Date Range Generator

Date Range Generator - Problem

JavaScript Medium

Given a start date start, an end date end, and a positive integer step, return a generator object that yields dates in the range from start to end inclusive.

The value of step indicates the number of days between consecutive yielded values. All yielded dates must be in the string format YYYY-MM-DD.

Note: The generator should yield dates chronologically and include both start and end dates if they fall on valid step intervals.

Input & Output

Example 1 — Basic Weekly Step

$ Input: start = "2023-01-01", end = "2023-01-15", step = 7

› Output: ["2023-01-01", "2023-01-08", "2023-01-15"]

💡 Note: Starting from 2023-01-01, add 7 days each time: 2023-01-01 → 2023-01-08 → 2023-01-15. All dates are within range.

Example 2 — Daily Step

$ Input: start = "2023-02-28", end = "2023-03-03", step = 1

› Output: ["2023-02-28", "2023-03-01", "2023-03-02", "2023-03-03"]

💡 Note: Step of 1 day across month boundary. February ends on 28th (2023 is not leap year), so next date is March 1st.

Example 3 — Large Step

$ Input: start = "2023-01-01", end = "2023-01-10", step = 15

› Output: ["2023-01-01"]

💡 Note: First date is 2023-01-01. Next would be 2023-01-16 which exceeds end date 2023-01-10, so only start date is yielded.

Constraints

start and end are valid dates in format YYYY-MM-DD
start ≤ end
1 ≤ step ≤ 1000
Date range will not exceed 10 years

Visualization

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Asked in

G Google 35 M Microsoft 28 A Amazon 22 A Apple 15

The key insight is to use built-in date libraries to handle complex date arithmetic automatically, avoiding manual month/year boundary calculations. Best approach uses generator functions with efficient date objects. Time: O(n), Space: O(1) where n is the number of dates to yield.

Common Approaches

✓ Brute Force Generator

⏱️ Time: O(n) Space: O(1)

Create a simple generator that starts from the start date and keeps adding the step number of days until it exceeds the end date. Use JavaScript's Date object to handle date arithmetic automatically.

Efficient Date Iterator

⏱️ Time: O(n) Space: O(1)

Optimize by calculating the total number of iterations needed and using more efficient date operations. This approach minimizes date object creation and uses built-in date libraries more effectively.

Brute Force Generator — Algorithm Steps

Parse start and end dates into Date objects
Create generator that yields current date and increments by step days
Continue until current date exceeds end date

Visualization

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Step-by-Step Walkthrough

Initialize

Parse start date, end date, and step value

Generate

Yield current date and add step days

Repeat

Continue until current date exceeds end date

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct {
    int year, month, day;
    int endYear, endMonth, endDay;
    int step;
} DateGenerator;

DateGenerator solution(char* start, char* end, int step) {
    DateGenerator gen;
    sscanf(start, "%d-%d-%d", &gen.year, &gen.month, &gen.day);
    sscanf(end, "%d-%d-%d", &gen.endYear, &gen.endMonth, &gen.endDay);
    gen.step = step;
    return gen;
}

int hasNext(DateGenerator* gen) {
    if (gen->year > gen->endYear) return 0;
    if (gen->year == gen->endYear && gen->month > gen->endMonth) return 0;
    if (gen->year == gen->endYear && gen->month == gen->endMonth && gen->day > gen->endDay) return 0;
    return 1;
}

void nextDate(DateGenerator* gen, char* buffer) {
    sprintf(buffer, "%04d-%02d-%02d", gen->year, gen->month, gen->day);
    
    gen->day += gen->step;
    
    int daysInMonth[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    if ((gen->year % 4 == 0 && gen->year % 100 != 0) || (gen->year % 400 == 0)) daysInMonth[1] = 29;
    
    while (gen->day > daysInMonth[gen->month - 1]) {
        gen->day -= daysInMonth[gen->month - 1];
        gen->month++;
        if (gen->month > 12) {
            gen->month = 1;
            gen->year++;
            if ((gen->year % 4 == 0 && gen->year % 100 != 0) || (gen->year % 400 == 0)) daysInMonth[1] = 29;
            else daysInMonth[1] = 28;
        }
    }
}

int main() {
    char start[20], end[20];
    int step;
    
    fgets(start, sizeof(start), stdin);
    start[strcspn(start, "\n")] = 0;
    fgets(end, sizeof(end), stdin);
    end[strcspn(end, "\n")] = 0;
    scanf("%d", &step);
    
    DateGenerator gen = solution(start, end, step);
    char dates[1000][12];
    int count = 0;
    
    while (hasNext(&gen) && count < 1000) {
        nextDate(&gen, dates[count]);
        count++;
    }
    
    printf("[");
    for (int i = 0; i < count; i++) {
        printf("\"%s\"", dates[i]);
        if (i < count - 1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Where n is the number of dates in the range divided by step

✓ Linear Growth

Space Complexity

O(1)

Only stores current date and a few variables

✓ Linear Space

34.7K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Generator — Algorithm Steps

Visualization

Code -

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