1122. Relative Sort Array

Description

Given two arrays arr1 and arr2, the elements of arr2 are distinct, and all elements in arr2 are also in arr1.

Sort the elements of arr1 such that the relative ordering of items in arr1 are the same as in arr2. Elements that do not appear in arr2 should be placed at the end of arr1 in ascending order.

 

Example 1:

Input: arr1 = [2,3,1,3,2,4,6,7,9,2,19], arr2 = [2,1,4,3,9,6]
Output: [2,2,2,1,4,3,3,9,6,7,19]

Example 2:

Input: arr1 = [28,6,22,8,44,17], arr2 = [22,28,8,6]
Output: [22,28,8,6,17,44]

 

Constraints:

• 1 <= arr1.length, arr2.length <= 1000
• 0 <= arr1[i], arr2[i] <= 1000
• All the elements of arr2 are distinct.
• Each arr2[i] is in arr1.

Solutions

Solution 1: Custom Sorting

First, we use a hash table $pos$ to record the position of each element in array $arr2$. Then, we map each element in array $arr1$ to a tuple $(pos.get(x, 1000 + x), x)$, and sort these tuples. Finally, we take out the second element of all tuples and return it.

The time complexity is $O(n \times \log n + m)$, and the space complexity is $O(n + m)$. Here, $n$ and $m$ are the lengths of arrays $arr1$ and $arr2$, respectively.

Python3JavaC++GoTypeScriptSwift

class Solution:
    def relativeSortArray(self, arr1: List[int], arr2: List[int]) -> List[int]:
        pos = {x: i for i, x in enumerate(arr2)}
        return sorted(arr1, key=lambda x: pos.get(x, 1000 + x))

class Solution {
    public int[] relativeSortArray(int[] arr1, int[] arr2) {
        Map<Integer, Integer> pos = new HashMap<>(arr2.length);
        for (int i = 0; i < arr2.length; ++i) {
            pos.put(arr2[i], i);
        }
        int[][] arr = new int[arr1.length][0];
        for (int i = 0; i < arr.length; ++i) {
            arr[i] = new int[] {arr1[i], pos.getOrDefault(arr1[i], arr2.length + arr1[i])};
        }
        Arrays.sort(arr, (a, b) -> a[1] - b[1]);
        for (int i = 0; i < arr.length; ++i) {
            arr1[i] = arr[i][0];
        }
        return arr1;
    }
}

class Solution {
public:
    vector<int> relativeSortArray(vector<int>& arr1, vector<int>& arr2) {
        unordered_map<int, int> pos;
        for (int i = 0; i < arr2.size(); ++i) {
            pos[arr2[i]] = i;
        }
        vector<pair<int, int>> arr;
        for (int i = 0; i < arr1.size(); ++i) {
            int j = pos.count(arr1[i]) ? pos[arr1[i]] : arr2.size();
            arr.emplace_back(j, arr1[i]);
        }
        sort(arr.begin(), arr.end());
        for (int i = 0; i < arr1.size(); ++i) {
            arr1[i] = arr[i].second;
        }
        return arr1;
    }
};

func relativeSortArray(arr1 []int, arr2 []int) []int {
    pos := map[int]int{}
    for i, x := range arr2 {
        pos[x] = i
    }
    arr := make([][2]int, len(arr1))
    for i, x := range arr1 {
        if p, ok := pos[x]; ok {
            arr[i] = [2]int{p, x}
        } else {
            arr[i] = [2]int{len(arr2), x}
        }
    }
    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 i, x := range arr {
        arr1[i] = x[1]
    }
    return arr1
}

function relativeSortArray(arr1: number[], arr2: number[]): number[] {
    const pos: Map<number, number> = new Map();
    for (let i = 0; i < arr2.length; ++i) {
        pos.set(arr2[i], i);
    }
    const arr: number[][] = [];
    for (const x of arr1) {
        const j = pos.get(x) ?? arr2.length;
        arr.push([j, x]);
    }
    arr.sort((a, b) => a[0] - b[0] || a[1] - b[1]);
    return arr.map(a => a[1]);
}

class Solution {
    func relativeSortArray(_ arr1: [Int], _ arr2: [Int]) -> [Int] {
        var pos = [Int: Int]()
        for (i, x) in arr2.enumerated() {
            pos[x] = i
        }
        var arr = [(Int, Int)]()
        for x in arr1 {
            let j = pos[x] ?? arr2.count
            arr.append((j, x))
        }
        arr.sort { $0.0 < $1.0 || ($0.0 == $1.0 && $0.1 < $1.1) }
        return arr.map { $0.1 }
    }
}

Solution 2: Counting Sort

We can use the idea of counting sort. First, count the occurrence of each element in array $arr1$. Then, according to the order in array $arr2$, put the elements in $arr1$ into the answer array $ans$ according to their occurrence. Finally, we traverse all elements in $arr1$ and put the elements that do not appear in $arr2$ in ascending order at the end of the answer array $ans$.

