SQL: Joining Person and Address Tables (with description)
Code\n\n# Write your MySQL query statement below\nSELECT p.firstName, p.lastName, a.city, a.state\nFROM person p\nLEFT JOIN address a\nON p.personId = a.personI
# Code\n```\n# Write your MySQL query statement below\nSELECT p.firstName, p.lastName, a.city, a.state\nFROM person p\nLEFT JOIN address a\nON p.personId = a.personId;\n```\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe approach involves using a LEFT JOIN operation in SQL to combine records from the \'person\' and \'address\' tables. The query selects the first name and last name from the \'person\' table and the city and state from the \'address\' table. The LEFT JOIN ensures that all records from the \'person\' table are included in the result set, even if there is no corresponding entry in the \'address\' table for a given \'personId\'.\n\n# Complexity\n- Time complexity: $$O(n*m)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity is dependent on the implementation of the SQL database. However, generally, a LEFT JOIN operation is 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.\n\n\n- Space complexity: $$O(n*m)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity is also O(n*m) as it depends on the data size of the \'person\' and \'address\' tables that are being joined.\n\n![LeetCode-Upvote.png](https://assets.leetcode.com/users/images/3cb2b75f-c9c6-4745-aec2-7e1b098bf8c7_1702963068.3517854.png)\n\n\n
['MySQL']