LeetCode 627: Swap Salary Solution in Python – A Step-by-Step Guide

What Is LeetCode 627: Swap Salary?

In LeetCode 627: Swap Salary, you’re given a table salary with columns id (integer), name (string), sex (char: 'm' or 'f'), and salary (integer). Your task is to swap all values in the sex column: every 'm' becomes 'f', and every 'f' becomes 'm', leaving other columns unchanged. The output is the modified table, typically unsorted as the original order is preserved in SQL contexts, but we’ll sort by id for clarity in Python. For example, with [1, "John", "m", 5000], you’d get [1, "John", "f", 5000]. This problem tests your ability to update table data with a simple transformation, usually posed as an SQL UPDATE with CASE, but we’ll solve it in Python for a coding twist.

Problem Statement

• Input:
• salary: A list of records with [id, name, sex, salary].
• Output: A list of [id, name, sex, salary] with sex values swapped.
• Rules:
• Swap 'm' to 'f' and 'f' to 'm' in sex column.
• Other columns remain unchanged.
• Typically, preserve order, but we’ll sort by id for consistency.

Constraints

• Table has at least 1 row.
• 1 ≤ id ≤ 10⁵.
• name is a string ≤ 100 characters.
• sex is 'm' or 'f'.
• 0 ≤ salary ≤ 10⁵.

Examples

• Input:
``` id | name | sex | salary 1 | John | m | 5000 2 | Jane | f | 6000 ```
• Output:
``` [[1, "John", "f", 5000], [2, "Jane", "m", 6000]] ```
• Input:
``` id | name | sex | salary 3 | Alice | f | 7000 ```
• Output:
``` [[3, "Alice", "m", 7000]] ```
• Input:
``` id | name | sex | salary 1 | Bob | m | 4000 2 | Eve | m | 4500 ```
• Output:
``` [[1, "Bob", "f", 4000], [2, "Eve", "f", 4500]] ```

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

Understanding the Problem: Flipping Genders

To solve LeetCode 627: Swap Salary in Python, we need to process a list of [id, name, sex, salary] records and flip the sex value in each row from 'm' to 'f' or 'f' to 'm', keeping other fields intact. A naive approach might overcomplicate with multiple passes, but this is a simple transformation. We’ll use:

• Best Solution (List-Based Conditional Swapping): O(n) time, O(n) space—fast and direct (n = number of rows).
• Alternative Solution (Dictionary-Based Mapping): O(n) time, O(n) space—explicit but unnecessary complexity.

Let’s start with the list-based solution, breaking it down for beginners.

Best Solution: List-Based Conditional Swapping

Why This Is the Best Solution

The list-based conditional swapping approach is the top choice for LeetCode 627 because it’s efficient—O(n) time with O(n) space—and intuitive, modifying the list in place or creating a new one with a simple condition in one pass. It fits constraints (≤ 10⁵ rows) and mirrors SQL’s CASE logic in a Python-friendly way. It’s like flipping a switch on each student’s gender marker as you go!

How It Works

Think of this as a payroll update:

• Step 1: Process Each Row:
• For each record, check sex and swap it.
• Step 2: Conditional Swap:
• If 'm', change to 'f'.
• If 'f', change to 'm'.
• Step 3: Preserve Other Fields:
• Keep id, name, salary unchanged.
• Step 4: Sort by ID (Optional):
• Ensure output is ordered (not required in SQL, added for clarity).
• Step 5: Return Result:
• Output the modified list.

It’s like a gender toggler—flip and move on!

Step-by-Step Example

Example:

Copy codeid | name  | sex | salary
1  | John  | m   | 5000
2  | Jane  | f   | 6000

• Step 1: Process rows:
• Row 1: [1, "John", "m", 5000].
• Row 2: [2, "Jane", "f", 6000].
• Step 2: Swap:
• "m" → "f": [1, "John", "f", 5000].
• "f" → "m": [2, "Jane", "m", 6000].
• Step 3: Other fields unchanged.
• Step 4: Already sorted by ID.
• Output: [[1, "John", "f", 5000], [2, "Jane", "m", 6000]].

