You are given a string `expr` representing an algebraic expression made of lowercase letters, the operators `+` (addition) and `*` (multiplication), and parentheses. Multiplication binds tighter than addition, and parentheses group sub-expressions. Each letter behaves like a variable, and concatenation denotes a product (so `a*b` is written `ab` after expansion). Return an equivalent expression that uses ONLY `+`, obtained by fully distributing every product over every sum (i.e., expand into a flat sum of monomials). Within each resulting monomial, keep the letters in the left-to-right order in which they were multiplied (do NOT sort letters). Keep the monomials in the natural left-to-right traversal order produced by distribution, and join them with `+`. For example: - `(a+b)*(c+d)` expands to `ac+ad+bc+bd`. - `a+b*c*(d+e+f)+k+m*(g+h)+i` expands to `a+bcd+bce+bcf+k+mg+mh+i`. The input is always a well-formed expression (balanced parentheses, no unary operators, no empty groups). The original problem first restricts `+` to appear only inside parentheses and asks for O(1) extra space; the general follow-up allows `+` anywhere (outside parentheses too). This solution handles the general case, which subsumes the restricted one, using a stack-based recursive parse (O(n) auxiliary space for the parse stack and the output list of monomials).