The time complexity is $O(n + m)$, and the space complexity is $O(n)$. Where $n$ and $m$ are the lengths of arrays $arr1$ and $arr2$ respectively.

Python3

class Solution:
    def relativeSortArray(self, arr1: List[int], arr2: List[int]) -> List[int]:
        cnt = Counter(arr1)
        ans = []
        for x in arr2:
            ans.extend([x] * cnt[x])
            cnt.pop(x)
        mi, mx = min(arr1), max(arr1)
        for x in range(mi, mx + 1):
            ans.extend([x] * cnt[x])
        return ans

Java

class Solution {
    public int[] relativeSortArray(int[] arr1, int[] arr2) {
        int[] cnt = new int[1001];
        int mi = 1001, mx = 0;
        for (int x : arr1) {
            ++cnt[x];
            mi = Math.min(mi, x);
            mx = Math.max(mx, x);
        }
        int m = arr1.length;
        int[] ans = new int[m];
        int i = 0;
        for (int x : arr2) {
            while (cnt[x] > 0) {
                --cnt[x];
                ans[i++] = x;
            }
        }
        for (int x = mi; x <= mx; ++x) {
            while (cnt[x] > 0) {
                --cnt[x];
                ans[i++] = x;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> relativeSortArray(vector<int>& arr1, vector<int>& arr2) {
        vector<int> cnt(1001);
        for (int x : arr1) {
            ++cnt[x];
        }
        auto [mi, mx] = minmax_element(arr1.begin(), arr1.end());
        vector<int> ans;
        for (int x : arr2) {
            while (cnt[x]) {
                ans.push_back(x);
                --cnt[x];
            }
        }
        for (int x = *mi; x <= *mx; ++x) {
            while (cnt[x]) {
                ans.push_back(x);
                --cnt[x];
            }
        }
        return ans;
    }
};

Go

func relativeSortArray(arr1 []int, arr2 []int) []int {
    cnt := make([]int, 1001)
    mi, mx := 1001, 0
    for _, x := range arr1 {
        cnt[x]++
        mi = min(mi, x)
        mx = max(mx, x)
    }
    ans := make([]int, 0, len(arr1))
    for _, x := range arr2 {
        for cnt[x] > 0 {
            ans = append(ans, x)
            cnt[x]--
        }
    }
    for x := mi; x <= mx; x++ {
        for cnt[x] > 0 {
            ans = append(ans, x)
            cnt[x]--
        }
    }
    return ans
}

TypeScript

function relativeSortArray(arr1: number[], arr2: number[]): number[] {
    const cnt = Array(1001).fill(0);
    let mi = Number.POSITIVE_INFINITY;
    let mx = Number.NEGATIVE_INFINITY;

    for (const x of arr1) {
        cnt[x]++;
        mi = Math.min(mi, x);
        mx = Math.max(mx, x);
    }

    const ans: number[] = [];
    for (const x of arr2) {
        while (cnt[x]) {
            cnt[x]--;
            ans.push(x);
        }
    }

    for (let i = mi; i <= mx; i++) {
        while (cnt[i]) {
            cnt[i]--;
            ans.push(i);
        }
    }

    return ans;
}

Swift

class Solution {
    func relativeSortArray(_ arr1: [Int], _ arr2: [Int]) -> [Int] {
        var cnt = [Int](repeating: 0, count: 1001)
        for x in arr1 {
            cnt[x] += 1
        }

        guard let mi = arr1.min(), let mx = arr1.max() else {
            return []
        }

        var ans = [Int]()
        for x in arr2 {
            while cnt[x] > 0 {
                ans.append(x)
                cnt[x] -= 1
            }
        }

        for x in mi...mx {
            while cnt[x] > 0 {
                ans.append(x)
                cnt[x] -= 1
            }
        }

        return ans
    }
}

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