DSA LAB | Minimum Swaps to Sort by Digit Sum

01 Understand the Question

You are given an integer array nums. You need to calculate the digit sum for each element and sort the array according to the following rules:

Sort primary by Digit Sum in ascending order.
If two numbers share the same digit sum, sort them by Numerical Value in ascending order.

Determine the minimum number of swaps required to transform the original array into this target order.

Example Walkthrough: `nums = [30, 10, 20]`

Original Index	Number	Digit Sum Calculation	Digit Sum
`0`	30	3 + 0	3
`1`	10	1 + 0	1
`2`	20	2 + 0	2

Sorting by Digit Sum yields the target array: [10, 20, 30].

Swapping 30 and 10 yields [10, 30, 20]. Swapping 30 and 20 yields [10, 20, 30]. Thus, 2 swaps are needed.

Try It: Interactive Digit Sum Calculator

Result: 9 + 8 + 7 = 24

02 What Hints Does the Problem Give Us?

Hint 1

The Sorting Rule is Deterministic

Every number has a fixed, predictable position once digit sums and tie-breaker rules are evaluated.

Hint 2

Sorting Tells Destination, Not Swaps

Sorting gives the final target array, but standard sorting routines perform arbitrary comparisons and swaps rather than finding the minimal sequence.

Hint 3

Original Indices Are Essential

If we retain each number's original index alongside its sorted position, we can construct a mapping of current_position → target_position.

Hint 4

Permutation Cycles Emerge

Mapping elements to their target positions forms closed cycles of misplaced elements. A disjoint cycle containing K elements requires exactly K - 1 swaps to resolve.

Hint 5

Avoid Exhaustive Simulation

By counting cycle lengths mathematically, we bypass brute-force search space explosions and solve the problem in O(N log N) time.

03 Possible Approaches

Infeasible

Approach A: Brute Force Swaps

Recursively simulate all possible element pair swaps to find the shortest path to the sorted state (BFS/DFS over permutations).

Time: O(N!) Space: O(N!)

Explodes exponentially for N > 10. Completely impractical for production constraints.

Sub-Optimal

Approach B: Greedy Position Swapping

Sort the target array. For index i, find where target[i] resides in the current array and swap it into place.

Time: O(N²) Space: O(N)

Repeated array searching takes quadratic time unless complex positional hash maps are maintained.

Recommended

Approach C: Sorting + Cycle Detection

Augment elements as (digitSum, value, originalIndex). Sort them to form permutation cycles, then count disjoint cycle lengths.

Time: O(N log N) Space: O(N)

Optimal, clean, and handles duplicate values deterministically via tie-breaking rules.

04 Pseudocode

Algorithm: Minimum Swaps via Cycle Decomposition

function minSwaps(nums):
    n = length(nums)
    arr = empty list

    // Step 1: Augment elements with digit sum & original index
    for i from 0 to n - 1:
        sum = calculateDigitSum(nums[i])
        arr.append((sum, nums[i], i))

    // Step 2: Sort based on problem criteria (digit sum, then value)
    sort arr ascending by arr[i].sum, then arr[i].value

    // Step 3: Track visited nodes and count cycle lengths
    visited = array of size n initialized to false
    totalSwaps = 0

    for i from 0 to n - 1:
        // Skip already visited elements or elements already in correct position
        if visited[i] or arr[i].originalIndex == i:
            continue

        cycleSize = 0
        j = i

        // Traverse permutation cycle
        while visited[j] is false:
            visited[j] = true
            j = arr[j].originalIndex
            cycleSize = cycleSize + 1

        if cycleSize > 1:
            totalSwaps = totalSwaps + (cycleSize - 1)

    return totalSwaps

05 Construction of Sorted Mapping

For input nums = [30, 10, 20], augmented tuple structures after sorting demonstrate the cycle relation:

Sorted Pos (i)	Number	Digit Sum	Original Index (arr[i].originalIndex)	Element Shift Mapping
`0`	10	1	`1`	Sorted Pos 0 needs element from Original Index 1
`1`	20	2	`2`	Sorted Pos 1 needs element from Original Index 2
`2`	30	3	`0`	Sorted Pos 2 needs element from Original Index 0

Following the i → arr[i].originalIndex links yields the single permutation cycle:

Index 0 → Index 1 → Index 2 → Index 0 (Loop Complete)

Cycle Length = 3. Swaps required for this cycle = 3 - 1 = 2.

06 Interactive Cycle Traversal Visualizer

Watch step-by-step how the while (!visited[j]) traversal marks elements and computes swaps.

Current Node (j) --

Visited State [F, F, F]

Cycle Size 0

Total Swaps 0

Click Next or Auto Play to begin visualization of input [30, 10, 20].

Q: If we set `visited[j] = true`, why does the while-loop keep running?

A: The loop condition while (!visited[j]) is evaluated at the start of each iteration. Inside the body, after setting visited[j] = true, we update j = arr[j].originalIndex to point to the next element in the cycle. The loop only terminates when j steps onto a node that was previously visited!

