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This approach utilizes a map or dictionary to streamline the merging process. By iterating over the two input arrays, we can store each object in a map with its id as the key. This allows constant-time access when merging objects sharing the same id. After processing both arrays, the map's entries can be retrieved and sorted by id to form the final result.
Time Complexity: O((n + m) log(n + m)) where n and m are the lengths of arr1 and arr2, due to the sorting operation.
Space Complexity: O(n + m) for the storage of objects in the dictionary.
1#include <stdio.h>
2#
In this C solution, we combine objects into the result array and use the function qsort to sort it by id. We iterate through both input arrays to add objects to result, checking an inner loop to handle overriding by arr2.
This method involves sorting both arrays prior to merging. After sorting, we can employ a two-pointer technique to iterate through the arrays simultaneously, aligning ids and merging objects as necessary. By maintaining an additional list, the sorted and merged result can be constructed efficiently.
Time Complexity: O(n log n + m log m) due to sorting each array.
Space Complexity: O(n + m) for the merged results.
1function joinArrays(arr1, arr2) {
2 arr1.sort((a, b) => a.id - b.id);
3 arr2.sort((a, b) => a.id - b.id);
4
5 let i = 0, j = 0;
6 const result = [];
7
8 while (i < arr1.length && j < arr2.length) {
9 if (arr1[i].id < arr2[j].id) {
10 result.push(arr1[i++]);
11 } else if (arr1[i].id > arr2[j].id) {
12 result.push(arr2[j++]);
13 } else {
14 result.push({ ...arr1[i++], ...arr2[j++] });
15 }
16 }
17
18 while (i < arr1.length) result.push(arr1[i++]);
19 while (j < arr2.length) result.push(arr2[j++]);
20
21 return result;
22}
23Using JavaScript, both arr1 and arr2 are sorted by id. We then iterate through both arrays using two pointers, merging overlapping objects where necessary by using the spread operator to ensure arr2's properties take precedence.