Given a string date representing a Gregorian calendar date formatted as YYYY-MM-DD, return the day number of the year.

Example 1:
Input: date = “2019-01-09”
Output: 9
Explanation: Given date is the 9th day of the year in 2019.

Example 2:
Input: date = “2019-02-10”
Output: 41

Example 3:
Input: date = “2003-03-01”
Output: 60

Example 4:
Input: date = “2004-03-01”
Output: 61

Constraints:
date.length == 10
date[4] == date[7] == ‘-‘, and all other date[i]’s are digits
date represents a calendar date between Jan 1st, 1900 and Dec 31, 2019.

C++ Algorithm to Count the Days Before the Date

To count the days given a date string of ISO 8601 format, we need to first parse/extract the string into three integers: year, month and day.

Then, we need to check if the year is a leap year (which we can use the little function described in How to Determine the Leap Year?).

For February, if it is leap year, we need to add one more day i.e. 29 days in a leap year. For January, March, May, July, August, October, December, there are 31 days, all others (except February which is 28 days in non-leap year), are 30 days.

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class Solution {
public:
    int dayOfYear(string date) {
        int year = std::stoi(date.substr(0, 4));
        int month = (date[5] - '0') * 10 +
            date[6] - '0';
        int day = (date[8] - '0') * 10 +
            date[9] - '0';
        int ans = 0;
        for (int i = 1; i < month; ++ i) {
            switch (i) {
                case 1:
                case 3:
                case 5:
                case 7:
                case 8:
                case 10:
                case 12:
                    ans += 31;
                    break;
                case 2:
                    if (isLeap(year)) {
                        ans += 29;
                    } else {
                        ans += 28;
                    }
                    break;
                default:
                    ans += 30;
            }
        }
        return ans + day;
    }
private:
    bool isLeap(int year) {
        if (year % 400 == 0) return true;
        if (year % 100 == 0) return false;
        if (year % 4 == 0) return true;
        return false;
    }
};

class Solution {
public:
    int dayOfYear(string date) {
        int year = std::stoi(date.substr(0, 4));
        int month = (date[5] - '0') * 10 +
            date[6] - '0';
        int day = (date[8] - '0') * 10 +
            date[9] - '0';
        int ans = 0;
        for (int i = 1; i < month; ++ i) {
            switch (i) {
                case 1:
                case 3:
                case 5:
                case 7:
                case 8:
                case 10:
                case 12:
                    ans += 31;
                    break;
                case 2:
                    if (isLeap(year)) {
                        ans += 29;
                    } else {
                        ans += 28;
                    }
                    break;
                default:
                    ans += 30;
            }
        }
        return ans + day;
    }
private:
    bool isLeap(int year) {
        if (year % 400 == 0) return true;
        if (year % 100 == 0) return false;
        if (year % 4 == 0) return true;
        return false;
    }
};

Then, the counting should be pretty straightforward, which can be done in O(1) constant, in both time and space.

–EOF (The Ultimate Computing & Technology Blog) —