Date Range Generator - Practice Coding Problems

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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Understanding the Visualization

Input

Start date, end date, and step size (days)

Process

Generate dates by adding step days repeatedly

Output

Array of date strings in YYYY-MM-DD format

Key Takeaway

🎯 Key Insight: Use generator functions with built-in date libraries to handle complex date arithmetic automatically

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

Approach	Time	Space	Notes
✓ Efficient Date Iterator	O(n)	O(1)	Pre-calculate date range and use efficient date arithmetic
Brute Force Generator	O(n)	O(1)	Generate all dates in range, yield every step-th date

Efficient Date Iterator — Algorithm Steps

Calculate total days between start and end
Determine exact number of iterations needed
Use optimized date arithmetic for each step

Visualization

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Step-by-Step Walkthrough

Convert

Parse dates to time stamps for efficient arithmetic

Calculate

Use millisecond/duration arithmetic for steps

Format

Convert back to YYYY-MM-DD format when yielding

Code -

solution.c — C

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

typedef struct {
    time_t current;
    time_t end;
    int step;
} OptimizedDateGenerator;

time_t dateToTimestamp(char* dateStr) {
    struct tm tm = {0};
    sscanf(dateStr, "%d-%d-%d", &tm.tm_year, &tm.tm_mon, &tm.tm_mday);
    tm.tm_year -= 1900;
    tm.tm_mon -= 1;
    return mktime(&tm);
}

OptimizedDateGenerator solution(char* start, char* end, int step) {
    OptimizedDateGenerator gen;
    gen.current = dateToTimestamp(start);
    gen.end = dateToTimestamp(end);
    gen.step = step;
    return gen;
}

int hasNextOpt(OptimizedDateGenerator* gen) {
    return gen->current <= gen->end;
}

void nextDateOpt(OptimizedDateGenerator* gen, char* buffer) {
    struct tm* tm = localtime(&gen->current);
    sprintf(buffer, "%04d-%02d-%02d", 
            tm->tm_year + 1900, 
            tm->tm_mon + 1, 
            tm->tm_mday);
    
    gen->current += gen->step * 24 * 60 * 60; // Add step days in seconds
}

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);
    
    OptimizedDateGenerator gen = solution(start, end, step);
    char dates[1000][12];
    int count = 0;
    
    while (hasNextOpt(&gen) && count < 1000) {
        nextDateOpt(&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 valid dates to yield

✓ Linear Growth

Space Complexity

O(1)

Only stores current date and iteration variables

✓ Linear Space

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~15 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Efficient Date Iterator — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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