Leetcode #2043: Simple Bank System
In this guide, we solve Leetcode #2043 Simple Bank System in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You have been tasked with writing a program for a popular bank that will automate all its incoming transactions (transfer, deposit, and withdraw). The bank has n accounts numbered from 1 to n.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Array, Hash Table, Simulation
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input
["Bank", "withdraw", "transfer", "deposit", "transfer", "withdraw"]
[[[10, 100, 20, 50, 30]], [3, 10], [5, 1, 20], [5, 20], [3, 4, 15], [10, 50]]
Output
[null, true, true, true, false, false]
Explanation
Bank bank = new Bank([10, 100, 20, 50, 30]);
bank.withdraw(3, 10); // return true, account 3 has a balance of $20, so it is valid to withdraw $10.
// Account 3 has $20 - $10 = $10.
bank.transfer(5, 1, 20); // return true, account 5 has a balance of $30, so it is valid to transfer $20.
// Account 5 has $30 - $20 = $10, and account 1 has $10 + $20 = $30.
bank.deposit(5, 20); // return true, it is valid to deposit $20 to account 5.
// Account 5 has $10 + $20 = $30.
bank.transfer(3, 4, 15); // return false, the current balance of account 3 is $10,
// so it is invalid to transfer $15 from it.
bank.withdraw(10, 50); // return false, it is invalid because account 10 does not exist.
Python Solution
class Bank:
def __init__(self, balance: List[int]):
self.balance = balance
self.n = len(balance)
def transfer(self, account1: int, account2: int, money: int) -> bool:
if account1 > self.n or account2 > self.n or self.balance[account1 - 1] < money:
return False
self.balance[account1 - 1] -= money
self.balance[account2 - 1] += money
return True
def deposit(self, account: int, money: int) -> bool:
if account > self.n:
return False
self.balance[account - 1] += money
return True
def withdraw(self, account: int, money: int) -> bool:
if account > self.n or self.balance[account - 1] < money:
return False
self.balance[account - 1] -= money
return True
# Your Bank object will be instantiated and called as such:
# obj = Bank(balance)
# param_1 = obj.transfer(account1,account2,money)
# param_2 = obj.deposit(account,money)
# param_3 = obj.withdraw(account,money)
Complexity
The time complexity is O(n). The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.