CF 1296E1 - String Coloring (easy version)
We have a string of lowercase letters. Every position must be assigned one of two colors, 0 or 1. After choosing the colors, we are allowed to swap neighboring characters only when their colors are different. We may perform such swaps as many times as we want.
CF 1296E1 - String Coloring (easy version)
Rating: 1800
Tags: constructive algorithms, dp, graphs, greedy, sortings
Solve time: 2m 3s
Verified: yes
Solution
Problem Understanding
We have a string of lowercase letters. Every position must be assigned one of two colors, 0 or 1.
After choosing the colors, we are allowed to swap neighboring characters only when their colors are different. We may perform such swaps as many times as we want. The goal is to determine whether some coloring allows the string to be rearranged into nondecreasing alphabetical order.
The output is not the sequence of swaps. We only need to print a valid coloring if one exists.
The key restriction is hidden inside the swap rule. Two adjacent characters of the same color can never be swapped directly. More importantly, their relative order can never change at all. Every swap exchanges characters of different colors, so characters belonging to the same color class always preserve their original order.
The length is at most 200. This is small enough that even an $O(n^2)$ or $O(n^3)$ solution is completely safe. The small constraint is a hint that we can explicitly build and analyze a graph on positions.
A common mistake is to think only about the characters themselves and ignore positions. The coloring is assigned to indices, not letter values. Two equal letters may receive different colors.
Consider the string:
ba
The sorted string is "ab". The two characters must change order. Since characters of the same color cannot change their relative order, they must receive different colors. A valid answer is:
YES
01
Another subtle case is:
cba
The sorted string is "abc". Every pair of positions is inverted. Position 1 conflicts with positions 2 and 3, and position 2 conflicts with position 3. This forms a triangle of conflicts, which cannot be colored with only two colors. The correct answer is:
NO
A careless greedy coloring based only on character values may incorrectly claim a solution exists.
One more edge case is when many letters are equal:
aaaa
The string is already sorted. There are no constraints at all, so every coloring works. The output can be:
YES
0000
Treating equal letters as inversions would incorrectly introduce unnecessary restrictions.
Approaches
A brute-force approach is to try every possible coloring. Since each position has two choices, there are $2^n$ colorings. For each coloring we could check whether the sorted string is reachable. Even for $n=200$, $2^{200}$ is astronomically large, so this is impossible.
To find something better, we need to understand what the swap operation actually allows.
Suppose two positions $i<j$ have the same color. Their relative order can never change because every allowed swap occurs between different colors. As a result, within each color class, the subsequence of characters must remain in its original order forever.
Now look at an inversion: positions $i<j$ with $s_i>s_j$.
In the final sorted string, these two characters must appear in the opposite order. Since same-colored positions can never reverse their relative order, the endpoints of every inversion must receive different colors.
This observation completely characterizes the problem. Create a graph whose vertices are string positions. Connect two positions if they form an inversion. Every edge requires its endpoints to have different colors.
The problem becomes: can the inversion graph be colored with two colors?
That is exactly the definition of a bipartite graph.
If the graph is bipartite, the bipartition gives a valid coloring. If it is not bipartite, some odd cycle forces contradictory color requirements and no solution exists.
The easy version is designed so that building the entire inversion graph is affordable. There are only $200$ positions, so checking all pairs takes $O(n^2)$.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | $O(2^n)$ | $O(n)$ | Too slow |
| Optimal | $O(n^2)$ | $O(n^2)$ | Accepted |
Algorithm Walkthrough
- Create a graph with one vertex for each string position.
- For every pair of indices $i<j$, check whether $s_i>s_j$.
- If an inversion exists, add an undirected edge between $i$ and $j$.
This edge means the two positions must receive different colors, otherwise their order could never be corrected.
4. Run a standard bipartite check using DFS or BFS.
5. Whenever we traverse an edge, assign the neighboring vertex the opposite color.
6. If we ever encounter an edge whose endpoints already have the same color, the graph is not bipartite. Print "NO".
7. If all connected components are successfully colored, print "YES" and output the assigned colors as a binary string.
Why it works
Every inversion must be resolved in the final sorted order. Two positions that share a color preserve their relative order forever, so the endpoints of every inversion must have different colors. The inversion graph records exactly these requirements.
If the graph is bipartite, assigning colors according to the bipartition satisfies every inversion constraint. Then each color class forms a subsequence that remains internally ordered. Since every inversion pair lies across the two color classes, all problematic pairs can be reordered through allowed swaps, and the string can be transformed into sorted order.
If the graph is not bipartite, some set of inversion constraints requires a position to be both colors simultaneously. No valid coloring exists. Thus the algorithm returns "YES" exactly when a solution exists.
Python Solution
import sys
input = sys.stdin.readline
n = int(input())
s = input().strip()
g = [[] for _ in range(n)]
for i in range(n):
for j in range(i + 1, n):
if s[i] > s[j]:
g[i].append(j)
g[j].append(i)
color = [-1] * n
def dfs(v, c):
color[v] = c
for to in g[v]:
if color[to] == -1:
if not dfs(to, c ^ 1):
return False
elif color[to] == c:
return False
return True
for i in range(n):
if color[i] == -1:
if not dfs(i, 0):
print("NO")
sys.exit()
print("YES")
print("".join(map(str, color)))
The first nested loop constructs the inversion graph. Every pair of positions is examined exactly once. Whenever the left character is larger than the right character, we create an edge because those positions must receive opposite colors.
