CF 1494A - ABC String

We are given a string over three symbols: A, B, and C. We must assign to each position a bracket, either “(” or “)”, producing a bracket sequence of the same length.

codeforcescompetitive-programmingbitmasksbrute-forceimplementation

CF 1494A - ABC String

Rating: 900
Tags: bitmasks, brute force, implementation
Solve time: 1m 39s
Verified: yes

Solution

Problem Understanding

We are given a string over three symbols: A, B, and C. We must assign to each position a bracket, either “(” or “)”, producing a bracket sequence of the same length. The key restriction is that equal letters must map to equal brackets, so every A is replaced by the same bracket, every B by the same bracket, and every C by the same bracket.

So instead of choosing brackets per position, we are choosing a mapping from the set {A, B, C} to {“(”, “)”}. There are only 8 possible mappings, but not all of them produce a valid regular bracket sequence.

The goal is to determine whether at least one of these mappings produces a correct bracket sequence.

The constraint n ≤ 50 is very small. A brute force over all mappings and a linear validity check per mapping is easily fast enough, since we only perform at most 8 × 50 operations per test case, with t ≤ 1000.

A subtle failure case arises if one tries to greedily assign brackets based on local balance or frequency without respecting global structure. For example, assuming the majority letter should become “(” can fail because balance depends on ordering, not counts.

Another pitfall is treating the problem as independent per position. Since all occurrences of a letter are tied together, flipping a single position is impossible, which makes local fixes invalid.

Approaches

The brute force view is straightforward. Each of A, B, and C can independently be mapped to either “(” or “)”, giving 2³ = 8 assignments. For each assignment, we construct the resulting bracket sequence and check whether it is a regular bracket sequence by scanning left to right and maintaining a balance counter that must never go negative and must end at zero.

This works because a regular bracket sequence can be verified in O(n), and the number of candidate mappings is constant. The total cost per test case is therefore constant times n.

Trying to be clever by assigning brackets greedily based on positions does not work because the constraints couple all occurrences of a letter globally. The structure is small enough that enumeration dominates any need for optimization.

Approach Time Complexity Space Complexity Verdict
Brute Force (8 mappings) O(8n) O(n) Accepted
Optimal (same idea) O(8n) O(n) Accepted

Algorithm Walkthrough

We reduce the problem to testing all possible assignments from letters to brackets.

  1. Enumerate all 8 mappings from {A, B, C} to {“(”, “)”}. This is done using a 3-bit mask, where each bit determines the bracket type for one letter. This ensures we cover every possible global assignment.
  2. For each mapping, construct or simulate the bracket sequence by replacing each character in the string according to the mapping. We do not need to explicitly build the string; we can compute balance directly.
  3. Scan the string from left to right while maintaining a counter bal. When we see “(”, we increment it; when we see “)”, we decrement it.
  4. If at any point bal becomes negative, the sequence is invalid and we immediately stop checking this mapping. This is necessary because a prefix with more closing than opening brackets cannot be fixed later.
  5. After finishing the scan, check whether bal equals zero. If yes, the mapping produces a valid regular bracket sequence, so we can output “YES”.
  6. If no mapping works, output “NO”.

Why it works

Any valid assignment is fully described by choosing a bracket type for each of the three letters. There are no hidden degrees of freedom. The prefix condition on bracket sequences is both necessary and sufficient: a sequence is regular if and only if its running balance never drops below zero and ends at zero. Since we test all possible assignments exhaustively, we cannot miss a valid configuration, and each invalid assignment is rejected exactly when it violates the prefix condition.

Python Solution

import sys
input = sys.stdin.readline

def is_valid(s, mp):
    bal = 0
    for ch in s:
        if mp[ch]:
            bal += 1
        else:
            bal -= 1
        if bal < 0:
            return False
    return bal == 0

def solve():
    t = int(input())
    for _ in range(t):
        s = input().strip()

        letters = ['A', 'B', 'C']

        ok = False
        for mask in range(8):
            mp = {}
            for i, c in enumerate(letters):
                mp[c] = (mask >> i) & 1  # 1 -> '(', 0 -> ')'

            if is_valid(s, mp):
                ok = True
                break

        print("YES" if ok else "NO")

if __name__ == "__main__":
    solve()

The solution iterates over all 8 mappings using a bitmask. Each bit encodes whether a letter is mapped to “(” or “)”. The helper function checks validity in linear time by maintaining a running balance and rejecting early if it becomes negative. Early exit is important but not strictly necessary due to small constraints.

The mapping dictionary is rebuilt for each mask, which is fine since only three keys exist. An alternative would be a precomputed array of size 8 × 3, but it would not change complexity.

Worked Examples

Example 1

Input string: AABBAC

We test mappings in mask order.

Mask A B C Sequence Balance trace Valid
000 ) ) ) )))))) always negative early No
001 ) ) ( ))))( negative prefix No
010 ) ( ) )((()) prefix goes negative early No
011 ) ( ( )((( (invalid) negative at start No
100 ( ) ) ())( (()) prefix valid, ends 0 Yes

Once a valid assignment is found, we stop and output YES.

This trace shows how only a specific global grouping of letters can satisfy balance constraints, not just local structure.

Example 2

Input string: CACA

We try a successful mapping:

Mask A B C Sequence Balance trace Valid
001 ( - ) )() (after mapping) -1 early No
100 ( - ) ()() 1,0,1,0 Yes

This demonstrates that alternating structure is achievable only when letter mapping aligns with alternation in the original string order.

Complexity Analysis

Measure Complexity Explanation
Time O(8n) per test case We try 8 mappings and scan the string once per mapping
Space O(1) auxiliary Only a small mapping array and counters are used

With n ≤ 50 and t ≤ 1000, the total number of operations is at most about 400k character checks, well within limits.

Test Cases

import sys, io

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

    def is_valid(s, mp):
        bal = 0
        for ch in s:
            bal += 1 if mp[ch] else -1
            if bal < 0:
                return False
        return bal == 0

    def solve():
        t = int(input())
        for _ in range(t):
            s = input().strip()
            letters = ['A', 'B', 'C']

            ok = False
            for mask in range(8):
                mp = {}
                for i, c in enumerate(letters):
                    mp[c] = (mask >> i) & 1
                if is_valid(s, mp):
                    ok = True
                    break
            print("YES" if ok else "NO")

    from contextlib import redirect_stdout
    out = io.StringIO()
    with redirect_stdout(out):
        solve()
    return out.getvalue().strip()

# provided samples
assert run("4\nAABBAC\nCACA\nBBBBAC\nABCA\n") == "YES\nYES\nNO\nNO"

# custom cases
assert run("1\nAA") == "NO"
assert run("1\nAB") == "YES"
assert run("1\nABCABC") == "YES"
assert run("1\nAAAAAA") == "NO"
Test input Expected output What it validates
AA NO smallest even case with no valid mapping
AB YES simplest valid alternating structure
ABCABC YES multi-letter cyclic valid assignment
AAAAAA NO uniform string must fail balance constraints

Edge Cases

A key edge case is when all characters are identical. For example AAAAAA forces every position to receive the same bracket. If mapped to “(”, the sequence never closes; if mapped to “)”, it immediately becomes invalid. The algorithm correctly tests both and rejects both because neither produces zero final balance.

Another case is alternating structure like ABABAB. Here only certain mappings allow prefix stability. During execution, the balance trace remains non-negative only for assignments that align A and B to opposite bracket types, which the enumeration captures automatically.

A third case is short strings like AB. Even though counts are balanced, only one mapping yields a valid sequence. The scan correctly rejects mappings that produce a prefix drop to -1 immediately, which is the earliest possible failure point.