Example:

Copy codeid | name  | sex | salary
3  | Alice | f   | 7000

• Step 1: [3, "Alice", "f", 7000].
• Step 2: "f" → "m": [3, "Alice", "m", 7000].
• Step 3: Unchanged.
• Step 4: Sorted.
• Output: [[3, "Alice", "m", 7000]].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codeclass Solution:
    def swapSalary(self, salary: List[List]) -> List[List]:
        # Step 1: Create result list (or modify in place)
        result = []

        # Step 2: Process each row and swap sex
        for row in salary:
            id_val, name, sex, sal = row
            # Swap sex value
            new_sex = 'f' if sex == 'm' else 'm'
            # Add updated row to result
            result.append([id_val, name, new_sex, sal])

        # Step 3: Sort by ID (optional for clarity)
        result.sort(key=lambda x: x[0])

        # Step 4: Return result
        return result

• Line 4: Initialize result list.
• Lines 7-11: Process and swap:
• Unpack row, flip sex with condition, rebuild row.
• Line 14: Sort by ID (not required in SQL context).
• Time Complexity: O(n)—one pass for swapping.
• Space Complexity: O(n)—result list.

This is like a payroll flipper—fast and simple!

Alternative Solution: Dictionary-Based Mapping

Why an Alternative Approach?

The dictionary-based mapping approach uses a dict to map IDs to swapped sex values—O(n) time, O(n) space. It’s more explicit, creating a lookup before updating, but adds unnecessary complexity for this straightforward task. It’s like building a roster before flipping genders!

How It Works

Picture this as a payroll reassignment:

• Step 1: Build initial dict with IDs and data.
• Step 2: Create swapped sex mapping.
• Step 3: Update list with new sex values.
• Step 4: Sort and return.

It’s like a gender mapper—detailed but overkill!

Step-by-Step Example

Example:

Copy codeid | name  | sex | salary
1  | John  | m   | 5000
2  | Jane  | f   | 6000

• Step 1: Dict = {1: ["John", "m", 5000], 2: ["Jane", "f", 6000]}.
• Step 2: Swapped sex:
• 1: "m" → "f".
• 2: "f" → "m".
• Step 3: Result = [[1, "John", "f", 5000], [2, "Jane", "m", 6000]].
• Step 4: Sorted.
• Output: [[1, "John", "f", 5000], [2, "Jane", "m", 6000]].

Code for Dictionary Approach

Copy codeclass Solution:
    def swapSalary(self, salary: List[List]) -> List[List]:
        # Step 1: Build dictionary
        data = {row[0]: [row[1], row[2], row[3]] for row in salary}

        # Step 2: Create swapped sex mapping
        swapped = {}
        for id_val, (name, sex, sal) in data.items():
            new_sex = 'f' if sex == 'm' else 'm'
            swapped[id_val] = [name, new_sex, sal]

        # Step 3: Build result list
        result = [[id_val, *values] for id_val, values in swapped.items()]

        # Step 4: Sort by ID
        result.sort(key=lambda x: x[0])

        # Step 5: Return result
        return result

• Line 4: Map ID to [name, sex, salary].
• Lines 7-10: Swap sex in new dict.
• Line 13: Build result with swapped values.
• Line 16: Sort by ID.
• Time Complexity: O(n)—mapping and building.
• Space Complexity: O(n)—dict and list.

It’s a payroll mapper—clear but heavier!

Comparing the Two Solutions

• List-Based (Best):
• Pros: O(n) time, O(n) space, simple and direct.
• Cons: Less flexible for complex updates.
• Dictionary (Alternative):
• Pros: O(n) time, O(n) space, explicit mapping.
• Cons: Extra dict step unnecessary.

List-based wins for simplicity.

Additional Examples and Edge Cases

• Input: [[1, "Eve", "f", 3000]]
• Output: [[1, "Eve", "m", 3000]].
• Input: [[2, "Bob", "m", 4000], [1, "Alice", "f", 5000]]
• Output: [[1, "Alice", "m", 5000], [2, "Bob", "f", 4000]].

Both handle these well.

Complexity Breakdown

• List-Based: Time O(n), Space O(n).
• Dictionary: Time O(n), Space O(n).

List-based is optimal in practice.

Key Takeaways

• List-Based: Quick flip—smart!
• Dictionary: Mapped flip—clear!
• Data: Updates are fun.
• Python Tip: Lists rock—see Python Basics.

Final Thoughts: Flip Those Genders

LeetCode 627: Swap Salary in Python is a neat data-update challenge. List-based swapping offers speed and clarity, while dictionary mapping provides an explicit alternative. Want more? Try LeetCode 175: Combine Two Tables or LeetCode 626: Exchange Seats. Ready to toggle? Head to Solve LeetCode 627 on LeetCode and swap those salaries today!