Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using core interview patterns strategy.
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 distinct salary from the Employee table. If there are less than n distinct salaries, 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 | +------------------------+
Problem summary: 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 distinct salary from the Employee table. If there are less than n distinct salaries, return null. The result format is in the following example.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
{"headers": {"Employee": ["id", "salary"]}, "argument": 2, "rows": {"Employee": [[1, 100], [2, 200], [3, 300]]}}{"headers": {"Employee": ["id", "salary"]}, "argument": 2, "rows": {"Employee": [[1, 100]]}}the-number-of-users-that-are-eligible-for-discount)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #177: Nth Highest Salary
// Auto-generated Java example from py.
class Solution {
public void exampleSolution() {
}
}
// Reference (py):
// # Accepted solution for LeetCode #177: Nth Highest Salary
// import pandas as pd
//
//
// import numpy as np
// def nth_highest_salary(employee: pd.DataFrame, N: int) -> pd.DataFrame:
// if N < 1:
// return pd.DataFrame({"getNthHighestSalary(" + str(N) + ")": [None]})
// unique_salaries = employee.salary.unique()
// if len(unique_salaries) < N:
// return pd.DataFrame([np.NaN], columns=[f"getNthHighestSalary({N})"])
// else:
// salary = sorted(unique_salaries, reverse=True)[N - 1]
// return pd.DataFrame([salary], columns=[f"getNthHighestSalary({N})"])
// Accepted solution for LeetCode #177: Nth Highest Salary
// Auto-generated Go example from py.
func exampleSolution() {
}
// Reference (py):
// # Accepted solution for LeetCode #177: Nth Highest Salary
// import pandas as pd
//
//
// import numpy as np
// def nth_highest_salary(employee: pd.DataFrame, N: int) -> pd.DataFrame:
// if N < 1:
// return pd.DataFrame({"getNthHighestSalary(" + str(N) + ")": [None]})
// unique_salaries = employee.salary.unique()
// if len(unique_salaries) < N:
// return pd.DataFrame([np.NaN], columns=[f"getNthHighestSalary({N})"])
// else:
// salary = sorted(unique_salaries, reverse=True)[N - 1]
// return pd.DataFrame([salary], columns=[f"getNthHighestSalary({N})"])
# Accepted solution for LeetCode #177: Nth Highest Salary
import pandas as pd
import numpy as np
def nth_highest_salary(employee: pd.DataFrame, N: int) -> pd.DataFrame:
if N < 1:
return pd.DataFrame({"getNthHighestSalary(" + str(N) + ")": [None]})
unique_salaries = employee.salary.unique()
if len(unique_salaries) < N:
return pd.DataFrame([np.NaN], columns=[f"getNthHighestSalary({N})"])
else:
salary = sorted(unique_salaries, reverse=True)[N - 1]
return pd.DataFrame([salary], columns=[f"getNthHighestSalary({N})"])
// Accepted solution for LeetCode #177: Nth Highest Salary
// Rust example auto-generated from py reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (py):
// # Accepted solution for LeetCode #177: Nth Highest Salary
// import pandas as pd
//
//
// import numpy as np
// def nth_highest_salary(employee: pd.DataFrame, N: int) -> pd.DataFrame:
// if N < 1:
// return pd.DataFrame({"getNthHighestSalary(" + str(N) + ")": [None]})
// unique_salaries = employee.salary.unique()
// if len(unique_salaries) < N:
// return pd.DataFrame([np.NaN], columns=[f"getNthHighestSalary({N})"])
// else:
// salary = sorted(unique_salaries, reverse=True)[N - 1]
// return pd.DataFrame([salary], columns=[f"getNthHighestSalary({N})"])
// Accepted solution for LeetCode #177: Nth Highest Salary
// Auto-generated TypeScript example from py.
function exampleSolution(): void {
}
// Reference (py):
// # Accepted solution for LeetCode #177: Nth Highest Salary
// import pandas as pd
//
//
// import numpy as np
// def nth_highest_salary(employee: pd.DataFrame, N: int) -> pd.DataFrame:
// if N < 1:
// return pd.DataFrame({"getNthHighestSalary(" + str(N) + ")": [None]})
// unique_salaries = employee.salary.unique()
// if len(unique_salaries) < N:
// return pd.DataFrame([np.NaN], columns=[f"getNthHighestSalary({N})"])
// else:
// salary = sorted(unique_salaries, reverse=True)[N - 1]
// return pd.DataFrame([salary], columns=[f"getNthHighestSalary({N})"])
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.