How do you solve Bank Account Summary II on LeetCode?

Bank Account Summary II (LeetCode #1587) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hash map allows for efficient balance calculations.

What is the brute force solution for Bank Account Summary II?

The brute force approach for Bank Account Summary II is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will calculate the balance for each user by iterating through all transactions. For each user, we sum up the amounts from the transactions associated with their account. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Bank Account Summary II?

Bank Account Summary II uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Bank Account Summary II?

Always clarify the requirements before coding. Think about edge cases, such as accounts with no transactions. Discuss your thought process and approach with the interviewer.

What are common mistakes when solving Bank Account Summary II?

Not using a hash map for balance calculations, leading to inefficiency. Overlooking the requirement to filter results based on balance.

#1587

Bank Account Summary II

Easy

DatabaseHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

This approach uses a single pass to calculate the balances while iterating through the transactions, which is more efficient than the brute-force method. We use a hash map to store balances, allowing for quick updates and lookups.

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Algorithm

4 steps

1Step 1: Create a hash map to store the balance for each account.
2Step 2: Iterate through the transactions and update the balances in the hash map.
3Step 3: After processing all transactions, filter the accounts with balances greater than 10000.
4Step 4: Join the results with the Users table to get the names.

solution.py14 lines

1# Full working Python code
2import pandas as pd
3
4# Sample data
5users = pd.DataFrame({'account': [900001, 900002, 900003], 'name': ['Alice', 'Bob', 'Charlie']})
6transactions = pd.DataFrame({'trans_id': [1, 2, 3, 4], 'account': [900001, 900001, 900002, 900003], 'amount': [15000, -2000, 5000, 20000]})
7
8# Optimal approach
9balances = transactions.groupby('account')['amount'].sum().reset_index()
10balances = balances[balances['amount'] > 10000]
11result = users[users['account'].isin(balances['account'])].merge(balances, on='account')
12result = result[['name', 'amount']].rename(columns={'amount': 'balance'})
13
14print(result)

ℹ

Complexity note: The time complexity is O(n) because we only pass through the transactions once to calculate balances, and the space complexity is O(n) due to the storage of balances in a hash map.

1Using a hash map allows for efficient balance calculations.
2Filtering results after processing all transactions minimizes unnecessary computations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.