Pandas4 Solutions

nth-highest-salary.py

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+"""
+
+Table: Employee
+
++-------------+------+
+| Column Name | Type |
++-------------+------+
+| id          | int  |
+| salary      | int  |
++-------------+------+
+id is the primary key (column with unique values) for this table.
+Each row of this table contains information about the salary of an employee.
+
+
+Write a solution to find the nth highest salary from the Employee table. If there is no nth highest salary, return null.
+
+The result format is in the following example.
+
+
+Example 1:
+
+Input: 
+Employee table:
++----+--------+
+| id | salary |
++----+--------+
+| 1  | 100    |
+| 2  | 200    |
+| 3  | 300    |
++----+--------+
+n = 2
+Output: 
++------------------------+
+| getNthHighestSalary(2) |
++------------------------+
+| 200                    |
++------------------------+
+Example 2:
+
+Input: 
+Employee table:
++----+--------+
+| id | salary |
++----+--------+
+| 1  | 100    |
++----+--------+
+n = 2
+Output: 
++------------------------+
+| getNthHighestSalary(2) |
++------------------------+
+| null                   |
++------------------------+
+
+"""
+
+# The function first removes duplicate salaries to ensure unique salary ranks. 
+# Then, it assigns a dense rank to each unique salary in descending order. 
+# Finally, it retrieves the salary corresponding to the Nth rank, returning None if N exceeds the number of unique salaries.
+
+
+import pandas as pd
+
+def nth_highest_salary(employee: pd.DataFrame, N: int) -> pd.DataFrame:
+
+    dist = employee.drop_duplicates(subset='salary')
+    dist['rnk'] = dist['salary'].rank(method='dense', ascending=False)
+    ans = dist[dist.rnk == N][['salary']]
+    if not len(ans):
+        return pd.DataFrame({f'getNthHighestSalary({N})': [None]})
+    ans = ans.rename(columns={'salary': f'getNthHighestSalary({N})'})
+
+    return ans

second-highest-salary.py

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+"""
+
+Table: Employee
+
++-------------+------+
+| Column Name | Type |
++-------------+------+
+| id          | int  |
+| salary      | int  |
++-------------+------+
+id is the primary key (column with unique values) for this table.
+Each row of this table contains information about the salary of an employee.
+
+
+Write a solution to find the second highest distinct salary from the Employee table. If there is no second highest salary, return null (return None in Pandas).
+
+The result format is in the following example.
+
+
+Example 1:
+
+Input: 
+Employee table:
++----+--------+
+| id | salary |
++----+--------+
+| 1  | 100    |
+| 2  | 200    |
+| 3  | 300    |
++----+--------+
+Output: 
++---------------------+
+| SecondHighestSalary |
++---------------------+
+| 200                 |
++---------------------+
+
+Example 2:
+
+Input: 
+Employee table:
++----+--------+
+| id | salary |
++----+--------+
+| 1  | 100    |
++----+--------+
+Output: 
++---------------------+
+| SecondHighestSalary |
++---------------------+
+| null                |
++---------------------+
+
+"""
+
+# This function finds the second highest salary from the given employee DataFrame.  
+# It removes duplicate salaries, checks if there are at least two unique values, and sorts them in descending order.  
+# Finally, it extracts and returns the second highest salary while renaming the column appropriately.  
+
+import pandas as pd
+import numpy as np
+
+def second_highest_salary(employee: pd.DataFrame) -> pd.DataFrame:
+
+    # 1. Drop any duplicate salaries.
+    employee = employee.drop_duplicates(["salary"])
+
+    # 2. If there are less than two unique salaries, return `np.NaN`.
+    if len(employee["salary"].unique()) < 2:
+        return pd.DataFrame({"SecondHighestSalary": [np.NaN]})
+
+    # 3. Sort the table by `salary` in descending order.
+    employee = employee.sort_values("salary", ascending=False)
+
+    # 4. Drop the `id` column.
+    employee.drop("id", axis=1, inplace=True)
+
+    # 5. Rename the `salary` column.
+    employee.rename({"salary": "SecondHighestSalary"}, axis=1, inplace=True)
+
+    # 6, 7. Return the second highest salary.
+    return employee.head(2).tail(1)