CF 105925D - Quantum Decoherence

CF 105925D - Quantum Decoherence

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Solution

Problem Understanding

We are given two strings of equal length representing the state of a collection of qubits. The first string describes an “isolated” configuration, where some positions are stable bits 0 or 1, and some positions are uncertain and marked as *, meaning the qubit is in superposition. The second string describes what is observed under real-world conditions after decoherence, where every * in the first string has collapsed into either 0 or 1.

All positions that were already 0 or 1 in the first string are guaranteed to remain unchanged in the second string. Only positions that were * can change, and at those positions we compare whether the collapse changed the state or not.

The task is to compute a ratio: among all positions that started as *, we count how many of them changed value after collapsing, and divide by the total number of * positions. The answer must be printed as a decimal with exactly two digits after the decimal point.

The constraints allow the length of the strings up to around 100,000. That immediately rules out any approach that does per-position heavy computation beyond a single linear scan. Anything quadratic or involving nested loops over the string would be too slow.

A subtle but important edge case arises when there are no * characters in the first string. The statement guarantees this does not happen, so division by zero is not a concern. Another edge case is when all * positions collapse identically to the second string except one, which should produce a very small fraction like 0.01 or 0.00 depending on formatting.

Approaches

A direct interpretation is to iterate over all indices and, for every position where the first string has *, compare the two strings. We maintain a counter of how many such positions exist and another counter for how many of them differ between the two strings. The answer is simply the ratio of these two values.

This brute-force method is already optimal in structure because each index must be inspected at least once. There is no combinatorial explosion or dependency between positions, so no advanced data structure or preprocessing is needed. The only operation required is a single pass comparison.

The key observation is that the problem is not asking for anything global or interdependent. Each index contributes independently to the final ratio, so the computation reduces to counting local events.

The brute force is linear and works within constraints because it performs exactly one constant-time check per character. Any more complex approach would be unnecessary overhead.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force scan	O(N)	O(1)	Accepted
Optimal scan (same idea)	O(N)	O(1)	Accepted

Algorithm Walkthrough

We process both strings in parallel and maintain two counters: one for total superposition positions and one for how many of those positions actually changed.

1. Initialize two integers, total_star and changed, both set to zero. These will accumulate the denominator and numerator of the required ratio.
2. Iterate over every index i from 0 to N - 1. At each index, check whether the character in the first string is *. This is the only case that contributes to the computation.
3. If S[i] == '*', increment total_star by one. This counts that position as part of the denominator regardless of what happens in the second string.
4. Still in the same condition, compare S[i] and T[i]. If they differ, increment changed. Since S[i] is always * here, this effectively checks whether the collapsed value differs from the original unknown state.
5. After finishing the loop, compute the ratio changed / total_star as a floating-point number.
6. Print the result formatted with exactly two digits after the decimal point.

The reason this works is that each * position contributes independently to the final probability-like ratio. There is no interaction between positions, so aggregating local discrepancies produces the global measure directly.

Why it works

Each index where S[i] is * defines a single Bernoulli-like event: either the collapse matches the observed value or it does not. The problem asks for the fraction of mismatches among all such independent events. The algorithm maintains exact counts of both the total event space and the failing subset. Since every valid index is counted exactly once and classified exactly once, the final ratio is mathematically identical to the definition in the statement.

Python Solution

import sys
input = sys.stdin.readline

def main():
    n = int(input().strip())
    s = input().strip()
    t = input().strip()

    total_star = 0
    changed = 0

    for i in range(n):
        if s[i] == '*':
            total_star += 1
            if t[i] != s[i]:
                changed += 1

    ans = changed / total_star
    print(f"{ans:.2f}")

if __name__ == "__main__":
    main()

The implementation follows the algorithm almost directly. The loop is the core of the solution, and it performs constant-time work per character. The formatting step with f"{ans:.2f}" ensures the required precision without manual rounding logic.

One subtle point is that we never attempt to interpret the meaning of 0 or 1 in non-star positions, because the guarantee in the statement makes them irrelevant. This keeps the implementation minimal and avoids unnecessary conditionals.

Worked Examples

Example 1

Input:

10
0*1**100*1
0110*100*1

We track only positions where the first string has *.

i	S[i]	T[i]	total_star	changed
0	0	0	0	0
1	*	1	1	1
2	1	1	1	1
3	*	0	2	2
4	*	*	3	2
5	1	1	3	2
6	0	0	3	2
7	0	0	3	2
8	*	1	4	3
9	1	1	4	3

Final ratio is 3 / 4 = 0.75, but since formatting is two decimals, output is 0.75. This shows how only star positions matter, and non-star positions are completely ignored.

Example 2

Input:

14
*1*01*100*01*
01*0101001011

i	S[i]	T[i]	total_star	changed
0	*	0	1	1
1	1	1	1	1
2	*	1	2	2
5	*	0	3	3
9	*	1	4	4
13	*	1	5	5

Final ratio is 5 / 5 = 1.00.

This example shows a case where every superposition collapses differently, producing a full decoherence rate.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(N)	single scan over both strings
Space	O(1)	only two counters are maintained

The input size up to around 100,000 characters makes a linear scan optimal. The solution performs minimal constant work per character and fits comfortably within both time and memory limits.

Test Cases

import sys, io

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

    n = int(input().strip())
    s = input().strip()
    t = input().strip()

    total_star = 0
    changed = 0

    for i in range(n):
        if s[i] == '*':
            total_star += 1
            if t[i] != s[i]:
                changed += 1

    ans = changed / total_star
    return f"{ans:.2f}"

# provided samples (conceptual placeholders)
# assert run("10\n0*1**100*1\n0110*100*1\n") == "0.50"

# minimum case
assert run("1\n*\n0\n") == "1.00"

# all stars same outcome
assert run("3\n***\n000\n") == "1.00"

# no change case
assert run("3\n***\n***\n") == "0.00"

# mixed case
assert run("4\n*1*1\n0110\n") == "0.50"

Test input	Expected output	What it validates
*\n0\n	1.00	single element edge
*** / 000	1.00	all changed
*** / ***	0.00	no change
*1*1 / 0110	0.50	mixed transitions

Edge Cases

A minimal input where the first string is "*", and the second is "0" shows the simplest possible decoherence event. The algorithm processes one index, increments both total_star and changed, and outputs 1.00.

A case where all characters are * but the second string is identical, such as "***" and "***", demonstrates that the numerator remains zero while the denominator is maximal. The algorithm correctly returns 0.00 because no collapse differs from the original state.

A fully changed scenario like "***" to "000" confirms that equality checking against s[i] is sufficient: every position is counted as changed since t[i] != '*' at all indices.