CF 1062471 - Beautiful Time

Rating: -
Tags: -
Solve time: 41s
Verified: yes

Solution

Problem Understanding

We are given two moments on a 24-hour digital clock, inclusive. Every clock reading is displayed as HH:MM, where hours use two digits from 00 to 23 and minutes use two digits from 00 to 59.

A time is considered beautiful if at least one of the following holds:

The hour value equals the minute value. Examples are 00:00 and 21:21.

The minute digits are exactly the reverse of the hour digits. For example, 21:12.

The four displayed digits form a sequence of consecutive increasing digits. For example, 01:23, whose digits are 0,1,2,3.

We must count how many beautiful readings appear between the given start time and end time, including both endpoints.

The constraints are tiny. A single day contains only 24 × 60 = 1440 possible clock readings. Even if we inspect every minute in the interval and test all three beauty conditions directly, the work is negligible. The intended solution is simple simulation.

The main source of mistakes is interpreting the beauty rules correctly.

Consider the time 12:12.

12
12

This is beautiful because hours equal minutes. A solution that checks only the mirrored condition would miss it.

Consider the time 20:02.

20
02

This is beautiful because the minute digits are the reverse of the hour digits. Treating the values numerically and dropping leading zeros would incorrectly compare 20 with 2.

Consider the time 01:23.

01
23

The four displayed digits are 0,1,2,3, which are consecutive increasing digits. The rule applies to digits on the display, not to the numeric values of hours and minutes separately.

A final edge case is when the start and end moments are identical. The statement explicitly says that only that single reading is considered.

Approaches

A brute-force idea is to generate every possible clock reading in the interval and test whether it is beautiful. For each time we can extract the four displayed digits and evaluate the three conditions directly.

Since there are at most 1440 readings in an entire day, even the brute-force approach performs only a few thousand operations. There is no need for combinatorics, dynamic programming, or any advanced observation.

The key realization is that the search space itself is tiny. Many clock problems require clever counting because times span large ranges, but here the entire universe of valid readings fits into a single day. Enumerating minute by minute is already optimal in practice.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Enumeration	O(1440)	O(1)	Accepted
Optimal Enumeration	O(1440)	O(1)	Accepted

In this problem, the brute-force and optimal solutions are effectively the same.

Algorithm Walkthrough

Convert the starting time into a single minute index:

start = h1 * 60 + m1. 2. Convert the ending time into a single minute index:

end = h2 * 60 + m2. 3. Iterate through every minute t from start to end, inclusive. 4. Recover the current hour and minute:

h = t // 60,

m = t % 60. 5. Extract the four displayed digits:

h_tens, h_ones, m_tens, m_ones. 6. Check whether hours equal minutes.

If h == m, the reading is beautiful. 7. Check whether the minute digits are the reverse of the hour digits.

This means:

h_tens == m_ones and h_ones == m_tens. 8. Check whether the four digits form consecutive increasing digits:

d1 + 1 == d2,

d2 + 1 == d3,

d3 + 1 == d4. 9. If at least one condition holds, increment the answer. 10. After processing the entire interval, print the count.

Why it works

The algorithm examines every clock reading that belongs to the required interval and applies exactly the three beauty conditions from the definition. A reading contributes to the answer if and only if at least one condition is satisfied. Since every eligible time is checked once and no time outside the interval is examined, the final count is exactly the number of beautiful readings.

Python Solution

import sys
input = sys.stdin.readline

h1 = int(input())
m1 = int(input())
h2 = int(input())
m2 = int(input())

start = h1 * 60 + m1
end = h2 * 60 + m2

ans = 0

for t in range(start, end + 1):
    h = t // 60
    m = t % 60

    beautiful = False

    if h == m:
        beautiful = True

    ht = h // 10
    ho = h % 10
    mt = m // 10
    mo = m % 10

    if ht == mo and ho == mt:
        beautiful = True

    if ht + 1 == ho and ho + 1 == mt and mt + 1 == mo:
        beautiful = True

    if beautiful:
        ans += 1

print(ans)

The first part converts the input times into minute indices. This makes iteration straightforward because advancing one minute simply means increasing an integer by one.

Inside the loop, the current hour and minute are reconstructed using division and modulo operations.

The mirrored condition must be checked using digits rather than integer reversal tricks. This avoids issues with leading zeros such as 20:02.

For the consecutive-digit condition, we compare the four displayed digits directly. The rule concerns the visible sequence on the clock, not the numeric values of the hour and minute fields.

The loop is inclusive on both ends, matching the statement exactly.

Worked Examples

Example 1

Input

The interval is from 20:00 to 20:30.

Relevant beautiful times:

Time	Equal	Mirror	Consecutive	Counted
20:00	No	No	No	No
20:02	No	Yes	No	Yes
20:20	Yes	No	No	Yes

Answer = 2.

This example shows that different beauty conditions can contribute independently.

Example 2

Input

Only one reading is examined.

Time	Equal	Mirror	Consecutive	Counted
17:31	No	No	No	No

Answer = 0.

This confirms that when the endpoints coincide, the algorithm processes exactly one reading.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1440)	At most one full day of clock readings is checked
Space	O(1)	Only a few integer variables are stored

A full day contains only 1440 minutes, so the running time is effectively constant. The solution easily fits within any reasonable time and memory limit.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def solve():
    h1 = int(input())
    m1 = int(input())
    h2 = int(input())
    m2 = int(input())

    start = h1 * 60 + m1
    end = h2 * 60 + m2

    ans = 0

    for t in range(start, end + 1):
        h = t // 60
        m = t % 60

        ok = False

        if h == m:
            ok = True

        ht, ho = divmod(h, 10)
        mt, mo = divmod(m, 10)

        if ht == mo and ho == mt:
            ok = True

        if ht + 1 == ho and ho + 1 == mt and mt + 1 == mo:
            ok = True

        if ok:
            ans += 1

    print(ans)

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    global input
    input = sys.stdin.readline

    out = io.StringIO()
    old = sys.stdout
    sys.stdout = out

    solve()

    sys.stdout = old
    return out.getvalue()

# provided samples
assert run("20\n0\n20\n30\n") == "2\n", "sample 1"
assert run("17\n31\n17\n31\n") == "0\n", "sample 2"

# custom cases
assert run("0\n0\n0\n0\n") == "1\n", "00:00 satisfies multiple conditions"
assert run("1\n23\n1\n23\n") == "1\n", "consecutive digits"
assert run("20\n2\n20\n2\n") == "1\n", "mirror with leading zero"
assert run("5\n5\n5\n5\n") == "1\n", "hours equal minutes"

Test input	Expected output	What it validates
`00:00 -> 00:00`	`1`	Equal and mirrored simultaneously
`01:23 -> 01:23`	`1`	Consecutive-digit condition
`20:02 -> 20:02`	`1`	Leading-zero mirror case
`05:05 -> 05:05`	`1`	Equality condition

Edge Cases

Consider the input

The displayed time is 20:02. The digits are (2,0,0,2). The hour digits reversed produce (0,2), which exactly matches the minute digits. The algorithm extracts digits explicitly and detects the mirror condition, producing answer 1.

Consider the input

The displayed digits are 0,1,2,3. The checks

0+1=1
1+1=2
2+1=3

all hold, so the time is counted. Treating hours and minutes as separate numbers would miss this case.

Consider the input

The interval contains exactly one reading. The loop runs once because it iterates from start to end inclusively. Since 00:00 satisfies the beauty rules, the answer is 1.

Consider the input

Hours equal minutes, so the reading is beautiful even though the mirror condition is not required. The algorithm checks all beauty rules independently and counts the time once.