The color array stores the bipartite coloring. A value of -1 means the vertex has not been visited yet.
The DFS assigns alternating colors along graph edges. If an already-colored neighbor has the same color as the current vertex, we have found a contradiction. At that point the graph is not bipartite and the answer is "NO".
The graph may have multiple connected components, so the outer loop starts a new DFS whenever an unvisited vertex is found.
One subtle detail is the use of c ^ 1. Since colors are only 0 and 1, XOR with one flips between them cleanly.
Another detail is that equal letters do not create edges. Only strict inversions matter. Positions containing the same character may keep their relative order in the sorted string, so no constraint is needed.
Worked Examples
Example 1
Input:
9
abacbecfd
Inversions create the following relevant constraints:
| Pair | Characters | Edge? |
|---|---|---|
| (1,3) | b,a | Yes |
| (2,5) | a,e | No |
| (4,7) | b,f | No |
| (5,7) | e,f | No |
| (6,8) | c,d | No |
Running DFS gives:
| Position | Character | Color |
|---|---|---|
| 0 | a | 0 |
| 1 | b | 0 |
| 2 | a | 1 |
| 3 | c | 0 |
| 4 | b | 1 |
| 5 | e | 0 |
| 6 | c | 1 |
| 7 | f | 0 |
| 8 | d | 1 |
Output:
YES
001010101
This example shows a graph that is bipartite. Every inversion edge connects opposite colors.
Example 2
Input:
3
cba
The inversion graph contains all three possible edges.
| Pair | Characters | Edge? |
|---|---|---|
| (0,1) | c,b | Yes |
| (0,2) | c,a | Yes |
| (1,2) | b,a | Yes |
The graph is:
0
|\
| \
| \
1---2
Trying to color it:
| Vertex | Assigned Color |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 1 |
The edge between 1 and 2 now connects equal colors, producing a contradiction.
Output:
NO
This trace demonstrates the odd-cycle obstruction that makes bipartite coloring impossible.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | $O(n^2)$ | Building the inversion graph dominates the running time |
| Space | $O(n^2)$ | In the worst case every pair forms an inversion edge |
With $n \le 200$, the graph contains at most $200 \cdot 199 / 2 = 19{,}900$ edges. Both time and memory usage are comfortably within the limits.
Test Cases
# helper: run solution on input string, return output string
import sys
import io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
input = sys.stdin.readline
n = int(input())
s = input().strip()
g = [[] for _ in range(n)]
for i in range(n):
for j in range(i + 1, n):
if s[i] > s[j]:
g[i].append(j)
g[j].append(i)
color = [-1] * n
def dfs(v, c):
color[v] = c
for to in g[v]:
if color[to] == -1:
if not dfs(to, c ^ 1):
return False
elif color[to] == c:
return False
return True
for i in range(n):
if color[i] == -1:
if not dfs(i, 0):
return "NO\n"
return "YES\n" + "".join(map(str, color)) + "\n"
# provided sample
out = run("9\nabacbecfd\n")
assert out.startswith("YES\n")
# minimum size
assert run("1\na\n") == "YES\n0\n", "single character"
# already sorted
assert run("4\naaaa\n") == "YES\n0000\n", "all equal"
# impossible triangle
assert run("3\ncba\n") == "NO\n", "complete graph K3"
# simple inversion
out = run("2\nba\n")
assert out.startswith("YES\n"), "one inversion"
# maximum-style stress pattern
s = "".join(chr(ord('z') - (i % 26)) for i in range(200))
out = run(f"200\n{s}\n")
assert out.startswith("YES\n") or out == "NO\n"
| Test input | Expected output | What it validates |
|---|---|---|
1 / a |
YES |
Minimum size |
4 / aaaa |
YES |
No inversion edges |
3 / cba |
NO |
Odd cycle in inversion graph |
2 / ba |
YES |
Single inversion must use opposite colors |
| Length 200 stress string | Valid result | Performance near limit |
Edge Cases
Single Character
Input:
1
a
No pair of positions exists, so the inversion graph has one isolated vertex.
The DFS colors it with 0.
Output:
YES
0
The algorithm naturally handles isolated vertices because every connected component is processed independently.
All Characters Equal
Input:
4
aaaa
Every comparison fails the condition s[i] > s[j], so no edges are added.
The graph consists of four isolated vertices.
The algorithm outputs:
YES
0000
Equal letters do not create inversions, which is exactly what we want.
Strictly Decreasing String
Input:
3
cba
Every pair is an inversion.
The graph becomes a triangle:
0-1
|/
2
A triangle is not bipartite. During DFS, vertices 1 and 2 receive the same color, but an edge exists between them. The conflict is detected immediately.
Output:
NO
This is the smallest example where inversion constraints become impossible to satisfy.
Repeated Letters with Real Inversions
Input:
4
baab
Inversion edges are:
(0,1), (0,2)
The graph is a simple path centered at vertex 0.
A valid coloring is:
0110
The repeated 'a' characters introduce no edge between themselves because equal letters are not inversions. The algorithm distinguishes correctly between equal values and genuine ordering conflicts.