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.
Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.
Table: Users
+--------------+---------+ | Column Name | Type | +--------------+---------+ | account | int | | name | varchar | +--------------+---------+ account is the primary key (column with unique values) for this table. Each row of this table contains the account number of each user in the bank. There will be no two users having the same name in the table.
Table: Transactions
+---------------+---------+ | Column Name | Type | +---------------+---------+ | trans_id | int | | account | int | | amount | int | | transacted_on | date | +---------------+---------+ trans_id is the primary key (column with unique values) for this table. Each row of this table contains all changes made to all accounts. amount is positive if the user received money and negative if they transferred money. All accounts start with a balance of 0.
Write a solution to report the name and balance of users with a balance higher than 10000. The balance of an account is equal to the sum of the amounts of all transactions involving that account.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input: Users table: +------------+--------------+ | account | name | +------------+--------------+ | 900001 | Alice | | 900002 | Bob | | 900003 | Charlie | +------------+--------------+ Transactions table: +------------+------------+------------+---------------+ | trans_id | account | amount | transacted_on | +------------+------------+------------+---------------+ | 1 | 900001 | 7000 | 2020-08-01 | | 2 | 900001 | 7000 | 2020-09-01 | | 3 | 900001 | -3000 | 2020-09-02 | | 4 | 900002 | 1000 | 2020-09-12 | | 5 | 900003 | 6000 | 2020-08-07 | | 6 | 900003 | 6000 | 2020-09-07 | | 7 | 900003 | -4000 | 2020-09-11 | +------------+------------+------------+---------------+ Output: +------------+------------+ | name | balance | +------------+------------+ | Alice | 11000 | +------------+------------+ Explanation: Alice's balance is (7000 + 7000 - 3000) = 11000. Bob's balance is 1000. Charlie's balance is (6000 + 6000 - 4000) = 8000.
Problem summary: Table: Users +--------------+---------+ | Column Name | Type | +--------------+---------+ | account | int | | name | varchar | +--------------+---------+ account is the primary key (column with unique values) for this table. Each row of this table contains the account number of each user in the bank. There will be no two users having the same name in the table. Table: Transactions +---------------+---------+ | Column Name | Type | +---------------+---------+ | trans_id | int | | account | int | | amount | int | | transacted_on | date | +---------------+---------+ trans_id is the primary key (column with unique values) for this table. Each row of this table contains all changes made to all accounts. amount is positive if the user received money and negative if they transferred money. All accounts start with a balance of 0. Write a solution to report the name and balance of users with a
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
{"headers": {"Users": ["account", "name"], "Transactions": ["trans_id", "account", "amount", "transacted_on"]}, "rows": {"Users": [[900001, "Alice"], [900002, "Bob"], [900003, "Charlie"]], "Transactions": [[1, 900001, 7000, "2020-08-01"], [2, 900001, 7000, "2020-09-01"], [3, 900001, -3000, "2020-09-02"], [4, 900002, 1000, "2020-09-12"], [5, 900003, 6000, "2020-08-07"], [6, 900003, 6000, "2020-09-07"], [7, 900003, -4000, "2020-09-11"]]}}Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1587: Bank Account Summary II
// Auto-generated Java example from rust.
class Solution {
public void exampleSolution() {
}
}
// Reference (rust):
// // Accepted solution for LeetCode #1587: Bank Account Summary II
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1587: Bank Account Summary II
// # Write your MySQL query statement below
// SELECT
// name,
// SUM(amount) AS balance
// FROM
// Users
// JOIN Transactions USING (account)
// GROUP BY account
// HAVING balance > 10000;
// "#
// }
// Accepted solution for LeetCode #1587: Bank Account Summary II
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1587: Bank Account Summary II
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1587: Bank Account Summary II
// # Write your MySQL query statement below
// SELECT
// name,
// SUM(amount) AS balance
// FROM
// Users
// JOIN Transactions USING (account)
// GROUP BY account
// HAVING balance > 10000;
// "#
// }
# Accepted solution for LeetCode #1587: Bank Account Summary II
# Auto-generated Python example from rust.
def example_solution() -> None:
return
# Reference (rust):
# // Accepted solution for LeetCode #1587: Bank Account Summary II
# pub fn sql_example() -> &'static str {
# r#"
# -- Accepted solution for LeetCode #1587: Bank Account Summary II
# # Write your MySQL query statement below
# SELECT
# name,
# SUM(amount) AS balance
# FROM
# Users
# JOIN Transactions USING (account)
# GROUP BY account
# HAVING balance > 10000;
# "#
# }
// Accepted solution for LeetCode #1587: Bank Account Summary II
pub fn sql_example() -> &'static str {
r#"
-- Accepted solution for LeetCode #1587: Bank Account Summary II
# Write your MySQL query statement below
SELECT
name,
SUM(amount) AS balance
FROM
Users
JOIN Transactions USING (account)
GROUP BY account
HAVING balance > 10000;
"#
}
// Accepted solution for LeetCode #1587: Bank Account Summary II
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1587: Bank Account Summary II
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1587: Bank Account Summary II
// # Write your MySQL query statement below
// SELECT
// name,
// SUM(amount) AS balance
// FROM
// Users
// JOIN Transactions USING (account)
// GROUP BY account
// HAVING balance > 10000;
// "#
// }
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.