LeetCode #1179 — EASY

Reformat Department Table

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

Table: Department

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| revenue     | int     |
| month       | varchar |
+-------------+---------+
In SQL,(id, month) is the primary key of this table.
The table has information about the revenue of each department per month.
The month has values in ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"].

Reformat the table such that there is a department id column and a revenue column for each month.

Return the result table in any order.

The result format is in the following example.

Example 1:

Input: 
Department table:
+------+---------+-------+
| id   | revenue | month |
+------+---------+-------+
| 1    | 8000    | Jan   |
| 2    | 9000    | Jan   |
| 3    | 10000   | Feb   |
| 1    | 7000    | Feb   |
| 1    | 6000    | Mar   |
+------+---------+-------+
Output: 
+------+-------------+-------------+-------------+-----+-------------+
| id   | Jan_Revenue | Feb_Revenue | Mar_Revenue | ... | Dec_Revenue |
+------+-------------+-------------+-------------+-----+-------------+
| 1    | 8000        | 7000        | 6000        | ... | null        |
| 2    | 9000        | null        | null        | ... | null        |
| 3    | null        | 10000       | null        | ... | null        |
+------+-------------+-------------+-------------+-----+-------------+
Explanation: The revenue from Apr to Dec is null.
Note that the result table has 13 columns (1 for the department id + 12 for the months).

Step 01

Brute Force Baseline

Problem summary: Table: Department +-------------+---------+ | Column Name | Type | +-------------+---------+ | id | int | | revenue | int | | month | varchar | +-------------+---------+ In SQL,(id, month) is the primary key of this table. The table has information about the revenue of each department per month. The month has values in ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"]. Reformat the table such that there is a department id column and a revenue column for each month. Return the result table in any order. The result format is in the following example.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

{"headers":{"Department":["id","revenue","month"]},"rows":{"Department":[[1,8000,"Jan"],[2,9000,"Jan"],[3,10000,"Feb"],[1,7000,"Feb"],[1,6000,"Mar"]]}}

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1179: Reformat Department Table
// Auto-generated Java example from rust.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (rust):
// // Accepted solution for LeetCode #1179: Reformat Department Table
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1179: Reformat Department Table
// # Write your MySQL query statement below
// SELECT
//     id,
//     SUM(
//         CASE month
//             WHEN 'Jan' THEN revenue
//         END
//     ) AS Jan_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Feb' THEN revenue
//         END
//     ) AS Feb_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Mar' THEN revenue
//         END
//     ) AS Mar_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Apr' THEN revenue
//         END
//     ) AS Apr_Revenue,
//     SUM(
//         CASE month
//             WHEN 'May' THEN revenue
//         END
//     ) AS May_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Jun' THEN revenue
//         END
//     ) AS Jun_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Jul' THEN revenue
//         END
//     ) AS Jul_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Aug' THEN revenue
//         END
//     ) AS Aug_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Sep' THEN revenue
//         END
//     ) AS Sep_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Oct' THEN revenue
//         END
//     ) AS Oct_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Nov' THEN revenue
//         END
//     ) AS Nov_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Dec' THEN revenue
//         END
//     ) AS Dec_Revenue
// FROM Department
// GROUP BY 1;
// "#
// }

// Accepted solution for LeetCode #1179: Reformat Department Table
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1179: Reformat Department Table
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1179: Reformat Department Table
// # Write your MySQL query statement below
// SELECT
//     id,
//     SUM(
//         CASE month
//             WHEN 'Jan' THEN revenue
//         END
//     ) AS Jan_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Feb' THEN revenue
//         END
//     ) AS Feb_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Mar' THEN revenue
//         END
//     ) AS Mar_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Apr' THEN revenue
//         END
//     ) AS Apr_Revenue,
//     SUM(
//         CASE month
//             WHEN 'May' THEN revenue
//         END
//     ) AS May_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Jun' THEN revenue
//         END
//     ) AS Jun_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Jul' THEN revenue
//         END
//     ) AS Jul_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Aug' THEN revenue
//         END
//     ) AS Aug_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Sep' THEN revenue
//         END
//     ) AS Sep_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Oct' THEN revenue
//         END
//     ) AS Oct_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Nov' THEN revenue
//         END
//     ) AS Nov_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Dec' THEN revenue
//         END
//     ) AS Dec_Revenue
// FROM Department
// GROUP BY 1;
// "#
// }

