2191. Sort the Jumbled Numbers

Description

You are given a 0-indexed integer array mapping which represents the mapping rule of a shuffled decimal system. mapping[i] = j means digit i should be mapped to digit j in this system.

The mapped value of an integer is the new integer obtained by replacing each occurrence of digit i in the integer with mapping[i] for all 0 <= i <= 9.

You are also given another integer array nums. Return the array nums sorted in non-decreasing order based on the mapped values of its elements.

Notes:

• Elements with the same mapped values should appear in the same relative order as in the input.
• The elements of nums should only be sorted based on their mapped values and not be replaced by them.

 

Example 1:

Input: mapping = [8,9,4,0,2,1,3,5,7,6], nums = [991,338,38]
Output: [338,38,991]
Explanation: 
Map the number 991 as follows:
1. mapping[9] = 6, so all occurrences of the digit 9 will become 6.
2. mapping[1] = 9, so all occurrences of the digit 1 will become 9.
Therefore, the mapped value of 991 is 669.
338 maps to 007, or 7 after removing the leading zeros.
38 maps to 07, which is also 7 after removing leading zeros.
Since 338 and 38 share the same mapped value, they should remain in the same relative order, so 338 comes before 38.
Thus, the sorted array is [338,38,991].

Example 2:

Input: mapping = [0,1,2,3,4,5,6,7,8,9], nums = [789,456,123]
Output: [123,456,789]
Explanation: 789 maps to 789, 456 maps to 456, and 123 maps to 123. Thus, the sorted array is [123,456,789].

 

Constraints:

• mapping.length == 10
• 0 <= mapping[i] <= 9
• All the values of mapping[i] are unique.
• 1 <= nums.length <= 3 * 104
• 0 <= nums[i] < 109

Solutions

Solution 1: Custom Sorting

We traverse each element $nums[i]$ in the array $nums$, store its mapped value $y$ and index $i$ into the array $arr$, then sort the array $arr$. Finally, we extract the index $i$ from the sorted array $arr$, convert it to the element $nums[i]$ in the original array $nums$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.

Python3JavaC++GoTypeScriptJavaScriptRustC#

class Solution:
    def sortJumbled(self, mapping: List[int], nums: List[int]) -> List[int]:
        def f(x: int) -> int:
            if x == 0:
                return mapping[0]
            y, k = 0, 1
            while x:
                x, v = divmod(x, 10)
                v = mapping[v]
                y = k * v + y
                k *= 10
            return y

        arr = sorted((f(x), i) for i, x in enumerate(nums))
        return [nums[i] for _, i in arr]

class Solution {
    private int[] mapping;

    public int[] sortJumbled(int[] mapping, int[] nums) {
        this.mapping = mapping;
        int n = nums.length;
        int[][] arr = new int[n][0];
        for (int i = 0; i < n; ++i) {
            arr[i] = new int[] {f(nums[i]), i};
        }
        Arrays.sort(arr, (a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            ans[i] = nums[arr[i][1]];
        }
        return ans;
    }

    private int f(int x) {
        if (x == 0) {
            return mapping[0];
        }
        int y = 0;
        for (int k = 1; x > 0; x /= 10) {
            int v = mapping[x % 10];
            y = k * v + y;
            k *= 10;
        }
        return y;
    }
}

class Solution {
public:
    vector<int> sortJumbled(vector<int>& mapping, vector<int>& nums) {
        auto f = [&](int x) {
            if (x == 0) {
                return mapping[0];
            }
            int y = 0;
            for (int k = 1; x; x /= 10) {
                int v = mapping[x % 10];
                y = k * v + y;
                k *= 10;
            }
            return y;
        };
        const int n = nums.size();
        vector<pair<int, int>> arr;
        for (int i = 0; i < n; ++i) {
            arr.emplace_back(f(nums[i]), i);
        }
        sort(arr.begin(), arr.end());
        vector<int> ans;
        for (const auto& [_, i] : arr) {
            ans.push_back(nums[i]);
        }
        return ans;
    }
};

func sortJumbled(mapping []int, nums []int) (ans []int) {
    n := len(nums)
    f := func(x int) (y int) {
        if x == 0 {
            return mapping[0]
        }
        for k := 1; x > 0; x /= 10 {
            v := mapping[x%10]
            y = k*v + y
            k *= 10
        }
        return
    }
    arr := make([][2]int, n)
    for i, x := range nums {
        arr[i] = [2]int{f(x), i}
    }
    sort.Slice(arr, func(i, j int) bool { return arr[i][0] < arr[j][0] || arr[i][0] == arr[j][0] && arr[i][1] < arr[j][1] })
    for _, p := range arr {
        ans = append(ans, nums[p[1]])
    }
    return
}

function sortJumbled(mapping: number[], nums: number[]): number[] {
    const n = nums.length;
    const f = (x: number): number => {
        if (x === 0) {
            return mapping[0];
        }
        let y = 0;
        for (let k = 1; x; x = (x / 10) | 0) {
            const v = mapping[x % 10];
            y += v * k;
            k *= 10;
        }
        return y;
    };
    const arr: number[][] = nums.map((x, i) => [f(x), i]);
    arr.sort((a, b) => (a[0] === b[0] ? a[1] - b[1] : a[0] - b[0]));
    return arr.map(x => nums[x[1]]);
}

/**
 * @param {number[]} mapping
 * @param {number[]} nums
 * @return {number[]}
 */
var sortJumbled = function (mapping, nums) {
    const n = nums.length;
    const f = x => {
        if (x === 0) {
            return mapping[0];
        }
        let y = 0;
        for (let k = 1; x; x = (x / 10) | 0) {
            const v = mapping[x % 10];
            y += v * k;
            k *= 10;
        }
        return y;
    };
    const arr = nums.map((x, i) => [f(x), i]);
    arr.sort((a, b) => (a[0] === b[0] ? a[1] - b[1] : a[0] - b[0]));
    return arr.map(x => nums[x[1]]);
};

impl Solution {
    pub fn sort_jumbled(mapping: Vec<i32>, nums: Vec<i32>) -> Vec<i32> {
        let f = |x: i32| -> i32 {
            if x == 0 {
                return mapping[0];
            }
            let mut y = 0;
            let mut k = 1;
            let mut num = x;
            while num != 0 {
                let v = mapping[(num % 10) as usize];
                y = k * v + y;
                k *= 10;
                num /= 10;
            }
            y
        };

        let n = nums.len();
        let mut arr: Vec<(i32, usize)> = Vec::with_capacity(n);
        for i in 0..n {
            arr.push((f(nums[i]), i));
        }
        arr.sort();

        let mut ans: Vec<i32> = Vec::with_capacity(n);
        for (_, i) in arr {
            ans.push(nums[i]);
        }
        ans
    }
}

public class Solution {
    public int[] SortJumbled(int[] mapping, int[] nums) {
        Func<int, int> f = (int x) => {
            if (x == 0) {
                return mapping[0];
            }
            int y = 0;
            int k = 1;
            int num = x;
            while (num != 0) {
                int v = mapping[num % 10];
                y = k * v + y;
                k *= 10;
                num /= 10;
            }
            return y;
        };

        int n = nums.Length;
        List<(int, int)> arr = new List<(int, int)>();
        for (int i = 0; i < n; ++i) {
            arr.Add((f(nums[i]), i));
        }
        arr.Sort();

        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            ans[i] = nums[arr[i].Item2];
        }
        return ans;
    }
}

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