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To solve this problem, we can utilize SQL's LEFT JOIN operation. The LEFT JOIN operation will allow us to join the Person table with the Address table on the personId. It returns all records from the left table (Person), and the matched records from the right table (Address). If there is no match, NULL values are returned for columns from the right table. This operation is suitable because we need to retrieve each person's details regardless of whether they have an address recorded. Thus, using LEFT JOIN will fulfill the requirement of including persons with NULL city and state when there is no matching entry in the Address table.
Time Complexity: O(n + m), where n is the number of rows in the Person table and m is the number of rows in the Address table.
Space Complexity: O(n), where n is the number of rows in the result set.
1SELECT p.firstName, p.lastName, A.city,Solve with full IDE support and test cases
This SQL query selects the first and last names from the Person table and combines them with the city and state fields from the Address table. We use a LEFT JOIN to ensure that even if a person's address doesn't exist, they still appear in the results, with NULL values for the city and state fields.
This approach involves using subqueries to fetch data from the Address table if it exists. This is done by selecting fields from the Person table and then using subqueries to attempt to find the corresponding city and state for each personId from the Address table. If there's no match, the subquery will result in NULL, which aligns with the problem's requirement of reporting NULL when no address is available.
Time Complexity: O(n * m), where n is the number of rows in the Person table and m is the number of rows in the Address table because each subquery runs a separate lookup operation for each person.
Space Complexity: O(n), where n is the number of rows in the result set.
1
2SELECT
3 p.firstName,
4 p.lastName,
5 (SELECT city FROM Address WHERE personId = p.personId) AS city,
6 (SELECT state FROM Address WHERE personId = p.personId) AS state
7FROM Person p;This solution fetches the first and last names directly from the Person table and utilizes subqueries to search for the corresponding city and state. Each subquery runs a lookup in the Address table for the personId. When no corresponding address is found, NULL is returned for both city and state.
This approach involves sorting the input data, which allows us to leverage properties of sorted sequences to efficiently solve the problem.
Time Complexity: O(n log n), Space Complexity: O(1) for in-place sorting
1import java.util.Arrays;
2
3public class Solution {
4 public static void solve(int[] arr) {
5 Arrays.sort(arr);
6 // Apply logic on sorted array
7 System.out.print("Sorted Array: ");
8 for (int num : arr) System.out.print(num + " ");
9 }
10
11 public static void main(String[] args) {
12 int[] arr = {3, 1, 4, 1, 5};
13 solve(arr);
14 }
15}Java provides Arrays.sort to handle sorting of arrays. Once sorted, the result can be probed for solving the problem efficiently.
This approach uses a hash map (or dict) to store elements for quick access. This is particularly useful if you need to quickly check for the existence of an element or store counts.
Time Complexity: O(1) average for insert/search, Space Complexity: O(n) where n is the number of elements.
1#include <iostream>
2#include <unordered_map>
3#include <vector>
4
5void solve(const std::vector<int>& vec) {
6 std::unordered_map<int, int> map;
7 for (int num : vec) {
8 map[num]++;
9 }
10 // Example to check presence
11 if (map.find(4) != map.end())
12 std::cout << "4 found!" << std::endl;
13 else
14 std::cout << "4 not found!" << std::endl;
15}
16
17int main() {
18 std::vector<int> vec = {3, 1, 4, 1, 5};
19 solve(vec);
20 return 0;
21}Using C++'s unordered_map, elements are stored with their counts or flags. This setup allows O(1) average time complexity for operations such as check existence, insert, and delete.