CF 1108C - Nice Garland
We are given a line of lamps, each painted in one of three colors: R, G, or B. We are allowed to repaint any lamp to any other color.
Rating: 1300
Tags: brute force, greedy, math
Solve time: 7m 4s
Verified: no
Solution
Problem Understanding
We are given a line of lamps, each painted in one of three colors: R, G, or B. We are allowed to repaint any lamp to any other color. The goal is to modify the sequence so that it becomes “periodic in a specific geometric sense”: any two lamps of the same final color must sit at positions whose indices differ by a multiple of 3.
This condition has a very rigid structural consequence. If we look at positions modulo 3, say indices 0, 1, 2 repeating, then all positions of the same color must fall into exactly one of these residue classes. In other words, each residue class mod 3 must be assigned exactly one color, and that assignment must be consistent across the whole string.
So the task becomes choosing a mapping from the three residue classes {0,1,2} to the three colors {R,G,B}, making sure every color is used exactly once, and then repainting each position to match the color assigned to its residue class. We want to minimize how many positions differ from the original string.
The input size can go up to 200,000. That immediately rules out anything quadratic or cubic over the number of positions. Any solution that tries all local repaint combinations per position independently without structure would be too slow or incorrect, because the constraint couples positions globally through modulo classes.
A subtle failure case for naive thinking is to try greedy local fixes like “paint each position to match the majority color among neighbors” or similar heuristics. For example, with input RGBRGB, a greedy approach might incorrectly alter a position early and break global consistency, even though the optimal solution is already valid.
Another subtle case is assuming a fixed pattern like “RGBRGB…” always works best. For BRB, such a fixed pattern may require more changes than another permutation of colors assigned to residues.
The core difficulty is that the problem is not about individual positions but about selecting the best permutation of a 3-color assignment over 3 residue classes.
Approaches
A brute-force view makes the structure obvious. Since every valid final garland is determined by assigning a permutation of {R, G, B} to residue classes 0, 1, and 2, we can try all 3! = 6 assignments. For each assignment, we compute how many positions already match and choose the best one.
For a fixed assignment, we scan the string once and count mismatches. This costs O(n). With 6 permutations, the total is O(6n), which is linear and very small in practice.
The key insight is recognizing that the “distance divisible by 3” condition is exactly equivalent to grouping indices by modulo 3. Once that is seen, the problem stops being about recoloring arbitrary structure and becomes a constant-size assignment problem over residue classes.
There is no need for dynamic programming or greedy simulation over the string; the entire optimization space collapses into six possibilities.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force (6 permutations) | O(n) | O(1) | Accepted |
| Optimal (same idea, structured) | O(n) | O(1) | Accepted |
Algorithm Walkthrough
- Split the indices of the string into three groups based on i mod 3.
This is the only structural constraint in the problem, so we isolate it immediately. 2. Precompute counts of how many R, G, and B appear in each residue class.
This avoids recomputing character comparisons repeatedly for each permutation. 3. Enumerate all permutations of the three colors.
Each permutation represents a full valid assignment of colors to residue classes. 4. For a given permutation, compute how many characters already match:
for each residue class r, add the count of target color assigned to r. 5. Track the permutation with the maximum number of matches.
Maximizing matches is equivalent to minimizing recolors. 6. Reconstruct the answer string using the best assignment by writing the chosen color for each position i based on i mod 3.
Why it works
Every valid “nice” string must assign exactly one color to each residue class modulo 3, because if two positions share a color, their indices must differ by a multiple of 3, which forces them into the same residue class. Since there are exactly three colors and three residue classes, this assignment is a bijection. The algorithm enumerates all bijections and selects the one that aligns best with the original string, guaranteeing optimality.
Python Solution
import sys
input = sys.stdin.readline
def solve():
n = int(input().strip())
s = input().strip()
colors = ['R', 'G', 'B']
# cnt[pos_mod][color_index]
cnt = [[0] * 3 for _ in range(3)]
idx = {'R': 0, 'G': 1, 'B': 2}
for i, ch in enumerate(s):
cnt[i % 3][idx[ch]] += 1
best_perm = None
best_keep = -1
# try all permutations
from itertools import permutations
for perm in permutations(range(3)):
keep = 0
for r in range(3):
keep += cnt[r][perm[r]]
if keep > best_keep:
best_keep = keep
best_perm = perm
res = list(s)
for i in range(n):
res[i] = colors[best_perm[i % 3]]
print(n - best_keep)
print("".join(res))
if __name__ == "__main__":
solve()
The implementation compresses the problem into counting matches per residue class. The cnt table stores how often each color appears in each modulo bucket. Each permutation represents a mapping from residue class to color index. The score of a permutation is simply how many positions already match that mapping.
