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Commit bc2a821

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committed
Oct 27
1 parent 8c0f9fe commit bc2a821

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+729
-8
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‎README.md‎

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// Tag: Math
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// Time: O(N)
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// Space: O(1)
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// Ref: -
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// Note: -
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// Video: https://youtu.be/fUbGNGqOi3w
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// Hercy wants to save money for his first car. He puts money in the Leetcode bank every day.
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// He starts by putting in 1ドル on Monday, the first day. Every day from Tuesday to Sunday, he will put in 1ドル more than the day before. On every subsequent Monday, he will put in 1ドル more than the previous Monday.
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// Given n, return the total amount of money he will have in the Leetcode bank at the end of the nth day.
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//
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// Example 1:
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//
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// Input: n = 4
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// Output: 10
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// Explanation: After the 4th day, the total is 1 +たす 2 +たす 3 +たす 4 =わ 10.
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//
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// Example 2:
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//
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// Input: n = 10
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// Output: 37
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// Explanation: After the 10th day, the total is (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4) = 37. Notice that on the 2nd Monday, Hercy only puts in 2ドル.
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//
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// Example 3:
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//
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// Input: n = 20
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// Output: 96
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// Explanation: After the 20th day, the total is (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4 + 5 + 6 + 7 + 8) + (3 + 4 + 5 + 6 + 7 + 8) = 96.
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//
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//
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// Constraints:
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//
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// 1 <= n <= 1000
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//
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//
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class Solution {
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public:
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int totalMoney(int n) {
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int base = 21;
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int week = n / 7;
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int day = n % 7;
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int res = 0;
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for (int i = 0; i < week; i++) {
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res += base + (i + 1) * 7;
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}
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for (int i = 0; i < day; i++) {
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res += i + (week + 1);
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}
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return res;
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}
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};
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# Tag: Math
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# Time: O(N)
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# Space: O(1)
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# Ref: -
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# Note: -
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# Video: https://youtu.be/fUbGNGqOi3w
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# Hercy wants to save money for his first car. He puts money in the Leetcode bank every day.
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# He starts by putting in 1ドル on Monday, the first day. Every day from Tuesday to Sunday, he will put in 1ドル more than the day before. On every subsequent Monday, he will put in 1ドル more than the previous Monday.
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# Given n, return the total amount of money he will have in the Leetcode bank at the end of the nth day.
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#
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# Example 1:
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#
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# Input: n = 4
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# Output: 10
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# Explanation: After the 4th day, the total is 1 +たす 2 +たす 3 +たす 4 =わ 10.
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#
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# Example 2:
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#
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# Input: n = 10
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# Output: 37
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# Explanation: After the 10th day, the total is (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4) = 37. Notice that on the 2nd Monday, Hercy only puts in 2ドル.
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#
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# Example 3:
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#
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# Input: n = 20
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# Output: 96
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# Explanation: After the 20th day, the total is (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4 + 5 + 6 + 7 + 8) + (3 + 4 + 5 + 6 + 7 + 8) = 96.
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#
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#
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# Constraints:
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#
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# 1 <= n <= 1000
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#
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#
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class Solution:
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def totalMoney(self, n: int) -> int:
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base = 21
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week = n // 7
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day = n % 7
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res = 0
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for i in range(week):
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res += base + (i + 1) * 7
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for i in range(day):
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res += i + (week + 1)
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return res

