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 <= 1041 <= typei.length, colori.length, namei.length, ruleValue.length <= 10ruleKey is equal to either "type", "color", or "name".When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The brute force method for this problem involves checking every single item to see if it matches the rule. It's like manually looking at each item and comparing its properties to what the rule requires. We count each time an item fits the description.
Here's how the algorithm would work step-by-step:
def count_items_matching_rule_brute_force(items, rule_key_index, rule_value):
number_of_matching_items = 0
for item in items:
# Access the correct property based on the rule key index
item_property = item[rule_key_index]
# Check if the property matches the given value
if item_property == rule_value:
#Increment the counter when properties match
number_of_matching_items += 1
return number_of_matching_itemsTo solve this problem efficiently, we need to check each item and see if it matches the rule. Instead of checking every single detail for each item, we'll focus on the specific detail the rule cares about and compare it directly.
Here's how the algorithm would work step-by-step:
def count_matches(items, rule_key, rule_value):
match_count = 0
for item in items:
# Determine the index based on the rule key.
if rule_key == "type":
index_to_check = 0
elif rule_key == "color":
index_to_check = 1
else:
index_to_check = 2
# Compare the item's value to the rule value.
if item[index_to_check] == rule_value:
# Increment the counter because we found a match
match_count += 1
return match_count| Case | How to Handle |
|---|---|
| Null or empty items list | Return 0 immediately as no items can match any rule. |
| Null or empty ruleKey or ruleValue | If either is null or empty, consider it an invalid input and return 0, or throw an exception depending on requirements. |
| items list containing null or empty inner lists | Handle null or empty inner lists by skipping them or returning an error, depending on desired behavior. |
| Rule key is not one of the expected strings (type, color, name) | Return 0 or throw an exception indicating an invalid ruleKey. |
| Large items list (performance considerations) | Ensure the chosen algorithm has a time complexity that scales reasonably with the size of the items list (e.g., O(n)). |
| Items list with very long strings (memory considerations) | Be mindful of memory usage if the strings are extremely long, potentially impacting performance or causing memory errors. |
| Case sensitivity of ruleValue (if applicable) | Convert both the item value and ruleValue to lowercase for case-insensitive comparison, if required. |
| Items list with mixed data types in the inner lists (e.g., numbers instead of strings) | Handle type checking of the values within the inner lists to ensure they are strings, and convert if necessary or return an error if conversion is not possible. |