CF 16A - Flag

CF 16A - Flag

Rating: 800
Tags: implementation
Solve time: 1m 8s
Verified: yes

Solution

Problem Understanding

We are given a rectangular grid representing a country’s flag. Each cell in the grid has a color, encoded as a digit from 0 to 9. The dimensions of the flag are $n$ rows by $m$ columns. The goal is to determine whether the flag is “striped” according to the new ISO standard. A flag is striped if every row contains cells of the same color, and consecutive rows do not share the same color.

The input is small, with $n, m \leq 100$. This means that even an $O(n \cdot m)$ solution is acceptable, because the total number of cells is at most 10,000, which is trivial for modern CPUs.

Some non-obvious edge cases include flags with a single row, where there is no previous row to compare to, and flags with a single column, where each row still needs to be uniform. For example, a flag with one row of three cells 111 should return YES because all cells in that row are identical, even though there is no row above or below to compare colors. Another tricky scenario is when two adjacent rows accidentally share the same color, like 111 followed by 111. This should return NO because the stripes are not distinct.

Approaches

A brute-force approach would iterate over every cell, checking two conditions: each row is uniform and each consecutive row is different from the previous one. This is correct because the problem explicitly defines the properties of a valid flag. The complexity is $O(n \cdot m)$, which is acceptable for this problem size.

The key insight is that we can simplify the check: instead of verifying each individual cell against every other cell in the row repeatedly, we only need to compare each cell to the first cell of the row. If any cell differs, the row is invalid. For checking consecutive rows, we just compare the first cell of the current row to the first cell of the previous row. This reduces unnecessary comparisons but does not change the asymptotic complexity.

The brute-force works because it checks every requirement explicitly, but the observation about row-first-cell comparison reduces the cognitive and implementation complexity, making the code cleaner and less error-prone.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n·m)	O(1)	Accepted
Optimized row-first check	O(n·m)	O(1)	Accepted

Algorithm Walkthrough

1. Read the dimensions $n$ and $m$ from input and store the flag as a list of strings, where each string represents a row. This preserves the row structure for easy checking.
2. Iterate over each row starting from the first. For each row, check that all cells are identical. Compare each character in the row to the first character. If any character differs, output NO and stop.
3. If the current row is not the first, compare its first character with the first character of the previous row. If they are the same, output NO and stop.
4. If all rows pass both checks, output YES.

Why it works: The algorithm ensures that every row is uniform by comparing all cells to the row’s first cell. By checking consecutive rows’ first cells, it ensures that adjacent rows are different. These two checks satisfy the definition of a striped flag. No other checks are necessary because color digits are guaranteed to be 0-9.

Python Solution

import sys
input = sys.stdin.readline

n, m = map(int, input().split())
flag = [input().strip() for _ in range(n)]

for i in range(n):
    row = flag[i]
    if any(c != row[0] for c in row):
        print("NO")
        sys.exit(0)
    if i > 0 and row[0] == flag[i - 1][0]:
        print("NO")
        sys.exit(0)

print("YES")

The code first reads the flag into a list of strings. The any(c != row[0] for c in row) construct checks row uniformity efficiently. The second condition ensures stripes are different. Using sys.exit(0) immediately terminates the program when a violation is found, which avoids unnecessary iteration over the remaining rows.

Worked Examples

Sample 1

Input:

3 3
000
111
222

Step	Row	Uniform?	First cell vs previous row?
0	000	Yes	N/A
1	111	Yes	1 != 0, OK
2	222	Yes	2 != 1, OK

Output: YES

Explanation: Each row is uniform and no two consecutive rows have the same color.

Custom Input 2

Input:

2 3
111
111

Step	Row	Uniform?	First cell vs previous row?
0	111	Yes	N/A
1	111	Yes	1 == 1, FAIL

Output: NO

Explanation: Consecutive rows share the same color.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n·m)	We scan each row entirely once.
Space	O(n·m)	Storing the grid as a list of strings requires one string per row.

For $n, m \leq 100$, the algorithm handles at most 10,000 cells, well within time and memory limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()
    n, m = map(int, input().split())
    flag = [input().strip() for _ in range(n)]
    for i in range(n):
        row = flag[i]
        if any(c != row[0] for c in row):
            print("NO")
            return sys.stdout.getvalue().strip()
        if i > 0 and row[0] == flag[i - 1][0]:
            print("NO")
            return sys.stdout.getvalue().strip()
    print("YES")
    return sys.stdout.getvalue().strip()

# Provided sample
assert run("3 3\n000\n111\n222\n") == "YES", "sample 1"

# Custom cases
assert run("1 5\n33333\n") == "YES", "single row"
assert run("5 1\n1\n2\n3\n4\n5\n") == "YES", "single column, distinct"
assert run("2 4\n7777\n7777\n") == "NO", "repeated color"
assert run("3 3\n123\n456\n789\n") == "NO", "row not uniform"
assert run("3 2\n11\n22\n22\n") == "NO", "last two rows same"

Test input	Expected output	What it validates
1 5\n33333	YES	Single row is valid
5 1\n1\n2\n3\n4\n5	YES	Single-column flag with distinct colors
2 4\n7777\n7777	NO	Adjacent rows share color
3 3\n123\n456\n789	NO	Row not uniform
3 2\n11\n22\n22	NO	Last two rows same color

Edge Cases

A flag with one row, such as:

1 4
4444

Algorithm checks the row uniformity: all(c == row[0] for c in row) passes. Since there is no previous row, the adjacent-row check is skipped. Output is YES.

A flag with one column:

3 1
1
2
2

Row uniformity trivially passes because each row has one cell. The adjacent-row check fails for the last two rows because 2 == 2. Output is NO.

These examples confirm that the algorithm handles minimal dimensions and edge comparisons correctly.