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1716. Calculate Money in Leetcode Bank

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:

\[ S_1 = \frac{k}{2} \times (28 + 28 + 7 \times (k - 1)) \]

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

\[ S_2 = \frac{b}{2} \times (k + 1 + k + 1 + b - 1) \]

The final total deposit amount is \(S = S_1 + S_2\).

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

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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