Problem Description

You are given an array nums containing distinct positive integers. Your task is to sort this array based on a specific rule and determine the minimum number of swaps needed to achieve this sorted order.

The sorting rule works as follows:

• Calculate the sum of digits for each number (e.g., for 123, the digit sum is 1 + 2 + 3 = 6)
• Sort the numbers in increasing order based on their digit sums
• If two numbers have the same digit sum, place the smaller number first

A swap means exchanging the values at two different positions in the array. You need to find the minimum number of such swaps required to transform the original array into the sorted arrangement.

For example, if nums = [312, 123, 456]:

• Digit sum of 312 = 3 + 1 + 2 = 6
• Digit sum of 123 = 1 + 2 + 3 = 6
• Digit sum of 456 = 4 + 5 + 6 = 15

Since 312 and 123 have the same digit sum (6), we place 123 before 312 (smaller value first). The sorted order would be [123, 312, 456].

The solution uses a cycle detection approach to find the minimum swaps. It first creates the target sorted arrangement, then maps each element to its target position. By identifying cycles in the permutation (where elements need to rotate positions), it calculates that each cycle of length k requires k-1 swaps. The total minimum swaps equals n - (number of cycles), where n is the array length.

Intuition

The key insight is recognizing that finding the minimum number of swaps to sort an array is fundamentally a problem about permutation cycles.

Think of it this way: we have the current array and we know where each element should end up in the sorted array. Each element wants to move to its target position. When we trace these movements, we discover cycles - chains of elements where element A needs to go to B's position, B needs to go to C's position, and eventually some element needs to go to A's position, forming a closed loop.

For example, if element at index 0 needs to go to index 2, element at index 2 needs to go to index 4, and element at index 4 needs to go back to index 0, we have a cycle of length 3. To arrange these 3 elements correctly, we need exactly 2 swaps (one less than the cycle length).

Why k-1 swaps for a cycle of length k? Consider a simple cycle: A → B → C → A. We can fix this with 2 swaps:

• First swap A and B: now B is in the right place
• Then swap A (now in B's old position) and C: now all three are correct

This pattern holds for any cycle length. Each swap fixes one element's position permanently, and the last element automatically falls into place when we've done k-1 swaps.

The beautiful realization is that the total minimum swaps equals n - (number of cycles). Each cycle of length k contributes k elements but needs k-1 swaps. If we have c cycles covering all n elements, we need (k₁-1) + (k₂-1) + ... + (kc-1) = n - c swaps total.

The solution first determines the target sorted order by computing digit sums, then maps each element to where it should go. By marking visited elements while traversing cycles, we count the total number of cycles and derive our answer.

Learn more about Sorting patterns.

Solution Approach

The implementation follows these steps to find the minimum number of swaps:

Step 1: Calculate digit sums

def f(x: int) -> int:
    s = 0
    while x:
        s += x % 10
        x //= 10
    return s

This helper function computes the sum of digits for any number by repeatedly extracting the last digit using modulo 10 and integer division.

Step 2: Create the sorted target arrangement

arr = sorted((f(x), x) for x in nums)

We create tuples of (digit_sum, value) for each number and sort them. Python's sort naturally handles our requirements: it sorts by digit sum first, and when digit sums are equal, it sorts by the actual value (smaller first).

Step 3: Build position mapping

d = {a[1]: i for i, a in enumerate(arr)}

We create a dictionary that maps each value to its target index in the sorted array. For each value in the original array, we can now look up where it should be positioned.

Step 4: Count cycles using graph traversal

ans = n
vis = [False] * n
for i in range(n):
    if not vis[i]:
        ans -= 1
        j = i
        while not vis[j]:
            vis[j] = True
            j = d[nums[j]]

• We initialize ans = n (total elements) and will subtract the number of cycles found
• The vis array tracks which positions we've already processed
• For each unvisited position i, we start tracing a cycle:
  • Mark the current position as visited
  • Look up where the current element should go: d[nums[j]]
  • Continue following this chain until we return to a visited position (completing the cycle)
• Each time we start a new cycle (when vis[i] is False), we decrement ans by 1

The final answer is n - (number of cycles), which gives us the minimum swaps needed.

Time Complexity: O(n log n) for sorting, plus O(n) for cycle detection Space Complexity: O(n) for the mapping dictionary and visited array

Example Walkthrough

Let's trace through the algorithm with nums = [312, 123, 456]:

Step 1: Calculate digit sums

• 312: digit sum = 3 + 1 + 2 = 6
• 123: digit sum = 1 + 2 + 3 = 6
• 456: digit sum = 4 + 5 + 6 = 15

Step 2: Create sorted target arrangement Creating tuples: [(6, 312), (6, 123), (15, 456)] After sorting: [(6, 123), (6, 312), (15, 456)] Target array: [123, 312, 456]

Step 3: Build position mapping

d = {123: 0, 312: 1, 456: 2}

This tells us:

• 123 should be at index 0
• 312 should be at index 1
• 456 should be at index 2

Step 4: Count cycles Initial state: nums = [312, 123, 456], vis = [False, False, False], ans = 3

Cycle detection process:

Starting at index 0:

• Current element: 312
• Where should 312 go? d[312] = 1
• Mark index 0 as visited: vis = [True, False, False]
• Move to index 1

