explosives in cyberspace

approach

part 1 actually builds the decompressed string: it scans the input, and at each (AxB) marker it copies the next A characters B times and jumps past them, so markers inside a repeated block are left as plain text. the answer is the length of the result.

part 2 never builds the string. the input is parsed recursively into a tree, where a marker becomes a node holding its length and repeat count, its children are whatever is found inside the span it covers, and a plain character is a leaf worth 1. a depth-first walk then adds up the lengths, multiplying each node's total by its repeat count.

the trick

only the length is needed, so nested markers become a tree of multipliers instead of an expanded string.

def dfs(node):
    if not node.children:
        return node.times * node.length
    s = 0
    for child in node.children:
        if type(child) is Node:
            r = dfs(child)
            s += node.times * r
        else:
            s += node.times * child
    return s