LeetCode 618: Students Report By Geography Solution in Python – A Step-by-Step Guide

What Is LeetCode 618: Students Report By Geography?

In LeetCode 618: Students Report By Geography, you’re given a table student with columns name and continent, listing students and their geographic origins (e.g., "America," "Asia," "Europe"). Your task is to transform this into a pivoted report where:

• Each row contains one student from each continent.
• Students are sorted alphabetically by name within their continent.
• All continents (America, Asia, Europe) appear in every row, with null if no student exists.

For example, if you have "Jack" from "America" and "John" from "Asia," the report might list them in one row with a null for Europe. This problem tests your ability to pivot and sort data, typically posed as an SQL query with PIVOT or conditional aggregation, but we’ll solve it in Python for a coding twist.

Problem Statement

• Input: A list of records (or table rows) with [name, continent].
• Output: A list of [america, asia, europe] where each element is a student name or null, sorted by name per continent.
• Rules:
• Columns: "America," "Asia," "Europe" in that order.
• Sort names alphabetically within each continent.
• Fill with null if no student exists for a continent in a row.

Constraints

• Table has at least 1 row.
• 1 ≤ name.length ≤ 100.
• continent is "America," "Asia," or "Europe."
• Number of rows ≤ 100.

Examples

• Input:
``` name | continent Jack | America John | Asia Juan | America ```
• Output:
``` ["Jack", "John", null], ["Juan", null, null] ```
• Input:
``` name | continent Alice | Europe Bob | America Carol | Asia ```
• Output:
``` ["Bob", "Carol", "Alice"] ```
• Input:
``` name | continent Eve | Asia ```
• Output:
``` [null, "Eve", null] ```

These examples show the pivot—let’s solve it!

Understanding the Problem: Pivoting Student Data

To solve LeetCode 618: Students Report By Geography in Python, we need to group students by continent, sort their names, and pivot the data into rows with one student per continent, filling gaps with nulls. A naive approach might manually pair students, but we can optimize with grouping. We’ll use:

• Best Solution (Pivot Table with Sorting): O(n log n) time, O(n) space—efficient and clean (n = number of students).
• Alternative Solution (Dictionary-Based Grouping): O(n) time, O(n) space—simple but less flexible.

Let’s start with the pivot table solution, breaking it down for beginners.

Best Solution: Pivot Table with Sorting

Why This Is the Best Solution

The pivot table approach is the top choice for LeetCode 618 because it’s efficient—O(n log n) time due to sorting—and elegantly transforms the data into the required format by grouping, sorting, and pivoting in a structured way. It mimics SQL’s PIVOT operation and scales well within constraints (≤ 100 rows). It’s like sorting students into continent buckets and lining them up across rows!

How It Works

Think of this as organizing a student roster:

• Step 1: Group by Continent:
• Create lists of names for America, Asia, Europe.
• Step 2: Sort Names:
• Alphabetically sort each continent’s list.
• Step 3: Determine Row Count:
• Find the max length of any continent’s list.
• Step 4: Pivot into Rows:
• For each row index, take one name (or null) from each continent’s sorted list.
• Step 5: Return Result:
• List of [america, asia, europe] rows.

It’s like a classroom seating chart—sorted and pivoted!

Step-by-Step Example

Example:

Copy codename  | continent
Jack  | America
John  | Asia
Juan  | America

• Step 1: Group:
• America: ["Jack", "Juan"].
• Asia: ["John"].
• Europe: [].
• Step 2: Sort:
• America: ["Jack", "Juan"] (already sorted).
• Asia: ["John"].
• Europe: [].
• Step 3: Max rows = 2 (America has 2).
• Step 4: Pivot:
• Row 0: ["Jack", "John", null].
• Row 1: ["Juan", null, null].
• Output: [["Jack", "John", null], ["Juan", null, null]].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codeclass Solution:
    def studentsReport(self, student: List[List[str]]) -> List[List[str]]:
        # Step 1: Group by continent
        america = []
        asia = []
        europe = []

        for name, continent in student:
            if continent == "America":
                america.append(name)
            elif continent == "Asia":
                asia.append(name)
            else:  # Europe
                europe.append(name)

