How do you solve Queries Quality and Percentage on LeetCode?

Queries Quality and Percentage (LeetCode #1211) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to aggregate data efficiently is crucial in database queries.

What is the brute force solution for Queries Quality and Percentage?

The brute force approach for Queries Quality and Percentage is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we will iterate through all the queries and calculate the quality and poor query percentage for each query_name. This method is straightforward but inefficient as it involves multiple passes over the data. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Queries Quality and Percentage?

Queries Quality and Percentage uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Queries Quality and Percentage?

Always clarify the requirements before jumping into coding. Think about edge cases, such as all queries being poor or none being poor. Practice writing SQL queries as they often come up in data-related problems.

What are common mistakes when solving Queries Quality and Percentage?

Not rounding the results correctly to two decimal places. Failing to handle cases where there are no poor queries, leading to division by zero.

#1211

Queries Quality and Percentage

Easy

DatabaseHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a single pass through the data to aggregate the necessary information for each query_name. This approach is efficient and avoids unnecessary computations by leveraging a single data structure.

⚙️

Algorithm

4 steps

1Step 1: Initialize a dictionary to store total quality, count of queries, and poor count for each query_name.
2Step 2: Loop through each row in the Queries table and update the dictionary with quality and counts.
3Step 3: After processing all queries, compute the average quality and poor query percentage for each query_name.
4Step 4: Round the results to 2 decimal places and prepare the final output.

solution.py22 lines

1# Full working Python code
2import pandas as pd
3
4def query_quality(queries):
5    result = {}
6    for _, row in queries.iterrows():
7        name = row['query_name']
8        rating = row['rating']
9        position = row['position']
10        quality = rating / position
11        if name not in result:
12            result[name] = {'total_quality': 0, 'count': 0, 'poor_count': 0}
13        result[name]['total_quality'] += quality
14        result[name]['count'] += 1
15        if rating < 3:
16            result[name]['poor_count'] += 1
17    output = []
18    for name, data in result.items():
19        avg_quality = data['total_quality'] / data['count']
20        poor_percentage = (data['poor_count'] / data['count']) * 100
21        output.append({'query_name': name, 'quality': round(avg_quality, 2), 'poor_query_percentage': round(poor_percentage, 2)})
22    return pd.DataFrame(output)

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the queries to gather all necessary data, making it efficient. The space complexity is O(n) due to the storage of results in a dictionary.

1Understanding how to aggregate data efficiently is crucial in database queries.
2Recognizing the importance of average calculations and percentage computations in data analysis.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.