Rating: -
Tags: -
Solve time: 50s
Verified: yes

Solution

Problem Understanding

The flag is made from n stars arranged in horizontal rows. The rows do not have to contain exactly the same number of stars, but every row size must differ from every other row size by at most one. If two different row sizes are used, rows of the same size cannot touch each other, so the two sizes must alternate. Among all possible arrangements, we need find the smallest possible difference between the number of rows and the maximum number of stars in any row.

The input contains a single integer n, the number of stars available. The output is the minimum possible absolute difference between the height and width of such a flag.

The value of n can be as large as 10^12, so iterating over every possible number of rows is impossible. A linear solution would require up to 10^12 checks, far beyond the available time. We need to exploit the mathematical structure of possible row counts. The number of different values of floor(n / x) is only about 2 * sqrt(n), which is small enough for this limit.

The tricky cases appear when the number of stars cannot be split into equal rows. For example, with n = 5, using two rows gives sizes 3 and 2, but those rows are allowed because there are only two of them. A careless implementation might reject it because the two row sizes are different. The correct answer is 1.

Output:

Another edge case is when all rows have the same length. For n = 16, four rows of four stars are valid. The answer is 0.

Output:

A solution that only checks alternating rows may miss this case because there are no two different row sizes to alternate.

Approaches

A direct approach is to try every possible number of rows r. For each choice, the maximum row length is ceil(n / r). We can check whether the remaining stars can be distributed among rows of sizes ceil(n / r) and floor(n / r). This is correct, because every valid flag must have exactly those two possible row sizes. However, trying all r from 1 to n performs O(n) checks. With n up to 10^12, this is impossible.

The key observation is that valid row counts can be described using the quotient q = floor(n / r). When q is fixed, all possible r values form one interval. There are only about 2 * sqrt(n) different quotients.

Suppose a flag has r rows and q = floor(n / r). Let rem = n mod r.

If rem = 0, every row has size q, so r must be n / q.

Otherwise, there are rem rows of size q + 1 and r - rem rows of size q. These two row sizes must alternate, so their counts can differ by at most one:

abs(rem - (r - rem)) <= 1

This gives three possible equations:

rem = r / 2
rem = (r - 1) / 2
rem = (r + 1) / 2

Substituting n = q * r + rem gives three direct candidates for r:

r = 2n / (2q + 1)
r = (2n + 1) / (2q + 1)
r = (2n - 1) / (2q + 1)

For every quotient interval, we only need to test these few candidates.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n)	O(1)	Too slow
Optimal	O(sqrt(n))	O(1)	Accepted

Algorithm Walkthrough

Iterate through all intervals of equal values of floor(n / r). For a starting row count l, compute q = n // l and the largest row count r with the same quotient. This skips many impossible row counts at once.
For the current quotient q, test the candidate n / q. This is the only possible row count where all rows have equal size.
Test the three formulas derived from alternating row counts:

2n / (2q + 1)
(2n + 1) / (2q + 1)
(2n - 1) / (2q + 1)

Only candidates inside the current quotient interval can actually have this value of q. 4. For every candidate row count, verify that the row distribution is valid and update the minimum value of abs(rows - maximum_row_size).

Why it works: every possible flag has some row count r. The algorithm visits the unique quotient interval containing this r. Inside that interval, the equations above cover every possible way the two row sizes can alternate. Equal-sized rows are handled separately. Since every valid arrangement is checked, the smallest recorded difference is the optimal one.

Python Solution

import sys
input = sys.stdin.readline

def solve_case(n):
    ans = n
    def check(r):
        nonlocal ans
        if r <= 0:
            return
        q = n // r
        rem = n % r
        if rem != 0 and abs(rem - (r - rem)) > 1:
            return
        ans = min(ans, abs(r - ((n + r - 1) // r)))

    l = 1
    while l <= n:
        q = n // l
        r = n // q

        if n % q == 0:
            check(n // q)

        d = 2 * q + 1
        for x in (2 * n, 2 * n + 1, 2 * n - 1):
            if x > 0 and x % d == 0:
                cand = x // d
                if l <= cand <= r:
                    check(cand)

        l = r + 1

    return ans

def main():
    n = int(input())
    print(solve_case(n))

if __name__ == "__main__":
    main()

The main loop jumps between quotient intervals instead of increasing the number of rows one by one. The interval [l, r] contains all row counts producing the same floor(n / rows) value.

The divisibility case is checked separately because all rows can have the same size. For the remaining cases, the three generated candidates are exactly the situations where the larger and smaller row counts can alternate.

The validation function recomputes the remainder and checks the alternating condition. This avoids relying on the algebra alone and prevents invalid candidates caused by integer division. Python integers handle the large values safely, so no overflow handling is needed.

Worked Examples

For n = 16:

rows	quotient	remainder	maximum row size	difference
4	4	0	4	0

The algorithm finds the equal-row arrangement with four rows of four stars, giving the optimal answer 0.

For n = 5:

rows	quotient	remainder	maximum row size	difference
2	2	1	3	1

The two rows contain three and two stars. Since there are only two rows, the different sizes alternate automatically.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(sqrt(n))	There are only O(sqrt(n)) different values of `floor(n / r)`, and each interval creates a constant number of checks.
Space	O(1)	The algorithm stores only a few counters and the current answer.

The largest input value is 10^12, so about two million quotient intervals are already a safe upper bound. The solution avoids any loop depending directly on n.

Test Cases

import sys, io

def solve_case(n):
    ans = n

    def check(r):
        nonlocal ans
        if r <= 0:
            return
        q = n // r
        rem = n % r
        if rem != 0 and abs(rem - (r - rem)) > 1:
            return
        ans = min(ans, abs(r - ((n + r - 1) // r)))

    l = 1
    while l <= n:
        q = n // l
        r = n // q

        if n % q == 0:
            check(n // q)

        d = 2 * q + 1
        for x in (2 * n, 2 * n + 1, 2 * n - 1):
            if x > 0 and x % d == 0:
                cand = x // d
                if l <= cand <= r:
                    check(cand)

        l = r + 1

    return ans

def run(inp: str) -> str:
    return str(solve_case(int(inp.strip()))) + "\n"

assert run("16") == "0\n", "equal rows"
assert run("5") == "1\n", "two alternating rows"
assert run("1") == "0\n", "single star"
assert run("1000000000000") == "0\n", "perfect square boundary"
assert run("50") == "3\n", "non-square large case"

Test input	Expected output	What it validates
`16`	`0`	Equal row sizes
`5`	`1`	Two different row sizes
`1`	`0`	Minimum size
`1000000000000`	`0`	Large input and integer handling
`50`	`3`	Non-perfect-square arrangement

Edge Cases

For n = 5, the algorithm reaches the quotient interval with q = 2. The candidate r = 2 is generated from the alternating formula. The remainder is 1, so the rows are sized 3 and 2. Their counts differ by one, meaning the flag is valid, and the answer becomes 1.

For n = 16, the candidate r = 4 comes from the equal-row case because 16 is divisible by 4. The remainder is zero, so every row contains four stars. The height and width are both four, producing answer 0.

For a large value such as n = 10^12, the algorithm never tries a trillion row counts. It only visits quotient intervals, and the same arithmetic works with large integers. This is the difference between a feasible solution and a brute-force search.

CF 102860E - Flag with Stars