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Formatted question description: https://leetcode.ca/all/1122.html

1122. Relative Sort Array (Easy)

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 don't 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]

 

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.

Related Topics:
Array, Sort

Solution 1.

  • class Solution {
    public int[] relativeSortArray(int[] arr1, int[] arr2) {
        Map<Integer, Integer> orderMap = new HashMap<Integer, Integer>();
        int length1 = arr1.length, length2 = arr2.length;
        for (int i = 0; i < length2; i++) {
            int num = arr2[i];
            orderMap.put(num, i);
        }
        for (int i = 1; i < length1; i++) {
            int num = arr1[i];
            int order = orderMap.getOrDefault(num, Integer.MAX_VALUE);
            int insertIndex = -1;
            for (int j = i - 1; j >= 0; j--) {
                int curOrder = orderMap.getOrDefault(arr1[j], Integer.MAX_VALUE);
                if (curOrder < order || curOrder == order && arr1[j] < num) {
                    insertIndex = j;
                    break;
                }
                arr1[j + 1] = arr1[j];
            }
            insertIndex++;
            arr1[insertIndex] = num;
        }
        return arr1;
    }
}

############

class Solution {
    public int[] relativeSortArray(int[] arr1, int[] arr2) {
        int[] mp = new int[1001];
        for (int x : arr1) {
            ++mp[x];
        }
        int i = 0;
        for (int x : arr2) {
            while (mp[x]-- > 0) {
                arr1[i++] = x;
            }
        }
        for (int j = 0; j < mp.length; ++j) {
            while (mp[j]-- > 0) {
                arr1[i++] = j;
            }
        }
        return arr1;
    }
}

  • // OJ: https://leetcode.com/problems/relative-sort-array/
// Time: O(AlogA)
// Space: O(B)
class Solution {
public:
    vector<int> relativeSortArray(vector<int>& A, vector<int>& B) {
        unordered_map<int, int> m;
        for (int i = 0; i < B.size(); ++i) m[B[i]] = i;
        sort(begin(A), end(A), [&](int a, int b) {
            if (m.count(a) && m.count(b)) return m[a] < m[b];
            if (m.count(a) == 0 && m.count(b) == 0) return a < b;
            return m.count(a) > 0;
        });
        return A;
    }
};

  • class Solution:
    def relativeSortArray(self, arr1: List[int], arr2: List[int]) -> List[int]:
        mp = [0] * 1001
        for x in arr1:
            mp[x] += 1
        i = 0
        for x in arr2:
            while mp[x] > 0:
                arr1[i] = x
                mp[x] -= 1
                i += 1
        for x, cnt in enumerate(mp):
            for _ in range(cnt):
                arr1[i] = x
                i += 1
        return arr1

############

# 1122. Relative Sort Array
# https://leetcode.com/problems/relative-sort-array/

class Solution:
    def relativeSortArray(self, arr1: List[int], arr2: List[int]) -> List[int]:
        mp = collections.Counter(arr1) 
        s = set(arr2)
        res = []
        
        for num in arr2:
            res += [num] * mp[num]
        
        leftover = []
        for num in arr1:
            if num not in s:
                leftover.append(num)

        return res + sorted(leftover)


  • func relativeSortArray(arr1 []int, arr2 []int) []int {
	mp := make([]int, 1001)
	for _, x := range arr1 {
		mp[x]++
	}
	i := 0
	for _, x := range arr2 {
		for mp[x] > 0 {
			arr1[i] = x
			mp[x]--
			i++
		}
	}
	for j, cnt := range mp {
		for cnt > 0 {
			arr1[i] = j
			i++
			cnt--
		}
	}
	return arr1
}

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