#3551

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Minimum Swaps to Sort by Digit Sum

Calculate the minimum swaps to sort an array by digit sum, ensuring correct order with tiebreaker values for efficiency.

array hash table sorting

Problem Statement

You are given an array nums containing distinct positive integers. Sort the array in increasing order based on the sum of digits of each number. If two numbers have the same digit sum, place the smaller number first. The goal is to transform the original array into this sorted order using the fewest swaps.

A swap exchanges the values at two distinct positions in the array. Return the minimum number of swaps required to rearrange nums into the sorted digit-sum order. Use efficient array scanning and mapping techniques to track positions and swap counts.

Examples

Example 1

Input: nums = [37,100]

Output: 1

Example 2

Input: nums = [22,14,33,7]

Output: 0

Example 3

Input: nums = [18,43,34,16]

Output: 2

Constraints

1 <= nums.length <= 105
1 <= nums[i] <= 109
nums consists of distinct positive integers.

Solution Approach

Compute Digit Sums and Sort

Calculate the digit sum for each number in nums. Sort the array based on digit sum, using the numeric value as a tiebreaker to maintain correct order.

Map Original Indices to Sorted Positions

Create a hash table that maps each original number to its index in the sorted array. This enables constant-time lookup for where each number should move, leveraging the array scanning plus hash lookup pattern.

Count Minimum Swaps

Traverse the original array and for each element not in its correct position, swap it with the element at its target index. Track visited positions to avoid redundant swaps. Sum all swaps to return the minimum required.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n log n) for sorting plus O(n) for mapping and swap counting, yielding overall O(n log n). Space complexity is O(n) for storing digit sums and the index map, as each element needs position tracking.

What Interviewers Usually Probe

Candidate sorts based on a computed key and tiebreaker, checking digit sum logic.
Using a hash map to track original indices shows recognition of array scanning plus hash lookup.
Counting swaps via visited positions indicates understanding of minimal swap strategies.

Common Pitfalls or Variants

Common pitfalls

Ignoring the numeric value tiebreaker when digit sums are equal, leading to wrong sort order.
Failing to track visited indices, which can double-count swaps or cause infinite loops.
Attempting swaps without mapping indices, resulting in inefficient O(n^2) operations.

Follow-up variants

Sort by sum of squares of digits instead of digit sum to test generalized key-based swaps.
Handle arrays with repeated numbers to test extensions beyond distinct values.
Compute maximum swaps allowed rather than minimum to explore constrained swap scenarios.

FAQ

What is the main pattern for Minimum Swaps to Sort by Digit Sum?

The problem follows an array scanning plus hash lookup pattern where digit sums guide the sorting key.

Can this approach handle arrays up to length 105?

Yes, sorting and mapping indices in O(n log n) with hash lookup scales efficiently for large arrays.

Do I need to consider ties in digit sums?

Yes, if two numbers share the same digit sum, the smaller number must come first to maintain correct order.

How does GhostInterview minimize swap counting errors?

It tracks visited indices and uses a mapping of original to sorted positions to avoid redundant or missed swaps.

What if the array contains repeated values?

This exact problem assumes distinct positive integers; repeated values require a modified approach to handle duplicates correctly.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def minSwaps(self, nums: List[int]) -> int:
        def f(x: int) -> int:
            s = 0
            while x:
                s += x % 10
                x //= 10
            return s

        n = len(nums)
        arr = sorted((f(x), x) for x in nums)
        d = {a[1]: i for i, a in enumerate(arr)}
        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]]
        return ans

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