‎leetcode/2043.simple-bank-system.cpp‎

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// Tag: Array, Hash Table, Design, Simulation
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// Time: O(1)
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// Space: O(N)
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// Ref: -
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// Note: -
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// Video: https://youtu.be/-ejSX-2Jznc
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// 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].
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// Execute all the valid transactions. A transaction is valid if:
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//
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// The given account number(s) are between 1 and n, and
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// The amount of money withdrawn or transferred from is less than or equal to the balance of the account.
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//
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// Implement the Bank class:
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//
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// Bank(long[] balance) Initializes the object with the 0-indexed integer array balance.
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// 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.
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// boolean deposit(int account, long money) Deposit money dollars into the account numbered account. Return true if the transaction was successful, false otherwise.
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// boolean withdraw(int account, long money) Withdraw money dollars from the account numbered account. Return true if the transaction was successful, false otherwise.
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//
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//
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// Example 1:
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//
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// Input
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// ["Bank", "withdraw", "transfer", "deposit", "transfer", "withdraw"]
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// [[[10, 100, 20, 50, 30]], [3, 10], [5, 1, 20], [5, 20], [3, 4, 15], [10, 50]]
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// Output
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// [null, true, true, true, false, false]
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//
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// Explanation
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// Bank bank = new Bank([10, 100, 20, 50, 30]);
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// bank.withdraw(3, 10); // return true, account 3 has a balance of 20,ドル so it is valid to withdraw 10ドル.
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// // Account 3 has 20ドル - 10ドル = 10ドル.
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// bank.transfer(5, 1, 20); // return true, account 5 has a balance of 30,ドル so it is valid to transfer 20ドル.
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// // Account 5 has 30ドル - 20ドル = 10,ドル and account 1 has 10ドル + 20ドル = 30ドル.
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// bank.deposit(5, 20); // return true, it is valid to deposit 20ドル to account 5.
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// // Account 5 has 10ドル + 20ドル = 30ドル.
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// bank.transfer(3, 4, 15); // return false, the current balance of account 3 is 10,ドル
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// // so it is invalid to transfer 15ドル from it.
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// bank.withdraw(10, 50); // return false, it is invalid because account 10 does not exist.
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//
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//
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// Constraints:
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//
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// n == balance.length
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// 1 <= n, account, account1, account2 <= 105
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// 0 <= balance[i], money <= 1012
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// At most 104 calls will be made to each function transfer, deposit, withdraw.
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//
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//
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class Bank {
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public:
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vector<long long> bal;
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Bank(vector<long long>& balance) {
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swap(bal, balance);
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}
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bool transfer(int account1, int account2, long long money) {
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int n = bal.size();
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if (account1 <= n && account2 <= n && bal[account1 - 1] >= money) {
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bal[account1 - 1] -= money;
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bal[account2 - 1] += money;
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return true;
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}
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return false;
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}
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bool deposit(int account, long long money) {
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int n = bal.size();
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if (account <= n) {
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bal[account - 1] += money;
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return true;
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}
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return false;
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}
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bool withdraw(int account, long long money) {
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int n = bal.size();
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if (account <= n && bal[account - 1] >= money) {
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bal[account - 1] -= money;
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return true;
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}
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return false;
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}
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};
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/**
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* Your Bank object will be instantiated and called as such:
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* Bank* obj = new Bank(balance);
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* bool param_1 = obj->transfer(account1,account2,money);
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* bool param_2 = obj->deposit(account,money);
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* bool param_3 = obj->withdraw(account,money);
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*/

