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Minimum Swaps to Sort by Digit Sum

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ArraysGreedy Algorithms

You are given an array nums of distinct positive integers. You need to sort the array in increasing order based on the sum of the digits of each number. If two numbers have the same digit sum, the smaller number appears first in the sorted order.

Return the minimum number of swaps required to rearrange nums into this sorted order.

A swap is defined as exchanging the values at two distinct positions in the array.

For example:

nums = [37,100]

  • Compute the digit sum for each integer: [3 + 7 = 10, 1 + 0 + 0 = 1] -> [10, 1]
  • Sort the integers based on digit sum: [100, 37]. Swap 37 with 100 to obtain the sorted order.
  • Thus, the minimum number of swaps required to rearrange nums is 1.

nums = [22,14,33,7]

  • Compute the digit sum for each integer: [2 + 2 = 4, 1 + 4 = 5, 3 + 3 = 6, 7 = 7] -> [4, 5, 6, 7]
  • Sort the integers based on digit sum: [22, 14, 33, 7]. The array is already sorted.
  • Thus, the minimum number of swaps required to rearrange nums is 0.

nums = [18,43,34,16]

  • Compute the digit sum for each integer: [1 + 8 = 9, 4 + 3 = 7, 3 + 4 = 7, 1 + 6 = 7] -> [9, 7, 7, 7]
  • Sort the integers based on digit sum: [16, 34, 43, 18]. Swap 18 with 16, and swap 43 with 34 to obtain the sorted order.
  • Thus, the minimum number of swaps required to rearrange nums is 2.

How would you implement an algorithm to find the minimum number of swaps to sort the array?

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 is the range of possible integer values within the input array?
  2. Can the input array be empty or contain null values?
  3. If multiple swap sequences achieve the minimum number of swaps, is any one acceptable?
  4. Are there any constraints on the size of the input array, such as a maximum number of elements?
  5. If two or more numbers have the same digit sum, how should their relative order be handled after sorting based on digit sum (e.g., should their original order be maintained)?

Brute Force Solution

Approach

The brute force approach to sorting with swaps involves trying every possible combination of swaps to see if we can sort the numbers based on their digit sums. We'll exhaustively check each possibility, keeping track of the smallest number of swaps we've seen that results in the correct sorted order.

Here's how the algorithm would work step-by-step:

  1. First, calculate the digit sum for each number in the list.
  2. Start by making a swap between the first two numbers.
  3. Check if the list is now sorted according to the digit sums.
  4. If it is sorted, record the number of swaps we've made (which is currently one).
  5. If it's not sorted, try swapping the first number with the third, then the first with the fourth, and so on, each time checking if the list becomes sorted after the swap.
  6. After trying all possible swaps with the first number, move on to the second number and try swapping it with all the numbers that come after it.
  7. Continue this process of swapping every pair of numbers and checking if the list becomes sorted.
  8. Keep track of the minimum number of swaps required to achieve a sorted list based on the digit sums.
  9. Consider all possible sequences of swaps, not just single swaps. This means trying a swap, then another swap, then another, and so on, until the list is sorted. Keep track of how many swaps each sequence takes.
  10. After exploring all possible combinations of swaps, pick the sequence that required the fewest swaps to sort the list by the digit sums.

Code Implementation

def minimum_swaps_to_sort_by_digit_sum(numbers):
    minimum_number_of_swaps = float('inf')
    number_of_numbers = len(numbers)

    def calculate_digit_sum(number):
        digit_sum = 0
        for digit in str(number):
            digit_sum += int(digit)
        return digit_sum

    def is_sorted_by_digit_sum(numbers):
        for index in range(number_of_numbers - 1):
            if calculate_digit_sum(numbers[index]) > calculate_digit_sum(numbers[index + 1]):
                return False
        return True

    def find_minimum_swaps(current_numbers, current_swaps):
        nonlocal minimum_number_of_swaps

        # If the current number of swaps exceeds the minimum, stop.
        if current_swaps >= minimum_number_of_swaps:
            return

        if is_sorted_by_digit_sum(current_numbers):
            minimum_number_of_swaps = current_swaps
            return

        # We iterate through all possible pairs of numbers and swap them
        for first_index in range(number_of_numbers):
            for second_index in range(first_index + 1, number_of_numbers):
                # Create a copy to avoid modifying the original list during recursion.
                new_numbers = current_numbers[:]
                new_numbers[first_index], new_numbers[second_index] = new_numbers[second_index], new_numbers[first_index]

                # Recursively check from this new state
                find_minimum_swaps(new_numbers, current_swaps + 1)

    # Start searching from the initial state with zero swaps
    find_minimum_swaps(numbers[:], 0)

    if minimum_number_of_swaps == float('inf'):
        return -1
    else:
        return minimum_number_of_swaps