07 C++ Source Solution

Clean & Efficient C++20 Implementation

Show Comments

#include <vector>
#include <numeric>
#include <algorithm>

using namespace std;

class Solution {
private:
    // Helper function to calculate sum of digits
    int getDigitSum(int val) {
        int sum = 0;
        while (val > 0) {
            sum += val % 10;
            val /= 10;
        }
        return sum;
    }

public:
    int minSwaps(vector<int>& nums) {
        int n = nums.size();
        
        // Element structure: {digitSum, numericalValue, originalIndex}
        struct Element {
            int digitSum;
            int val;
            int origIdx;
        };
        
        vector<Element> arr(n);
        for (int i = 0; i < n; ++i) {
            arr[i] = {getDigitSum(nums[i]), nums[i], i};
        }
        
        // Custom comparator for deterministic tie-breaking
        sort(arr.begin(), arr.end(), [](const Element& a, const Element& b) {
            if (a.digitSum != b.digitSum) {
                return a.digitSum < b.digitSum;
            }
            return a.val < b.val;
        });
        
        vector<bool> visited(n, false);
        int totalSwaps = 0;
        
        // Cycle decomposition
        for (int i = 0; i < n; ++i) {
            // Skip visited elements or elements already in target position
            if (visited[i] || arr[i].origIdx == i) {
                continue;
            }
            
            int cycleSize = 0;
            int j = i;
            
            // Traverse the permutation cycle
            while (!visited[j]) {
                visited[j] = true;
                j = arr[j].origIdx;
                cycleSize++;
            }
            
            if (cycleSize > 1) {
                totalSwaps += (cycleSize - 1);
            }
        }
        
        return totalSwaps;
    }
};

08 Execution Trace / Dry Run Table

Dry run table for nums = [30, 10, 20] after sorting:

Loop i	Start j	Visited Array [0, 1, 2]	Cycle Traversal	Cycle Size	Swaps Added	Total Swaps
`i = 0`	`j = 0`	`[T, T, T]`	0 → 1 → 2 → 0	3	3 - 1 = 2	2
`i = 1`	--	`[T, T, T]`	Skipped (visited[1] == true)	--	0	2
`i = 2`	--	`[T, T, T]`	Skipped (visited[2] == true)	--	0	2

09 Complexity Analysis

Time Complexity

Digit Sum Computation: O(N × log₁₀(M)) where M is the maximum value in nums. Since M ≤ 10⁹, digit sum calculation is O(1) per number.
Sorting: O(N log N) to sort the augmented elements.
Cycle Detection: O(N) as every element is visited at most once across all cycle traversals.

Overall Time Complexity: O(N log N)

Space Complexity

Augmented Array: O(N) auxiliary space to store digit sums and original indices.
Visited Array: O(N) boolean array.

Overall Auxiliary Space: O(N)

10 Interactive Practice Playground

Test the algorithm with your own custom input array!

Comma-separated array:

11 Knowledge Check

1. What is the primary purpose of the visited array during cycle decomposition?

2. If a disjoint permutation cycle contains 5 elements, how many swaps are needed to place all 5 elements in their correct positions?

3. Why do we store the original indices before sorting the array?

Minimum Swaps to Sort by Digit Sum

3551. Minimum Swaps to Sort by Digit Sum

01 Understand the Question

Example Walkthrough: `nums = [30, 10, 20]`

Try It: Interactive Digit Sum Calculator

02 What Hints Does the Problem Give Us?

The Sorting Rule is Deterministic

Sorting Tells Destination, Not Swaps

Original Indices Are Essential

Permutation Cycles Emerge

Avoid Exhaustive Simulation

03 Possible Approaches

Approach A: Brute Force Swaps

Approach B: Greedy Position Swapping

Approach C: Sorting + Cycle Detection

04 Pseudocode

05 Construction of Sorted Mapping

06 Interactive Cycle Traversal Visualizer

Q: If we set `visited[j] = true`, why does the while-loop keep running?

07 C++ Source Solution

08 Execution Trace / Dry Run Table

09 Complexity Analysis

Time Complexity

Space Complexity

10 Interactive Practice Playground

Detected Permutation Cycles:

11 Knowledge Check

Minimum Swaps to Sort by Digit Sum

3551. Minimum Swaps to Sort by Digit Sum

01 Understand the Question

Example Walkthrough: nums = [30, 10, 20]

Try It: Interactive Digit Sum Calculator

02 What Hints Does the Problem Give Us?

The Sorting Rule is Deterministic

Sorting Tells Destination, Not Swaps

Original Indices Are Essential

Permutation Cycles Emerge

Avoid Exhaustive Simulation

03 Possible Approaches

Approach A: Brute Force Swaps

Approach B: Greedy Position Swapping

Approach C: Sorting + Cycle Detection

04 Pseudocode

05 Construction of Sorted Mapping

06 Interactive Cycle Traversal Visualizer

Q: If we set visited[j] = true, why does the while-loop keep running?

07 C++ Source Solution

08 Execution Trace / Dry Run Table

09 Complexity Analysis

Time Complexity

Space Complexity

10 Interactive Practice Playground

Detected Permutation Cycles:

11 Knowledge Check

Example Walkthrough: `nums = [30, 10, 20]`

Q: If we set `visited[j] = true`, why does the while-loop keep running?