At index 1:

• Current element: 123
• Where should 123 go? d[123] = 0
• Mark index 1 as visited: vis = [True, True, False]
• Move to index 0
• Index 0 is already visited → cycle complete!
• This cycle has elements at positions 0 and 1
• Decrement ans: ans = 3 - 1 = 2

Starting at index 2:

• Current element: 456
• Where should 456 go? d[456] = 2
• Mark index 2 as visited: vis = [True, True, True]
• Move to index 2
• Index 2 is already visited → cycle complete!
• This is a self-loop (element already in correct position)
• Decrement ans: ans = 2 - 1 = 1

Result: We found 2 cycles, so minimum swaps = 3 - 2 = 1

Verification: Indeed, we only need 1 swap (swap positions 0 and 1) to transform [312, 123, 456] into [123, 312, 456].

Solution Implementation

Time and Space Complexity

Time Complexity: O(n log n + n * d) where n is the length of the input array and d is the maximum number of digits in any element.

• The function f(x) calculates the sum of digits, which takes O(d) time where d is the number of digits in x.
• Creating the arr list with digit sums requires O(n * d) time as we call f(x) for each of the n elements.
• Sorting the arr takes O(n log n) time.
• Building the dictionary d takes O(n) time.
• The cycle detection loop visits each element exactly once, taking O(n) time.
• Overall: O(n * d + n log n + n + n) = O(n log n + n * d)

Since the number of digits d in an integer is typically O(log m) where m is the maximum value, and assuming m is bounded by a constant or polynomial in n, the time complexity simplifies to O(n log n).

Space Complexity: O(n)

• The arr list stores n tuples: O(n) space.
• The dictionary d stores n key-value pairs: O(n) space.
• The vis boolean array has size n: O(n) space.
• The sorting algorithm typically uses O(log n) to O(n) auxiliary space depending on implementation.
• Overall: O(n + n + n) = O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Digit Sum Calculation for Edge Cases

A common mistake is not handling the digit sum calculation properly for single-digit numbers or zero values. While the current solution handles positive integers correctly, developers often write buggy implementations:

Incorrect Implementation:

def calculate_digit_sum(number: int) -> int:
    digit_sum = 0
    while number > 0:  # Bug: doesn't handle number = 0 properly
        digit_sum += number % 10
        number = number / 10  # Bug: using float division instead of integer division
    return digit_sum

Correct Implementation:

def calculate_digit_sum(number: int) -> int:
    # Handle zero explicitly if needed
    if number == 0:
        return 0
    digit_sum = 0
    while number:
        digit_sum += number % 10
        number //= 10  # Use integer division
    return digit_sum

2. Misunderstanding the Sorting Order

Developers often make mistakes in the sorting logic, especially when handling ties in digit sums:

Incorrect Approach:

# Wrong: Sorting by value first, then digit sum
sorted_pairs = sorted((num, calculate_digit_sum(num)) for num in nums)

Correct Approach:

# Correct: Sort by (digit_sum, value) - Python naturally sorts tuples element by element
sorted_pairs = sorted((calculate_digit_sum(num), num) for num in nums)

3. Building the Wrong Position Mapping

A critical error is mapping positions incorrectly, which leads to wrong cycle detection:

Incorrect Mapping:

# Wrong: Maps index to value instead of value to target index
value_to_target_position = {i: pair[1] for i, pair in enumerate(sorted_pairs)}

Correct Mapping:

# Correct: Maps each value to its target position in sorted array
value_to_target_position = {pair[1]: i for i, pair in enumerate(sorted_pairs)}

4. Cycle Detection Logic Errors

The cycle traversal can go wrong if we don't properly track visited positions or follow the wrong chain:

Common Mistake:

for i in range(n):
    if not visited[i]:
        min_swaps_needed -= 1
        # Wrong: Not properly traversing the cycle
        visited[i] = True
        j = value_to_target_position[nums[i]]
        visited[j] = True  # Only marks two positions, missing the rest of the cycle

Correct Implementation:

for i in range(n):
    if not visited[i]:
        min_swaps_needed -= 1
        current_position = i
        # Traverse the entire cycle
        while not visited[current_position]:
            visited[current_position] = True
            current_position = value_to_target_position[nums[current_position]]

5. Assuming Array Indices Instead of Values

Since the problem states the array contains "distinct positive integers" (not necessarily 0 to n-1), a common mistake is treating array values as if they were indices:

Incorrect Assumption:

# Wrong: Assumes nums contains values 0 to n-1
for i in range(n):
    if not visited[i]:
        j = i
        while not visited[j]:
            visited[j] = True
            j = nums[j]  # Bug: Using value as index directly

Correct Approach:

# Correct: Use the mapping to find target positions
for i in range(n):
    if not visited[i]:
        j = i
        while not visited[j]:
            visited[j] = True
            j = value_to_target_position[nums[j]]  # Look up where nums[j] should go

Prevention Tips:

1. Always test with edge cases like single-digit numbers and numbers with trailing zeros
2. Verify the sorting order with a small example before implementing cycle detection
3. Draw out the permutation cycles on paper to understand the mapping relationships
4. Remember that the visited array tracks indices, not values
5. Test with arrays containing large numbers (not just 0 to n-1) to catch index/value confusion