My first instinct was a recursive parser and I actually think that's fine here since you can do it in a single left-to-right pass with no backtracking.
Recognize that the recursive definition implies a specific structural property: every valid string has exactly one more '0' than '1', and every prefix has at least as many '0's as '1's. Use a single pass with a counter, incrementing for '0' and decrementing for '1', ensuring the counter never goes negative and ends at 1. This yields an O(n) time and O(1) space solution.
Pro tip: After presenting the counter approach, briefly mention that a recursive descent parser would also work but uses O(n) stack space, and that the counter method is essentially a space-optimized version of that parser. This shows you understand the trade-off between clarity and efficiency.
Parse the definition: a valid string is either '0' or '1' followed by two valid strings. This implies a binary tree structure where '1' is an internal node with two children, and '0' is a leaf.
From the tree interpretation, deduce that the number of leaves ('0's) is always one more than the number of internal nodes ('1's), and that in any prefix, the number of '0's seen so far must be at least the number of '1's plus one? Actually, careful: For a valid string, at any point, the number of '0's minus the number of '1's must be positive? Let's derive: Start with balance = 0. For each '0', balance increases by 1; for each '1', balance decreases by 1. For a valid string, the balance must never drop below 1? Wait, check '0': balance after '0' is 1. For '10100': process: '1' -> balance -1? That would be negative. So maybe we need to think differently. Actually, the recursive definition: '0' is valid. '1' followed by two valid strings is valid. So a valid string is a full binary tree with leaves as '0' and internal nodes as '1'. The preorder traversal of such a tree gives the string. In preorder, we visit node, then left subtree, then right subtree. For a leaf, we output '0'. For an internal node, we output '1' and then recursively output left and right. So the string is the preorder traversal. Now, in any prefix of a preorder traversal of a full binary tree, the number of internal nodes seen so far must be less than the number of leaves seen so far? Actually, consider the stack of nodes. When we see a '1', we push a node that expects two children. When we see a '0', we complete a leaf. The number of pending internal nodes (nodes that have not yet had both children processed) must be such that we don't have more internal nodes than available leaves? Let's simulate: Start with a stack of size 1 representing the root that needs to be processed. Actually, a simpler way: The string is valid iff it can be reduced by repeatedly replacing '100' with '0'? No, that's not right. Let's think of the grammar: S -> '0' | '1' S S. This is a context-free grammar. The language is the set of strings with equal? Actually, it's the set of strings where the number of '0's is one more than the number of '1's, and every prefix has at least as many '0's as '1's? Let's test: '0' -> 0:1, 1:0, prefix '0' has 0>=1? No, 0 is not >=1. So that condition is wrong. Let's test '10100': counts: 0:3, 1:2, so 0 = 1+1. Prefixes: '1': 0:0,1:1 -> 0<1. So the condition 'every prefix has at least as many 0s as 1s' fails. So what is the correct condition? Let's derive from the grammar. The grammar is S -> '0' | '1' S S. This is the grammar for full binary trees in preorder. There is a known characterization: A string is valid iff it has length 2k+1 for some k, and the number of '0's is k+1 and '1's is k, and for every prefix, the number of '1's is less than the number of '0's? Let's check '10100': prefix '1': 1s=1, 0s=0 -> 1>0, so fails. So that's not it. Actually, the condition is that if we process the string and maintain a counter that starts at 0, and for each '0' we increment, for each '1' we decrement, then the counter must never go negative? Let's test '0': start 0, '0' -> 1, never negative, end 1. '10100': start 0, '1' -> -1, negative! So fails. So that condition is also wrong. Wait, maybe we need to think of it as a stack of expected children. Let's simulate a parser: We can parse by maintaining a stack of integers representing the number of children still needed for each open node. Initially, we need to parse one valid string, so we can think of a stack with one element: 1 (meaning we need to complete one valid string). When we see '0', we complete a valid string, so we pop one from the stack. When we see '1', we start a new internal node that needs two children, so we push 2 onto the stack? But careful: The '1' itself is part of the string, and then we need to parse two valid strings. So after seeing '1', we need to parse two valid strings. So we can push 2 onto the stack. But we also need to account for the fact that the '1' itself is not a complete valid string; it's the start of one. So the stack represents the number of valid strings we still need to complete. Initially, we need 1 valid string. When we see '0', we complete one valid string, so we decrement the top of the stack? Actually, a simpler way: Use a counter that starts at 1 (we need one valid string). For each character: if '0', we complete one valid string, so we decrement the counter by 1? But then we also need to consider that when we complete a valid string, it might be a child of a '1', which then reduces the need for another child. Let's think recursively: The string is valid if it can be reduced to '0' by repeatedly replacing '1' followed by two valid strings with a single '0'? That's not a simple reduction. Alternatively, we can use a stack of counts. Let's derive the condition from the grammar. The grammar is S -> '0' | '1' S S. This is a deterministic context-free language? Actually, it's not deterministic because we don't know when to reduce? But we can parse it with a simple counter if we think of it as a tree. Consider the string as a preorder traversal. In a full binary tree, the number of leaves is one more than the number of internal nodes. So total length = 2*internal + 1. So length must be odd. Also, if we assign +1 to '0' and -1 to '1', the sum is (leaves) - (internal) = 1. So sum = 1. But that's not sufficient. For example, '11000' has sum: 0s=3, 1s=2, sum=1, but is it valid? Let's check: '11000' -> can we parse? Try: '1' then two valid strings. The first valid string could be '1' followed by two valid strings? But we only have '100' left? Actually, '11000': after first '1', we need two valid strings. The remaining is '1000'. Can we split '1000' into two valid strings? Possible splits: first valid string could be '10'? But '10' is not valid because '1' must be followed by two valid strings, so '10' is incomplete. First valid string could be '100'? Is '100' valid? '100': '1' followed by '0' and '0' -> yes, '1' + '0' + '0' is valid. So '100' is valid. Then the second valid string would be '0'? But we have '1000' -> if first is '100', second is '0', that works. So '11000' = '1' + '100' + '0' -> valid. So '11000' is valid. So sum=1 is necessary but not sufficient? Actually, '11000' is valid. What about '10100'? We already saw it's valid. What about '10010'? Sum: 0s=3, 1s=2, sum=1. Is it valid? Try: '1' + two valid strings. Remaining '0010'. Can we split? First valid string could be '0', then second would be '010' which is not valid (starts with '0' but length
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