai and index bi (0-indexed) of array source. Note that you can swap elements at a specific pair of indices multiple times and in any order.
The Hamming distance of two arrays of the same length, source and target, is the number of positions where the elements are different. Formally, it is the number of indices i for 0 <= i <= n-1 where source[i] != target[i] (0-indexed).
Return the minimum Hamming distance of source and target after performing any amount of swap operations on array source.
Example 1:
Input: source = [1,2,3,4], target = [2,1,4,5], allowedSwaps = [[0,1],[2,3]] Output: 1 Explanation: source can be transformed the following way: - Swap indices 0 and 1: source = [2,1,3,4] - Swap indices 2 and 3: source = [2,1,4,3] The Hamming distance of source and target is 1 as they differ in 1 position: index 3.
Example 2:
Input: source = [1,2,3,4], target = [1,3,2,4], allowedSwaps = [] Output: 2 Explanation: There are no allowed swaps. The Hamming distance of source and target is 2 as they differ in 2 positions: index 1 and index 2.
Example 3:
Input: source = [5,1,2,4,3], target = [1,5,4,2,3], allowedSwaps = [[0,4],[4,2],[1,3],[1,4]] Output: 0
Constraints:
n == source.length == target.length1 <= n <= 1051 <= source[i], target[i] <= 1050 <= allowedSwaps.length <= 105allowedSwaps[i].length == 20 <= ai, bi <= n - 1ai != biWhen you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The problem asks us to minimize the difference between two lists of numbers by swapping numbers within connected groups. The brute force approach is to try every possible combination of swaps within each group to see which one gives the smallest difference.
Here's how the algorithm would work step-by-step:
def minimize_hamming_distance_brute_force(source_array, target_array, allowed_swaps):
number_of_elements = len(source_array)
minimum_hamming_distance = number_of_elements
def calculate_hamming_distance(first_array, second_array):
hamming_distance = 0
for index in range(number_of_elements):
if first_array[index] != second_array[index]:
hamming_distance += 1
return hamming_distance
def find_connected_components(edges, number_of_nodes):
visited = [False] * number_of_nodes
connected_components = []
def depth_first_search(node, component):
visited[node] = True
component.append(node)
for neighbor in adjacency_list[node]:
if not visited[neighbor]:
depth_first_search(neighbor, component)
adjacency_list = [[] for _ in range(number_of_nodes)]
for first, second in edges:
adjacency_list[first].append(second)
adjacency_list[second].append(first)
for node in range(number_of_nodes):
if not visited[node]:
component = []
depth_first_search(node, component)
connected_components.append(component)
return connected_components
connected_components = find_connected_components(allowed_swaps, number_of_elements)
def generate_permutations(group, current_permutation, all_permutations):
if not group:
all_permutations.append(current_permutation)
return
for index in range(len(group)):
remaining_elements = group[:index] + group[index+1:]
generate_permutations(remaining_elements, current_permutation + [group[index]], all_permutations)
# Enumerate all possible permutations for each connected component
all_group_permutations = []
for group in connected_components:
all_permutations_for_group = []
generate_permutations(list(range(len(group))), [], all_permutations_for_group)
all_group_permutations.append((group, all_permutations_for_group))
# Iterate through all possible combinations of permutations across groups
number_of_groups = len(connected_components)
group_permutation_indices = [0] * number_of_groups
while True:
temp_source_array = source_array[:]
# Apply the current combination of permutations
for group_index in range(number_of_groups):
group, all_permutations = all_group_permutations[group_index]
current_permutation = all_permutations[group_permutation_indices[group_index]]
for index_in_permutation, original_index_in_group in enumerate(current_permutation):
temp_source_array[group[index_in_permutation]] = source_array[group[original_index_in_group]]
hamming_distance = calculate_hamming_distance(temp_source_array, target_array)
minimum_hamming_distance = min(minimum_hamming_distance, hamming_distance)
# Increment the permutation indices and check for termination
group_index = number_of_groups - 1
while group_index >= 0:
group_permutation_indices[group_index] += 1
if group_permutation_indices[group_index] == len(all_group_permutations[group_index][1]):
group_permutation_indices[group_index] = 0
group_index -= 1
else:
break
else:
break
return minimum_hamming_distanceThe most efficient way to minimize the differences between two lists after performing swaps is to group together positions that can be swapped. Then, we can freely rearrange the elements within each group to minimize the differences between the lists at those positions.
Here's how the algorithm would work step-by-step:
def minimize_hamming_distance(source_list, target_list, allowed_swaps):
number_of_elements = len(source_list)
parent = list(range(number_of_elements))
def find(element_index):
if parent[element_index] != element_index:
parent[element_index] = find(parent[element_index])
return parent[element_index]
def union(element_index_x, element_index_y):
root_x = find(element_index_x)
root_y = find(element_index_y)
if root_x != root_y:
parent[root_x] = root_y
# Group swappable positions using the Union-Find algorithm.
for element_index_x, element_index_y in allowed_swaps:
union(element_index_x, element_index_y)
swap_groups = {}
for element_index in range(number_of_elements):
root = find(element_index)
if root not in swap_groups:
swap_groups[root] = []
swap_groups[root].append(element_index)
# Collect elements for each group from source and target lists.
for group_root, group_indices in swap_groups.items():
source_group_elements = [source_list[element_index] for element_index in group_indices]
target_group_elements = [target_list[element_index] for element_index in group_indices]
source_group_elements.sort()
target_group_elements.sort()
# Assign the sorted values back to the original positions in the lists.
for element_index_within_group, original_element_index in enumerate(group_indices):
source_list[original_element_index] = source_group_elements[element_index_within_group]
target_list[original_element_index] = target_group_elements[element_index_within_group]
# Count differences after optimal swaps.
hamming_distance = 0
for element_index in range(number_of_elements):
# Counting the differences is the hamming distance
if source_list[element_index] != target_list[element_index]:
hamming_distance += 1
return hamming_distance| Case | How to Handle |
|---|---|
| nums1 and nums2 are null or empty | Return 0 immediately since there are no elements to compare. |
| nums1 and nums2 have different lengths | Throw an IllegalArgumentException or return an error code since hamming distance is undefined for arrays of differing lengths. |
| All values in nums1 and nums2 are identical | The hamming distance will be 0 initially and remain 0 regardless of swap operations, so the algorithm should still work correctly. |
| Swaps is null or empty | Treat it as no swaps are allowed, calculate the initial hamming distance. |
| Swaps contains invalid indices (out of bounds) | Ignore the invalid swap or throw an exception based on the problem's error-handling requirements. |
| Very large input arrays (memory constraints) | Consider using a more memory-efficient data structure than union-find or using an alternative algorithm with lower memory usage. |
| Integer overflow when calculating Hamming distance if values are large | Ensure intermediate Hamming distance calculations are performed using a data type that can accommodate large values (e.g., long). |
| Cycles in the swaps array causing infinite loops | Union-find handles cycles gracefully, but ensure the implementation is correct to avoid stack overflow or infinite loops. |