A common pitfall is rebuilding the candidate string during evaluation of each permutation. That is unnecessary; we only compute a score. Reconstruction is done once using the best permutation.
Another subtle point is ensuring that the mapping is applied by index modulo 3, not by position in the permutation loop. Mixing those two interpretations is a frequent source of wrong answers.
Worked Examples
Example 1
Input:
3
BRB
We compute residue classes:
| i mod 3 | characters |
|---|---|
| 0 | B |
| 1 | R |
| 2 | B |
Counts:
| mod | R | G | B |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 1 | 1 | 0 | 0 |
| 2 | 0 | 0 | 1 |
Now try a permutation, for example (G, R, B):
| mod | assigned | matches |
|---|---|---|
| 0 | G | 0 |
| 1 | R | 1 |
| 2 | B | 1 |
Total matches = 2, so only 1 change needed.
Result becomes:
GRB
This matches the sample and confirms that optimal assignment depends on distribution across residue classes.
Example 2
Input:
6
RRGGBB
Residue classes:
| mod | chars |
|---|---|
| 0 | R, G |
| 1 | R, B |
| 2 | G, B |
A good permutation is (R, G, B):
| mod | assigned | matches |
|---|---|---|
| 0 | R | 1 |
| 1 | G | 0 |
| 2 | B | 1 |
Total matches = 2, so 4 changes needed.
This example shows that even symmetric-looking strings may require multiple recolors because the distribution across residue classes is not aligned with a clean cyclic pattern.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n) | We scan the string once to build counts, then test 6 permutations, each O(1) work over fixed 3 classes |
| Space | O(1) | Only a 3x3 frequency table and constant arrays are used |
The linear scan fits easily within constraints for n up to 200,000. Memory usage is constant and negligible.
Test Cases
import sys, io
from itertools import permutations
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
import sys as _sys
from subprocess import Popen, PIPE
# simple inline execution via import trick
# assume solution is in solve()
# we redefine here for testing convenience
def solve():
n = int(input().strip())
s = input().strip()
colors = ['R', 'G', 'B']
idx = {'R':0,'G':1,'B':2}
cnt = [[0]*3 for _ in range(3)]
for i,ch in enumerate(s):
cnt[i%3][idx[ch]] += 1
best = -1
bestp = None
for p in permutations(range(3)):
keep = sum(cnt[r][p[r]] for r in range(3))
if keep > best:
best = keep
bestp = p
res = []
for i in range(n):
res.append(colors[bestp[i%3]])
print(n-best)
print("".join(res))
old_stdout = sys.stdout
sys.stdout = io.StringIO()
solve()
out = sys.stdout.getvalue().strip()
sys.stdout = old_stdout
return out
# provided sample
assert run("3\nBRB\n") == "1\nGRB"
# all same
assert run("4\nRRRR\n") in ["2\nRGRG", "2\nGRGR"]
# already optimal
assert run("3\nRGB\n") == "0\nRGB"
# alternating
assert run("6\nRGRGRG\n") in ["0\nRGRGRG"]
# mixed
assert run("5\nBRRGB\n")
| Test input | Expected output | What it validates |
|---|---|---|
| 3 BRB | 1 GRB | basic correctness |
| 4 RRRR | 2 RGRG or GRGR | symmetry and multiple optimal answers |
| 3 RGB | 0 RGB | already valid configuration |
| 6 RGRGRG | 0 RGRGRG | periodic alignment case |
Edge Cases
A key edge case is when the string is already perfectly periodic. For input RGBRGB, each residue class already matches a distinct color. The algorithm counts full matches for one permutation, resulting in zero changes. Since we only compare permutations, we naturally preserve the original structure.
Another edge case is uniform input like RRRRRR. Here each residue class has identical distributions, so multiple permutations yield the same score. The algorithm picks any permutation consistently, producing a valid alternating pattern such as RGRGBB or another equivalent optimal arrangement depending on implementation order. The correctness does not depend on which tie is chosen.
A third case is when one residue class dominates a particular color heavily but not exclusively. The algorithm still correctly assigns that color to the corresponding modulo class because the permutation scoring captures the global maximum over all assignments rather than making local decisions.