Big(O) Analysis

Time Complexity
O(n!)The provided brute force approach explores all possible swap combinations to sort the numbers based on their digit sums. In the worst-case scenario, we might need to try every possible permutation of the input array of size n. Generating all permutations of an array of size n takes O(n!) time. Therefore, the time complexity is dominated by the permutation generation, leading to O(n!) complexity.
Space Complexity
O(N)The provided brute force solution explores all possible swap combinations which implies creating copies of the original list of N numbers to test different swap sequences. The recursion depth for exploring all swap combinations can reach N in the worst-case, leading to N copies of the list being stored. Therefore, the auxiliary space used is proportional to the input size N. The space complexity is O(N).

Optimal Solution

Approach

The key idea is to arrange the numbers in the correct order based on their digit sums with the fewest possible swaps. We achieve this by identifying cycles of misplaced numbers and resolving each cycle with a specific number of swaps.

Here's how the algorithm would work step-by-step:

  1. First, calculate the digit sum for each number.
  2. Determine the sorted order of the numbers based on their digit sums.
  3. Create a mapping that shows where each number should ideally be located after sorting.
  4. Look for cycles. A cycle exists when a number is not in its correct position and by following the chain of misplaced numbers, you eventually return to the original number.
  5. For each cycle, the number of swaps needed to correctly place the numbers is one less than the number of elements in the cycle.
  6. Sum the number of swaps needed for each cycle to get the minimum number of swaps required to sort the original list.

Code Implementation

def minimum_swaps_to_sort_by_digit_sum(numbers):
    def calculate_digit_sum(number):
        digit_sum = 0
        for digit in str(number):
            digit_sum += int(digit)
        return digit_sum

    digit_sums = [calculate_digit_sum(number) for number in numbers]
    # Create pairs of (digit_sum, index) to preserve original positions.
    digit_sum_index_pairs = list(enumerate(digit_sums))
    # Sort based on digit sums, keeping track of original indices.
    digit_sum_index_pairs.sort(key=lambda x: x[1])
    sorted_indices = [index for index, _ in digit_sum_index_pairs]
    number_of_swaps = 0
    visited = [False] * len(numbers)

    # Iterate through each index to find and process cycles.
    for i in range(len(numbers)):
        # Skip if already visited or correctly placed.
        if visited[i] or sorted_indices[i] == i:
            continue

        cycle_size = 0
        current_index = i

        # Traverse the cycle until we return to the starting index.
        while not visited[current_index]:
            visited[current_index] = True
            current_index = sorted_indices[current_index]
            cycle_size += 1

        # Each cycle needs cycle_size - 1 swaps to be resolved.
        if cycle_size > 0:
            number_of_swaps += (cycle_size - 1)

    return number_of_swaps

Big(O) Analysis

Time Complexity
O(n)Calculating the digit sum for each of the n numbers takes O(n) time, assuming the number of digits in each number is relatively small and constant. Determining the sorted order and creating the mapping also involves iterating through the n numbers, resulting in O(n) time. The cycle detection process iterates through the n numbers and follows chains, but each number is visited at most once within a cycle, and the overall number of iterations is bounded by n. Therefore, the dominant factor is the initial iteration and subsequent cycle traversals which are both linearly proportional to the number of elements n in the input array; thus the time complexity is O(n).
Space Complexity
O(N)The algorithm uses several auxiliary data structures. It calculates the digit sum for each of the N numbers, which might be stored in an array of size N. It determines the sorted order, implying an auxiliary array of size N to store indices or the sorted numbers themselves. The mapping of original index to sorted index also requires O(N) space. These data structures contribute to a space complexity directly proportional to the input size, N. Therefore, the overall auxiliary space complexity is O(N).

Edge Cases

CaseHow to Handle
Null or empty input arrayReturn 0, as no swaps are needed for an empty or null array.
Input array with only one elementReturn 0, as a single-element array is already sorted.
Input array with all elements having the same digit sumReturn 0, as the array is effectively already sorted based on the defined criteria.
Large input array exceeding memory limitsEnsure the algorithm utilizes memory efficiently and consider using iterative approaches to avoid stack overflow errors.
Input array containing very large numbers (potential integer overflow when calculating digit sum)Use appropriate data types (e.g., long) to store the digit sum to prevent integer overflow.
Input array already sorted according to digit sumThe algorithm should correctly identify this scenario and return 0 swaps.
Duplicate numbers in the array with differing original positionsUse an appropriate data structure (e.g., hash map) to maintain the correct index associations after sorting based on digit sum.
Arrays of numbers that have the same digit sumHandle these numbers based on their original array position to avoid incorrect swaps.