> For the complete documentation index, see [llms.txt](https://liuzhenglaichn.gitbook.io/algorithm/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://liuzhenglaichn.gitbook.io/algorithm/date.md).

# Date

Implement a function to get the number of days from `0000-00-00` to `yyyy-mm-dd`.

## Accumulate

```cpp
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
// Ref: https://leetcode.com/problems/number-of-days-between-two-dates/discuss/517605/Similar-to-day-of-the-year
const int days[12] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
bool isLeap(int year) {
    return year % 4 == 0 && (year % 100 != 0 || year % 400 == 0);
}
int getDays(string date) {
    int y = stoi(date.substr(0, 4)), m = stoi(date.substr(5, 2)), d = stoi(date.substr(8));
    for (int yr = 0; yr < y; ++yr) d += isLeap(yr) ? 366 : 365;
    return d + (m > 2 && isLeap(y)) + accumulate(begin(days), begin(days) + m - 1, 0);
}
```

## Magic Formula

When `m=1` or `m=2` (January or February), we let `m=13` or `m=14` and let `y` decreased by 1. Imagine it is 13th or 14th month of the last year. By doing that, we let the magical formula also work for those two months. `(153 * m + 8) // 5` is just a carefully designed way to record the days of each month. More specifically, it is designed to record the difference of days between two months. Suppose we have March 1st and April 1st, `(153 * 3 + 8) // 5 = 93` while `(153 * 4 + 8) // 5 = 124`, the difference is 31 which is the number of days in March. Suppose we have April 1st to May 1st, `(153 * 4 + 8) // 5 = 124` and `(153 * 5 + 8) // 5 = 154`, the difference is now 30 which is the number of days in April. You can also check other months.

```cpp
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
// Ref: https://leetcode.com/problems/number-of-days-between-two-dates/discuss/517582/Python-Magical-Formula
int days(string date) {
    int y = stoi(date.substr(0, 4)), m = stoi(date.substr(5, 2)), d = stoi(date.substr(8));
    if (m < 3) {
        m += 12;
        y -= 1;
    }
    return 365 * y + y / 4 + y / 400 - y / 100 + d + (153 * m + 8) / 5;
}
```

## Problems

* [1360. Number of Days Between Two Dates (Easy)](https://leetcode.com/problems/number-of-days-between-two-dates/)
* [1154. Day of the Year (Easy)](https://leetcode.com/problems/day-of-the-year/)


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