Construct a Turing machine that can determine if an input bit string has more $1\text{'}$s than $0\text{'}$s$.$

If the string contains more $1\text{'}$s than $0\text{'}$s$,$ output a $\text{'}{Y}^{\text{'}}$ and halt. If the string contains more $0\text{'}$s than $1\text{'}$s output an $\text{'}N\text{'}$ and halt. If the string contains an equal number of $1\text{'}$s and $0\text{'}$s$,$ output an $\text{'}{E}^{\text{'}}$ and halt.

You can let $B$ represent the blank symbol.

You can assume that the Turing machine starts at the beginning of the string. The machine can halt anywhere along the tape.

You can assume that the string is always at least one bit in length $($no empty strings$).$

Your answer should define the state transition table for the Turing machine. That is$,f(s_i,x_i)=(s_j,x_j,d)$ where $s_i$ is the current state, $s_j$ is the next state, $x_i$ is the input scanned symbol, $x_j$ is the output symbol written on tape, and $d$ indicates a direction of either $L$ indicating moving left, $R$ indicating moving right, or $H$ indicating halting$).$