# Solving Leetcode Interviews in Seconds with AI: Count Items Matching a Rule


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1773" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given an array items, where each items[i] = [typei, colori, namei] describes the type, color, and name of the ith item. You are also given a rule represented by two strings, ruleKey and ruleValue. The ith item is said to match the rule if one of the following is true:  ruleKey == "type" and ruleValue == typei. ruleKey == "color" and ruleValue == colori. ruleKey == "name" and ruleValue == namei.  Return the number of items that match the given rule.   Example 1:  Input: items = [["phone","blue","pixel"],["computer","silver","lenovo"],["phone","gold","iphone"]], ruleKey = "color", ruleValue = "silver" Output: 1 Explanation: There is only one item matching the given rule, which is ["computer","silver","lenovo"].  Example 2:  Input: items = [["phone","blue","pixel"],["computer","silver","phone"],["phone","gold","iphone"]], ruleKey = "type", ruleValue = "phone" Output: 2 Explanation: There are only two items matching the given rule, which are ["phone","blue","pixel"] and ["phone","gold","iphone"]. Note that the item ["computer","silver","phone"] does not match.   Constraints:  1 <= items.length <= 104 1 <= typei.length, colori.length, namei.length, ruleValue.length <= 10 ruleKey is equal to either "type", "color", or "name". All strings consist only of lowercase letters.  

	# Explanation
	Here's the breakdown of the solution:

*   **Direct Mapping:** Create a mapping (dictionary) from `ruleKey` values ("type", "color", "name") to the corresponding index in the `items` array (0, 1, 2 respectively). This avoids multiple `if/else` checks, improving efficiency.
*   **Iterate and Compare:** Iterate through the `items` array, and for each item, use the mapping to directly access the relevant element (type, color, or name) based on the `ruleKey`. Compare this element with the `ruleValue`.
*   **Count Matches:** Increment a counter each time a match is found.

*   **Runtime Complexity:** O(n), where n is the number of items. **Storage Complexity:** O(1) - The dictionary has a fixed size.

	
	# Code
	```python
	def countMatches(items, ruleKey, ruleValue):
    """
    Counts the number of items that match the given rule.

    Args:
        items: A list of lists, where each inner list represents an item.
        ruleKey: The key to match against ("type", "color", or "name").
        ruleValue: The value to match against.

    Returns:
        The number of items that match the rule.
    """

    rule_map = {
        "type": 0,
        "color": 1,
        "name": 2
    }

    index = rule_map[ruleKey]
    count = 0

    for item in items:
        if item[index] == ruleValue:
            count += 1

    return count
	```