# Accepted solution for LeetCode #1179: Reformat Department Table
# Auto-generated Python example from rust.
def example_solution() -> None:
    return
# Reference (rust):
# // Accepted solution for LeetCode #1179: Reformat Department Table
# pub fn sql_example() -> &'static str {
#     r#"
# -- Accepted solution for LeetCode #1179: Reformat Department Table
# # Write your MySQL query statement below
# SELECT
#     id,
#     SUM(
#         CASE month
#             WHEN 'Jan' THEN revenue
#         END
#     ) AS Jan_Revenue,
#     SUM(
#         CASE month
#             WHEN 'Feb' THEN revenue
#         END
#     ) AS Feb_Revenue,
#     SUM(
#         CASE month
#             WHEN 'Mar' THEN revenue
#         END
#     ) AS Mar_Revenue,
#     SUM(
#         CASE month
#             WHEN 'Apr' THEN revenue
#         END
#     ) AS Apr_Revenue,
#     SUM(
#         CASE month
#             WHEN 'May' THEN revenue
#         END
#     ) AS May_Revenue,
#     SUM(
#         CASE month
#             WHEN 'Jun' THEN revenue
#         END
#     ) AS Jun_Revenue,
#     SUM(
#         CASE month
#             WHEN 'Jul' THEN revenue
#         END
#     ) AS Jul_Revenue,
#     SUM(
#         CASE month
#             WHEN 'Aug' THEN revenue
#         END
#     ) AS Aug_Revenue,
#     SUM(
#         CASE month
#             WHEN 'Sep' THEN revenue
#         END
#     ) AS Sep_Revenue,
#     SUM(
#         CASE month
#             WHEN 'Oct' THEN revenue
#         END
#     ) AS Oct_Revenue,
#     SUM(
#         CASE month
#             WHEN 'Nov' THEN revenue
#         END
#     ) AS Nov_Revenue,
#     SUM(
#         CASE month
#             WHEN 'Dec' THEN revenue
#         END
#     ) AS Dec_Revenue
# FROM Department
# GROUP BY 1;
# "#
# }

// Accepted solution for LeetCode #1179: Reformat Department Table
pub fn sql_example() -> &'static str {
    r#"
-- Accepted solution for LeetCode #1179: Reformat Department Table
# Write your MySQL query statement below
SELECT
    id,
    SUM(
        CASE month
            WHEN 'Jan' THEN revenue
        END
    ) AS Jan_Revenue,
    SUM(
        CASE month
            WHEN 'Feb' THEN revenue
        END
    ) AS Feb_Revenue,
    SUM(
        CASE month
            WHEN 'Mar' THEN revenue
        END
    ) AS Mar_Revenue,
    SUM(
        CASE month
            WHEN 'Apr' THEN revenue
        END
    ) AS Apr_Revenue,
    SUM(
        CASE month
            WHEN 'May' THEN revenue
        END
    ) AS May_Revenue,
    SUM(
        CASE month
            WHEN 'Jun' THEN revenue
        END
    ) AS Jun_Revenue,
    SUM(
        CASE month
            WHEN 'Jul' THEN revenue
        END
    ) AS Jul_Revenue,
    SUM(
        CASE month
            WHEN 'Aug' THEN revenue
        END
    ) AS Aug_Revenue,
    SUM(
        CASE month
            WHEN 'Sep' THEN revenue
        END
    ) AS Sep_Revenue,
    SUM(
        CASE month
            WHEN 'Oct' THEN revenue
        END
    ) AS Oct_Revenue,
    SUM(
        CASE month
            WHEN 'Nov' THEN revenue
        END
    ) AS Nov_Revenue,
    SUM(
        CASE month
            WHEN 'Dec' THEN revenue
        END
    ) AS Dec_Revenue
FROM Department
GROUP BY 1;
"#
}

// Accepted solution for LeetCode #1179: Reformat Department Table
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1179: Reformat Department Table
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1179: Reformat Department Table
// # Write your MySQL query statement below
// SELECT
//     id,
//     SUM(
//         CASE month
//             WHEN 'Jan' THEN revenue
//         END
//     ) AS Jan_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Feb' THEN revenue
//         END
//     ) AS Feb_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Mar' THEN revenue
//         END
//     ) AS Mar_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Apr' THEN revenue
//         END
//     ) AS Apr_Revenue,
//     SUM(
//         CASE month
//             WHEN 'May' THEN revenue
//         END
//     ) AS May_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Jun' THEN revenue
//         END
//     ) AS Jun_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Jul' THEN revenue
//         END
//     ) AS Jul_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Aug' THEN revenue
//         END
//     ) AS Aug_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Sep' THEN revenue
//         END
//     ) AS Sep_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Oct' THEN revenue
//         END
//     ) AS Oct_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Nov' THEN revenue
//         END
//     ) AS Nov_Revenue,
//     SUM(
//         CASE month
//             WHEN 'Dec' THEN revenue
//         END
//     ) AS Dec_Revenue
// FROM Department
// GROUP BY 1;
// "#
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

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.

OPTIMIZED

O(n) time

O(1) space

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.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.