Taro Logo

Minimize Hamming Distance After Swap Operations

Medium
Asked by:
Profile picture
24 views
Topics:
ArraysGraphsDynamic Programming

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.length
  • 1 <= n <= 105
  • 1 <= source[i], target[i] <= 105
  • 0 <= allowedSwaps.length <= 105
  • allowedSwaps[i].length == 2
  • 0 <= ai, bi <= n - 1
  • ai != bi

Solution


Clarifying Questions

When 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:

  1. What are the size constraints for `nums1`, `nums2`, and `allowedSwaps`? What is the maximum value for elements in `nums1` and `nums2`?
  2. Can `nums1` and `nums2` be empty or null? Can `allowedSwaps` be empty?
  3. Are the swap pairs in `allowedSwaps` guaranteed to be valid indices within the bounds of the arrays?
  4. If no swaps are possible (e.g., `allowedSwaps` is empty) or if no swaps can improve the Hamming distance, should I return the original Hamming distance?
  5. Are the indices in `allowedSwaps` zero-based or one-based?

Brute Force Solution

Approach

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:

  1. First, find the connected groups of positions where swaps are allowed based on the given connections.
  2. For each of those groups, consider every possible arrangement (permutation) of the numbers in the first list corresponding to the positions within that group.
  3. After generating all possible arrangements within all groups, calculate the difference between the first list and the second list based on the current arrangement of elements in the first list.
  4. Keep track of the minimum difference found so far and the arrangement that produced it.
  5. Repeat the process of considering all arrangements for all groups and calculating the difference.
  6. After checking every possible arrangement, the minimum difference you found is the answer.

Code Implementation

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_distance

Big(O) Analysis

Time Complexity
O(n! * m)Identifying connected components using a graph algorithm like Depth-First Search or Union-Find takes O(n) time, where n is the length of the input array. However, the dominant factor is generating permutations for each connected component. In the worst-case scenario, a single connected component could contain all n elements. Generating all permutations for n elements takes O(n!) time. If there are m connected components, and we assume each component's permutation affects the overall result (requiring recalculation of Hamming distance), and assuming the permutation generation dominates everything else, the cost is O(n! * m) where m is the number of connected components, but m <= n. Recalculating the Hamming distance after each permutation takes O(n). Therefore, the overall time complexity is approximately O(n! * m), where m is the number of connected components.
Space Complexity
O(N!)The algorithm generates all possible arrangements (permutations) of numbers within each connected group. In the worst-case scenario, a single connected group might contain all N elements, requiring storage for all N! permutations. Although intermediate arrangements are calculated, the dominant space factor stems from the permutations. Therefore, the auxiliary space is dictated by the need to, at least conceptually, represent the arrangements to find the minimal Hamming Distance. Consequently, the space complexity is O(N!).

Optimal Solution

Approach

The 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:

  1. First, identify which positions are connected to each other via possible swap operations. Think of each position as belonging to a group of swappable locations.
  2. For each group of swappable positions, gather the elements from both lists that correspond to those positions.
  3. Within each group, sort the elements from both lists independently.
  4. Now, reassign the sorted elements from both lists back to their original positions within the group. This ensures the closest possible match within the swap group.
  5. By reassigning elements within each group in this manner, we reduce the overall number of mismatched positions between the two lists to the absolute minimum.

Code Implementation

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

Big(O) Analysis

Time Complexity
O(n log n)The dominant operations are identifying swappable positions, sorting the elements within each connected component, and iterating through lists. Identifying connected components using a disjoint set union find data structure takes nearly O(n α(n)) time where α(n) is the inverse Ackermann function, which grows very slowly and can be considered O(1) for practical input sizes. Sorting each connected component of size k takes O(k log k) time. Since the total size of all components is n, the overall sorting time is O(n log n). Iterating through the lists to collect and reassign elements takes O(n) time. Therefore the total runtime is dominated by the sorting step, resulting in O(n log n) complexity.
Space Complexity
O(N)The algorithm uses a disjoint-set data structure (Union-Find) to identify connected components, which requires space proportional to the number of positions, N. For each group of swappable positions, temporary lists are created to store the elements from both input lists, also taking O(N) space in the worst case where all positions are swappable. Sorting these temporary lists uses O(N) space in the worst case, depending on the sorting algorithm. Thus, the auxiliary space used by the algorithm is O(N).

Edge Cases

nums1 and nums2 are null or empty
How to Handle:
Return 0 immediately since there are no elements to compare.
nums1 and nums2 have different lengths
How to Handle:
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
How to Handle:
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
How to Handle:
Treat it as no swaps are allowed, calculate the initial hamming distance.
Swaps contains invalid indices (out of bounds)
How to Handle:
Ignore the invalid swap or throw an exception based on the problem's error-handling requirements.
Very large input arrays (memory constraints)
How to Handle:
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
How to Handle:
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
How to Handle:
Union-find handles cycles gracefully, but ensure the implementation is correct to avoid stack overflow or infinite loops.