LeetCode 1716. Calculate Money in Leetcode Bank Shell Solution | Explanation + Code

Description

Hercy wants to save money for his first car. He puts money in the Leetcode bank every day.

He starts by putting in 1 on Monday, the first day. Every day from Tuesday to Sunday, he will put in 1 more than the day before. On every subsequent Monday, he will put in $1 more than the previous Monday.

Given n, return the total amount of money he will have in the Leetcode bank at the end of the nth day.

 

Example 1:

Input: n = 4
Output: 10
Explanation: After the 4th day, the total is 1 + 2 + 3 + 4 = 10.

Example 2:

Input: n = 10
Output: 37
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.

Example 3:

Input: n = 20
Output: 96
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.

 

Constraints:

• 1 <= n <= 1000

Solutions

Solution 1: Math

According to the problem description, the deposit situation for each week is as follows:

Week 1: 1, 2, 3, 4, 5, 6, 7
Week 2: 2, 3, 4, 5, 6, 7, 8
Week 3: 3, 4, 5, 6, 7, 8, 9
...
Week k: k, k+1, k+2, k+3, k+4, k+5, k+6

Given n days of deposits, the number of complete weeks is k = \lfloor n / 7 \rfloor, and the remaining days is b = n \mod 7.

The total deposit for the complete k weeks can be calculated using the arithmetic sequence sum formula:

S₁ = ^k⁄₂ × (28 + 28 + 7 × (k - 1))

The total deposit for the remaining b days can also be calculated using the arithmetic sequence sum formula:

S₂ = ^b⁄₂ × (k + 1 + k + 1 + b - 1)

The final total deposit amount is S = S₁ + S₂.

The time complexity is O(1) and the space complexity is O(1).

PythonJavaC++GoTypeScript

class Solution: def totalMoney(self, n: int) -> int: k, b = divmod(n, 7) s1 = (28 + 28 + 7 * (k - 1)) * k // 2 s2 = (k + 1 + k + 1 + b - 1) * b // 2 return s1 + s2(code-box)

class Solution { public int totalMoney(int n) { int k = n / 7, b = n % 7; int s1 = (28 + 28 + 7 * (k - 1)) * k / 2; int s2 = (k + 1 + k + 1 + b - 1) * b / 2; return s1 + s2; } }(code-box)

class Solution { public: int totalMoney(int n) { int k = n / 7, b = n % 7; int s1 = (28 + 28 + 7 * (k - 1)) * k / 2; int s2 = (k + 1 + k + 1 + b - 1) * b / 2; return s1 + s2; } };(code-box)

func totalMoney(n int) int { k, b := n/7, n%7 s1 := (28 + 28 + 7*(k-1)) * k / 2 s2 := (k + 1 + k + 1 + b - 1) * b / 2 return s1 + s2 }(code-box)

function totalMoney(n: number): number { const k = (n / 7) | 0; const b = n % 7; const s1 = (((28 + 28 + 7 * (k - 1)) * k) / 2) | 0; const s2 = (((k + 1 + k + 1 + b - 1) * b) / 2) | 0; return s1 + s2; }(code-box)