‎leetcode/2043.simple-bank-system.py‎

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# Tag: Array, Hash Table, Design, Simulation
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# Time: O(1)
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# Space: O(N)
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# Ref: -
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# Note: -
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# Video: https://youtu.be/-ejSX-2Jznc
7+
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# 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].
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# Execute all the valid transactions. A transaction is valid if:
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#
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# The given account number(s) are between 1 and n, and
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# The amount of money withdrawn or transferred from is less than or equal to the balance of the account.
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#
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# Implement the Bank class:
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#
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# Bank(long[] balance) Initializes the object with the 0-indexed integer array balance.
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# 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.
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# boolean deposit(int account, long money) Deposit money dollars into the account numbered account. Return true if the transaction was successful, false otherwise.
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# boolean withdraw(int account, long money) Withdraw money dollars from the account numbered account. Return true if the transaction was successful, false otherwise.
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#
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#
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# Example 1:
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#
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# Input
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# ["Bank", "withdraw", "transfer", "deposit", "transfer", "withdraw"]
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# [[[10, 100, 20, 50, 30]], [3, 10], [5, 1, 20], [5, 20], [3, 4, 15], [10, 50]]
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# Output
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# [null, true, true, true, false, false]
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#
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# Explanation
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# Bank bank = new Bank([10, 100, 20, 50, 30]);
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# bank.withdraw(3, 10); // return true, account 3 has a balance of 20,ドル so it is valid to withdraw 10ドル.
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# // Account 3 has 20ドル - 10ドル = 10ドル.
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# bank.transfer(5, 1, 20); // return true, account 5 has a balance of 30,ドル so it is valid to transfer 20ドル.
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# // Account 5 has 30ドル - 20ドル = 10,ドル and account 1 has 10ドル + 20ドル = 30ドル.
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# bank.deposit(5, 20); // return true, it is valid to deposit 20ドル to account 5.
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# // Account 5 has 10ドル + 20ドル = 30ドル.
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# bank.transfer(3, 4, 15); // return false, the current balance of account 3 is 10,ドル
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# // so it is invalid to transfer 15ドル from it.
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# bank.withdraw(10, 50); // return false, it is invalid because account 10 does not exist.
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#
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#
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# Constraints:
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#
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# n == balance.length
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# 1 <= n, account, account1, account2 <= 105
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# 0 <= balance[i], money <= 1012
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# At most 104 calls will be made to each function transfer, deposit, withdraw.
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#
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#
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class Bank:
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def __init__(self, balance: List[int]):
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self.bal = balance
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def transfer(self, account1: int, account2: int, money: int) -> bool:
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n = len(self.bal)
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if account1 <= n and account2 <= n and self.bal[account1 - 1] >= money:
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self.bal[account1 - 1] -= money
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self.bal[account2 - 1] += money
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return True
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return False
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def deposit(self, account: int, money: int) -> bool:
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n = len(self.bal)
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if account <= n:
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self.bal[account - 1] += money
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return True
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return False
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def withdraw(self, account: int, money: int) -> bool:
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n = len(self.bal)
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if account <= n and self.bal[account - 1] >= money:
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self.bal[account - 1] -= money
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return True
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return False
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# Your Bank object will be instantiated and called as such:
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# obj = Bank(balance)
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# param_1 = obj.transfer(account1,account2,money)
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# param_2 = obj.deposit(account,money)
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# param_3 = obj.withdraw(account,money)
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// Tag: Hash Table, Math, Backtracking, Counting, Enumeration
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// Time: O(N)
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// Space: O(N)
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// Ref: -
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// Note: -
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// Video: https://youtu.be/yMjRNt9h1nQ
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// An integer x is numerically balanced if for every digit d in the number x, there are exactly d occurrences of that digit in x.
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// Given an integer n, return the smallest numerically balanced number strictly greater than n.
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//
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// Example 1:
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//
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// Input: n = 1
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// Output: 22
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// Explanation:
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// 22 is numerically balanced since:
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// - The digit 2 occurs 2 times.
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// It is also the smallest numerically balanced number strictly greater than 1.
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//
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// Example 2:
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//
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// Input: n = 1000
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// Output: 1333
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// Explanation:
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// 1333 is numerically balanced since:
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// - The digit 1 occurs 1 time.
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// - The digit 3 occurs 3 times.
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// It is also the smallest numerically balanced number strictly greater than 1000.
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// Note that 1022 cannot be the answer because 0 appeared more than 0 times.
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//
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// Example 3:
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//
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// Input: n = 3000
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// Output: 3133
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// Explanation:
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// 3133 is numerically balanced since:
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// - The digit 1 occurs 1 time.
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// - The digit 3 occurs 3 times.
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// It is also the smallest numerically balanced number strictly greater than 3000.
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//
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//
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// Constraints:
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//
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// 0 <= n <= 106
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//
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//
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class Solution {
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public:
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int nextBeautifulNumber(int n) {
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vector<int> counter = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
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int size = to_string(n).size();
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return dfs(n, 0, size, counter) ?: dfs(n, 0, size + 1, counter);
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}
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int dfs(int n, int val, int size, vector<int> &counter) {
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if (size == 0) {
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for (int i = 1; i <= 9; i++) {
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if (counter[i] == i) {
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continue;
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}
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if (counter[i] != 0) {
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return 0;
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}
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}
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return val > n ? val : 0;
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}
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int res = 0;
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for (int i = 1; i <= 9 && res == 0; i++) {
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if (counter[i] > 0 && counter[i] <= size) {
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counter[i] -= 1;
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res = dfs(n, val * 10 + i, size - 1, counter);
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counter[i] += 1;
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}
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}
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return res;
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}
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};

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