How do you solve Minimum Swaps to Sort by Digit Sum on LeetCode?

Minimum Swaps to Sort by Digit Sum (LeetCode #3551) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Understanding digit sums is crucial for sorting.

What is the time complexity of Minimum Swaps to Sort by Digit Sum?

The optimal solution for Minimum Swaps to Sort by Digit Sum runs in O(n log n) time and uses O(n) space. The sorting step dominates the complexity, making it O(n log n).

What is the brute force solution for Minimum Swaps to Sort by Digit Sum?

The brute force approach for Minimum Swaps to Sort by Digit Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try all possible swaps to sort the array based on digit sums. This is inefficient but straightforward. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Swaps to Sort by Digit Sum?

Minimum Swaps to Sort by Digit Sum uses the following concepts: Array, Hash Table, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Swaps to Sort by Digit Sum?

Explain your thought process clearly. Practice cycle detection problems. Be ready to discuss time and space complexities.

What are common mistakes when solving Minimum Swaps to Sort by Digit Sum?

Forgetting to handle cycles in the optimal approach. Not using a stable sort for tie-breaking.

#3551

Minimum Swaps to Sort by Digit Sum

Medium

ArrayHash TableSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

By sorting based on digit sums and using a cycle detection method, we can calculate swaps efficiently.

⚙️

Algorithm

3 steps

1Step 1: Calculate the digit sums and create a list of tuples (digit sum, original index).
2Step 2: Sort this list based on digit sums and values to get the target positions.
3Step 3: Use cycle detection to count the number of swaps needed.

solution.py16 lines

1def minSwaps(nums):
2    def digit_sum(x): return sum(int(d) for d in str(x))
3    indexed_nums = [(digit_sum(num), i) for i, num in enumerate(nums)]
4    indexed_nums.sort()
5    visited = [False] * len(nums)
6    swaps = 0
7    for i in range(len(nums)):
8        if visited[i] or indexed_nums[i][1] == i:
9            continue
10        cycle_size = 0
11        while not visited[i]:
12            visited[i] = True
13            i = indexed_nums[i][1]
14            cycle_size += 1
15        swaps += cycle_size - 1
16    return swaps

ℹ

Complexity note: The sorting step dominates the complexity, making it O(n log n).

1Understanding digit sums is crucial for sorting.
2Cycle detection efficiently counts swaps.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.