Introduction

In this blog post, we will explore how to solve the LeetCode problem "2043" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. The initial balance of each account is stored in a 0-indexed integer array balance, with the (i + 1)th account having an initial balance of balance[i]. Execute all the valid transactions. A transaction is valid if: The given account number(s) are between 1 and n, and The amount of money withdrawn or transferred from is less than or equal to the balance of the account. Implement the Bank class: Bank(long[] balance) Initializes the object with the 0-indexed integer array balance. boolean transfer(int account1, int account2, long money) Transfers money dollars from the account numbered account1 to the account numbered account2. Return true if the transaction was successful, false otherwise. boolean deposit(int account, long money) Deposit money dollars into the account numbered account. Return true if the transaction was successful, false otherwise. boolean withdraw(int account, long money) Withdraw money dollars from the account numbered account. Return true if the transaction was successful, false otherwise. Example 1: 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. Constraints: n == balance.length 1 <= n, account, account1, account2 <= 105 0 <= balance[i], money <= 1012 At most 104 calls will be made to each function transfer, deposit, withdraw.

Explanation

• Use a list to store account balances: The balance array provided in the constructor is directly stored as a list within the Bank class. This allows efficient access to account balances using their index.
  • Index adjustment: Since account numbers are 1-indexed but the balance list is 0-indexed, we adjust account numbers by subtracting 1 before accessing the balance list.
  • Validity checks: Before performing any transaction, we validate the account numbers to make sure they are within range and that the account has sufficient balance for withdrawals or transfers.

• Runtime Complexity: O(1) for all operations (transfer, deposit, withdraw).
• Storage Complexity: O(n) where n is the number of accounts.

Code

    class Bank:
    def __init__(self, balance: list[int]):
        self.balance = balance

    def transfer(self, account1: int, account2: int, money: int) -> bool:
        n = len(self.balance)
        if 1 <= account1 <= n and 1 <= account2 <= n:
            if self.balance[account1 - 1] >= money:
                self.balance[account1 - 1] -= money
                self.balance[account2 - 1] += money
                return True
            else:
                return False
        else:
            return False

    def deposit(self, account: int, money: int) -> bool:
        n = len(self.balance)
        if 1 <= account <= n:
            self.balance[account - 1] += money
            return True
        else:
            return False

    def withdraw(self, account: int, money: int) -> bool:
        n = len(self.balance)
        if 1 <= account <= n:
            if self.balance[account - 1] >= money:
                self.balance[account - 1] -= money
                return True
            else:
                return False
        else:
            return False