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Solution
def checkStrings(binary, arr):
# Write your code here
ans = []
n = len(binary)
// ... rest of solution available after purchaseExplanation
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To solve this problem, it's essential to deeply understand the operation allowed: selecting any **subsequence** of the binary string and sorting it. The operation can be done any number of times, and it is not restricted to contiguous substrings. This is powerful because a subsequence can skip characters and still maintain the original relative order of the characters not selected.
The main challenge lies in determining whether, starting from a given binary string, it's possible to perform this operation any number of times to reach a target string (after replacing the wildcard characters '?').
Let’s break down the process and the thought direction you should follow:
First, think about what this "subsequence sort" really allows us to do. You can select any number of positions and sort those bits. This means you can effectively move ‘0’s to the left and ‘1’s to the right as long as you carefully choose your subsequences. Importantly, through multiple such operations, the overall binary string can be converted into a state where **all the ‘0’s are to the left and all the ‘1’s to the right**. This is because you can keep sorting combinations of indices that progressively bubble the 0s leftward.
This leads to the key insight: the only constraint that exists is that you **cannot move a ‘1’ to the left of a ‘0’ in the original binary** beyond a certain point. In other words, even though the operation is flexible, it’s not omnipotent. It can't move characters past each other arbitrarily; it only works through sorting, and sorting preserves relative order among same elements but reorders different elements.
So you can interpret the full set of operations as allowing you to "compress" the 0s to the left as much as possible, and "push" the 1s to the right, but without violating the inherent ordering potential of the original string. This defines a **partial order**: no ‘1’ in the target string should appear before a ‘1’ in the original that cannot be moved to that position via sorting a subsequence.
The question then becomes: given a string from `arr` (after replacing '?' with all possible binary combinations), can this string be achieved by sorting subsequences of the original `binary` string?
Here’s a strategic path to answer that:
1. For each string in `arr`, generate all possible versions by replacing '?' with '0' and '1'. This might seem computationally expensive, but it’s manageable since the string lengths are usually moderate. However, you don't need to literally generate all permutations if you can analyze character-by-character possibilities.
2. For each of these concrete candidate strings, determine if it can be reached from the original binary string. The clever way to think about this is through **greedy matching**. You can simulate the sorting effect by trying to align the original string to the target using a two-pointer approach.
3. Start from the left and try to "match" the required number of 0s and 1s in the target string from the available 0s and 1s in the binary string, keeping in mind that you can’t move a ‘1’ past a ‘0’ if the relative order in the original doesn’t allow it. In essence, a ‘1’ can’t leap over a ‘0’ unless there's a sort operation that can make that happen—so you have to track how many 0s and 1s are before a certain index.
4. Consider keeping a prefix sum or position list of all 0s and 1s in the original binary string. When you attempt to construct a target string, try to greedily consume characters from the original, in the order they appear. If at any point the target requires a 1 but only 0s are left in the matching positions (or vice versa), then that version of the target string is not reachable.
5. If at least one of the '?' combinations leads to a reachable target string, return "YES" for that element in the array. If none do, return "NO".
6. Edge cases to think about: completely wildcard strings (e.g., "????"), strings with no wildcards but already different from the original in an irreversible way, and cases where the original binary string is already sorted.
To summarize:
- Understand the operation as a powerful but not unlimited way of reordering the string.
- Realize that the ultimate state the original string can be transformed into is one where all 0s are on the left and all 1s are on the right.
- Compare each target string (after wildcard replacement) to see if it can be derived from the original via valid subsequence sorts.
- Use greedy matching to simulate this transformation, respecting the original order constraints.
- Return "YES" if at least one feasible replacement exists that makes the transformation possible, otherwise return "NO".
This approach ensures you're reasoning through the constraints without relying on brute force or code-level details, and helps frame the problem in terms of transformation potential within allowed operations.
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