1122. Relative Sort Array

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.

From: LeetCode
Link: 1122. Relative Sort Array

Solution:

Ideas:

count numbers first, output numbers in arr2 order, then output remaining numbers from small to large.

Code:

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* relativeSortArray(int* arr1, int arr1Size, int* arr2, int arr2Size, int* returnSize) {
    int count[1001] = {0};

    for (int i = 0; i < arr1Size; i++) {
        count[arr1[i]]++;
    }

    int* ans = (int*)malloc(sizeof(int) * arr1Size);
    int idx = 0;

    for (int i = 0; i < arr2Size; i++) {
        int num = arr2[i];
        while (count[num] > 0) {
            ans[idx++] = num;
            count[num]--;
        }
    }

    for (int num = 0; num <= 1000; num++) {
        while (count[num] > 0) {
            ans[idx++] = num;
            count[num]--;
        }
    }

    *returnSize = arr1Size;
    return ans;
}