Deterministic Pushdown Automata that accepts #a = #b

Question

I am trying to create a DPDA that accepts words from the following Language:

$$ L = \{wx \; | \;w \in \{a,b\}^*, \#a = \#b \} $$

My intuition was to initially put an $x$ on the stack and then write an unlimited amount of $a$ or $b$, while both operations put an $B$ or $A$ on the stack. Then, in the second last state we can write an limited amount of $a$ and $b$ for popping an $A$ or $B$ from the stack.This ensures that the amount of $a$ and $b$ are equal before we can read the initial $x$ from the stack. As a last step, we read $x$ as the last element of the stack. However, building this determinstic seems too complex. Any hints?

Answer 1

You have to keep track of $\#a - \#b$ at each point of the input. This number can become negative, so say the number of excess $a$ is represented by that many $A$ on the stack, and the number of excess $b$ by $B$s. If at the end of the string there is no excess (empty stack), accept.

Details of the construction are left to the gentle reader.

Answer 2

Reading a $a$ and a $b$ should "compensate" each other. For example, reading a $a$ could push a $A$ or remove a $B$ depending on the stack, and conversely when reading a $b$.

You can then accept when the stack is empty (or contains the initial symbol).