        # Step 2: Sort names
        america.sort()
        asia.sort()
        europe.sort()

        # Step 3: Determine number of rows
        max_rows = max(len(america), len(asia), len(europe))

        # Step 4: Pivot into rows
        result = []
        for i in range(max_rows):
            row = [
                america[i] if i < len(america) else None,
                asia[i] if i < len(asia) else None,
                europe[i] if i < len(europe) else None
            ]
            result.append(row)

        # Step 5: Return result
        return result

• Lines 4-6: Initialize lists for each continent.
• Lines 8-14: Group names by continent.
• Lines 17-19: Sort each list alphabetically.
• Line 22: Max rows based on longest list.
• Lines 25-30: Build rows with nulls for missing entries.
• Time Complexity: O(n log n)—sorting dominates.
• Space Complexity: O(n)—store all names.

This is like a student organizer—sorted and pivoted!

Alternative Solution: Dictionary-Based Grouping with Manual Iteration

Why an Alternative Approach?

The dictionary-based approach uses a dictionary to group students—O(n) time, O(n) space. It’s simpler to implement without sorting overhead if order isn’t critical (though we sort for the problem), but it requires manual iteration to build rows, making it less streamlined. It’s like keeping a student logbook and flipping through it!

How It Works

Picture this as a student ledger:

• Step 1: Group into dictionary by continent.
• Step 2: Sort names in each group.
• Step 3: Iterate to max length, building rows.
• Step 4: Return result list.

It’s like a manual student lineup!

Step-by-Step Example

Example:

Copy codename  | continent
Jack  | America
John  | Asia
Juan  | America

• Step 1: students = {"America": ["Jack", "Juan"], "Asia": ["John"], "Europe": []}.
• Step 2: Sort:
• {"America": ["Jack", "Juan"], "Asia": ["John"], "Europe": []}.
• Step 3: Max length = 2:
• Row 0: ["Jack", "John", null].
• Row 1: ["Juan", null, null].
• Output: [["Jack", "John", null], ["Juan", null, null]].

Code for Dictionary Approach

Copy codefrom collections import defaultdict

class Solution:
    def studentsReport(self, student: List[List[str]]) -> List[List[str]]:
        # Step 1: Group by continent using dictionary
        students = defaultdict(list)
        for name, continent in student:
            students[continent].append(name)

        # Step 2: Sort names in each group
        for continent in students:
            students[continent].sort()

        # Step 3: Determine max rows and build result
        max_rows = max(len(students["America"]), len(students["Asia"]), len(students["Europe"]))
        result = []
        for i in range(max_rows):
            row = [
                students["America"][i] if i < len(students["America"]) else None,
                students["Asia"][i] if i < len(students["Asia"]) else None,
                students["Europe"][i] if i < len(students["Europe"]) else None
            ]
            result.append(row)

        # Step 4: Return result
        return result

• Lines 6-8: Group into dict.
• Lines 11-12: Sort each group.
• Lines 15-21: Build rows with nulls.
• Time Complexity: O(n log n)—sorting.
• Space Complexity: O(n)—dict storage.

It’s a student log—clear but less elegant!

Comparing the Two Solutions

• Pivot Table (Best):
• Pros: O(n log n) time, O(n) space, structured and clean.
• Cons: Slightly more setup.
• Dictionary (Alternative):
• Pros: O(n log n) time, O(n) space, straightforward grouping.
• Cons: Less intuitive row-building.

Pivot table wins for clarity.

Additional Examples and Edge Cases

• Input: [["Eve", "Asia"]]
• Output: [[null, "Eve", null]].
• Input: [["Bob", "America"], ["Alice", "America"]]
• Output: [["Alice", null, null], ["Bob", null, null]].

Both handle these well.

Complexity Breakdown

• Pivot Table: Time O(n log n), Space O(n).
• Dictionary: Time O(n log n), Space O(n).

Pivot table is optimal in practice.

Key Takeaways

• Pivot Table: Sorted pivot—smart!
• Dictionary: Grouped iteration—clear!
• Data: Pivoting is fun.
• Python Tip: Lists rock—see Python Basics.

Final Thoughts: Craft That Report

LeetCode 618: Students Report By Geography in Python is a neat data transformation challenge. The pivot table approach offers efficiency and structure, while dictionary grouping keeps it simple. Want more? Try LeetCode 177: Nth Highest Salary or LeetCode 185: Department Top Three Salaries. Ready to report? Head to Solve LeetCode 618 on LeetCode and pivot those students today!