3551. Minimum Swaps to Sort by Digit Sum

Description

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.

Solutions

Solution 1

Thinking

The target order is uniquely given by \((\textit{digitSum}(x), x)\). The minimum number of swaps in a permutation is \(n\) minus the number of cycles.

Map each value to its index in the sorted array and walk those pointers, marking visits. Each cycle of length \(\ell\) needs \(\ell-1\) swaps, so the total is \(n\) minus the cycle count.

Python3JavaC++GoTypeScript

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

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;
    }
}

class Solution {
public:
    int f(int x) {
        int s = 0;
        while (x) {
            s += x % 10;
            x /= 10;
        }
        return s;
    }

    int minSwaps(vector<int>& nums) {
        int n = nums.size();
        vector<pair<int, int>> arr(n);
        for (int i = 0; i < n; ++i) arr[i] = {f(nums[i]), nums[i]};
        sort(arr.begin(), arr.end());
        unordered_map<int, int> d;
        for (int i = 0; i < n; ++i) d[arr[i].second] = i;
        vector<char> vis(n, 0);
        int ans = n;
        for (int i = 0; i < n; ++i) {
            if (!vis[i]) {
                --ans;
                int j = i;
                while (!vis[j]) {
                    vis[j] = 1;
                    j = d[nums[j]];
                }
            }
        }
        return ans;
    }
};

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
}

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;
}

3551. Minimum Swaps to Sort by Digit Sum

Description

Solutions

Solution 1

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