LeetCode #3551 — MEDIUM

Minimum Swaps to Sort by Digit Sum

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Example 1:

Input: nums = [37,100]

Output: 1

Explanation:

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.

Example 2:

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

Output: 0

Explanation:

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.

Example 3:

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

Output: 2

Explanation:

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.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[37,100]

Example 2

[22,14,33,7]

Example 3

[18,43,34,16]

Step 02

Core Insight

What unlocks the optimal approach

First, sort the array based on digit sum (and value as a tiebreaker). Then, map each original value to its index in the sorted array.
Now, the problem reduces to finding the minimum number of swaps to sort this mapped array. The answer is <code>n - number_of_cycles</code> in the permutation.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3551: Minimum Swaps to Sort by Digit Sum
class Solution {
    public int minSwaps(int[] nums) {
        int n = nums.length;
        int[][] arr = new int[n][2];
        for (int i = 0; i < n; i++) {
            arr[i][0] = f(nums[i]);
            arr[i][1] = nums[i];
        }
        Arrays.sort(arr, (a, b) -> {
            if (a[0] != b[0]) return Integer.compare(a[0], b[0]);
            return Integer.compare(a[1], b[1]);
        });
        Map<Integer, Integer> d = new HashMap<>();
        for (int i = 0; i < n; i++) {
            d.put(arr[i][1], i);
        }
        boolean[] vis = new boolean[n];
        int ans = n;
        for (int i = 0; i < n; i++) {
            if (!vis[i]) {
                ans--;
                int j = i;
                while (!vis[j]) {
                    vis[j] = true;
                    j = d.get(nums[j]);
                }
            }
        }
        return ans;
    }

    private int f(int x) {
        int s = 0;
        while (x != 0) {
            s += x % 10;
            x /= 10;
        }
        return s;
    }
}

// Accepted solution for LeetCode #3551: Minimum Swaps to Sort by Digit Sum
func minSwaps(nums []int) int {
	n := len(nums)
	arr := make([][2]int, n)
	for i := 0; i < n; i++ {
		arr[i][0] = f(nums[i])
		arr[i][1] = nums[i]
	}
	sort.Slice(arr, func(i, j int) bool {
		if arr[i][0] != arr[j][0] {
			return arr[i][0] < arr[j][0]
		}
		return arr[i][1] < arr[j][1]
	})
	d := make(map[int]int, n)
	for i := 0; i < n; i++ {
		d[arr[i][1]] = i
	}
	vis := make([]bool, n)
	ans := n
	for i := 0; i < n; i++ {
		if !vis[i] {
			ans--
			j := i
			for !vis[j] {
				vis[j] = true
				j = d[nums[j]]
			}
		}
	}
	return ans
}

func f(x int) int {
	s := 0
	for x != 0 {
		s += x % 10
		x /= 10
	}
	return s
}

# Accepted solution for LeetCode #3551: Minimum Swaps to Sort by Digit Sum
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

// Accepted solution for LeetCode #3551: Minimum Swaps to Sort by Digit Sum
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3551: Minimum Swaps to Sort by Digit Sum
// class Solution {
//     public int minSwaps(int[] nums) {
//         int n = nums.length;
//         int[][] arr = new int[n][2];
//         for (int i = 0; i < n; i++) {
//             arr[i][0] = f(nums[i]);
//             arr[i][1] = nums[i];
//         }
//         Arrays.sort(arr, (a, b) -> {
//             if (a[0] != b[0]) return Integer.compare(a[0], b[0]);
//             return Integer.compare(a[1], b[1]);
//         });
//         Map<Integer, Integer> d = new HashMap<>();
//         for (int i = 0; i < n; i++) {
//             d.put(arr[i][1], i);
//         }
//         boolean[] vis = new boolean[n];
//         int ans = n;
//         for (int i = 0; i < n; i++) {
//             if (!vis[i]) {
//                 ans--;
//                 int j = i;
//                 while (!vis[j]) {
//                     vis[j] = true;
//                     j = d.get(nums[j]);
//                 }
//             }
//         }
//         return ans;
//     }
// 
//     private int f(int x) {
//         int s = 0;
//         while (x != 0) {
//             s += x % 10;
//             x /= 10;
//         }
//         return s;
//     }
// }

// Accepted solution for LeetCode #3551: Minimum Swaps to Sort by Digit Sum
function f(x: number): number {
    let s = 0;
    while (x !== 0) {
        s += x % 10;
        x = Math.floor(x / 10);
    }
    return s;
}

function minSwaps(nums: number[]): number {
    const n = nums.length;
    const arr: [number, number][] = new Array(n);
    for (let i = 0; i < n; i++) {
        arr[i] = [f(nums[i]), nums[i]];
    }
    arr.sort((a, b) => (a[0] !== b[0] ? a[0] - b[0] : a[1] - b[1]));
    const d = new Map<number, number>();
    for (let i = 0; i < n; i++) {
        d.set(arr[i][1], i);
    }
    const vis: boolean[] = new Array(n).fill(false);
    let ans = n;
    for (let i = 0; i < n; i++) {
        if (!vis[i]) {
            ans--;
            let j = i;
            while (!vis[j]) {
                vis[j] = true;
                j = d.get(nums[j])!;
            }
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.