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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.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  };
  

• 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))
  
  

• 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]);
  }
  
  

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