LeetCode #1393 — MEDIUM

Capital Gain/Loss

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

Table: Stocks

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| stock_name    | varchar |
| operation     | enum    |
| operation_day | int     |
| price         | int     |
+---------------+---------+
(stock_name, operation_day) is the primary key (combination of columns with unique values) for this table.
The operation column is an ENUM (category) of type ('Sell', 'Buy')
Each row of this table indicates that the stock which has stock_name had an operation on the day operation_day with the price.
It is guaranteed that each 'Sell' operation for a stock has a corresponding 'Buy' operation in a previous day. It is also guaranteed that each 'Buy' operation for a stock has a corresponding 'Sell' operation in an upcoming day.

Write a solution to report the Capital gain/loss for each stock.

The Capital gain/loss of a stock is the total gain or loss after buying and selling the stock one or many times.

Return the result table in any order.

The result format is in the following example.

Example 1:

Input: 
Stocks table:
+---------------+-----------+---------------+--------+
| stock_name    | operation | operation_day | price  |
+---------------+-----------+---------------+--------+
| Leetcode      | Buy       | 1             | 1000   |
| Corona Masks  | Buy       | 2             | 10     |
| Leetcode      | Sell      | 5             | 9000   |
| Handbags      | Buy       | 17            | 30000  |
| Corona Masks  | Sell      | 3             | 1010   |
| Corona Masks  | Buy       | 4             | 1000   |
| Corona Masks  | Sell      | 5             | 500    |
| Corona Masks  | Buy       | 6             | 1000   |
| Handbags      | Sell      | 29            | 7000   |
| Corona Masks  | Sell      | 10            | 10000  |
+---------------+-----------+---------------+--------+
Output: 
+---------------+-------------------+
| stock_name    | capital_gain_loss |
+---------------+-------------------+
| Corona Masks  | 9500              |
| Leetcode      | 8000              |
| Handbags      | -23000            |
+---------------+-------------------+
Explanation: 
Leetcode stock was bought at day 1 for 1000$ and was sold at day 5 for 9000$. Capital gain = 9000 - 1000 = 8000$.
Handbags stock was bought at day 17 for 30000$ and was sold at day 29 for 7000$. Capital loss = 7000 - 30000 = -23000$.
Corona Masks stock was bought at day 1 for 10$ and was sold at day 3 for 1010$. It was bought again at day 4 for 1000$ and was sold at day 5 for 500$. At last, it was bought at day 6 for 1000$ and was sold at day 10 for 10000$. Capital gain/loss is the sum of capital gains/losses for each ('Buy' --> 'Sell') operation = (1010 - 10) + (500 - 1000) + (10000 - 1000) = 1000 - 500 + 9000 = 9500$.

Step 01

Brute Force Baseline

Problem summary: Table: Stocks +---------------+---------+ | Column Name | Type | +---------------+---------+ | stock_name | varchar | | operation | enum | | operation_day | int | | price | int | +---------------+---------+ (stock_name, operation_day) is the primary key (combination of columns with unique values) for this table. The operation column is an ENUM (category) of type ('Sell', 'Buy') Each row of this table indicates that the stock which has stock_name had an operation on the day operation_day with the price. It is guaranteed that each 'Sell' operation for a stock has a corresponding 'Buy' operation in a previous day. It is also guaranteed that each 'Buy' operation for a stock has a corresponding 'Sell' operation in an upcoming day. Write a solution to report the Capital gain/loss for each stock. The Capital gain/loss of a stock is the total gain or loss after buying and selling the stock one

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

{"headers":{"Stocks":["stock_name","operation","operation_day","price"]},"rows":{"Stocks":[["Leetcode","Buy",1,1000],["Corona Masks","Buy",2,10],["Leetcode","Sell",5,9000],["Handbags","Buy",17,30000],["Corona Masks","Sell",3,1010],["Corona Masks","Buy",4,1000],["Corona Masks","Sell",5,500],["Corona Masks","Buy",6,1000],["Handbags","Sell",29,7000],["Corona Masks","Sell",10,10000]]}}

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 #1393: Capital Gain/Loss
// Auto-generated Java example from rust.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (rust):
// // Accepted solution for LeetCode #1393: Capital Gain/Loss
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1393: Capital Gain/Loss
// # Write your MySQL query statement below
// SELECT
//     stock_name,
//     SUM(IF(operation = 'Buy', -price, price)) AS capital_gain_loss
// FROM Stocks
// GROUP BY 1;
// "#
// }

// Accepted solution for LeetCode #1393: Capital Gain/Loss
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1393: Capital Gain/Loss
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1393: Capital Gain/Loss
// # Write your MySQL query statement below
// SELECT
//     stock_name,
//     SUM(IF(operation = 'Buy', -price, price)) AS capital_gain_loss
// FROM Stocks
// GROUP BY 1;
// "#
// }

# Accepted solution for LeetCode #1393: Capital Gain/Loss
# Auto-generated Python example from rust.
def example_solution() -> None:
    return
# Reference (rust):
# // Accepted solution for LeetCode #1393: Capital Gain/Loss
# pub fn sql_example() -> &'static str {
#     r#"
# -- Accepted solution for LeetCode #1393: Capital Gain/Loss
# # Write your MySQL query statement below
# SELECT
#     stock_name,
#     SUM(IF(operation = 'Buy', -price, price)) AS capital_gain_loss
# FROM Stocks
# GROUP BY 1;
# "#
# }

// Accepted solution for LeetCode #1393: Capital Gain/Loss
pub fn sql_example() -> &'static str {
    r#"
-- Accepted solution for LeetCode #1393: Capital Gain/Loss
# Write your MySQL query statement below
SELECT
    stock_name,
    SUM(IF(operation = 'Buy', -price, price)) AS capital_gain_loss
FROM Stocks
GROUP BY 1;
"#
}

// Accepted solution for LeetCode #1393: Capital Gain/Loss
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1393: Capital Gain/Loss
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1393: Capital Gain/Loss
// # Write your MySQL query statement below
// SELECT
//     stock_name,
//     SUM(IF(operation = 'Buy', -price, price)) AS capital_gain_loss
// FROM Stocks
// 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.