We will determine the runtime of the following recursive We will determine the runtime of the following recursive

algorithm.

This algorithm takes two $n-$digit integers a and $b($as strings$)$

and determines if

algorithm.

This algorithm takes two n$-$digit integers a and $b($as strings$)$

and determines if $T(n)=a$

LTRec$(a1a2...an,b1b2...bn)$:

$Ifn==1$:

$ifa1<mtext\; id="EditorElement\_1889778"></mtext><mi\; id="EditorElement\_1889779">b</mi><mn\; id="EditorElement\_1889780">1</mn>$:

return TRUE

else

return FALSE

$ifa1<mtext\; id="EditorElement\_1889849"></mtext><mi\; id="EditorElement\_1889850">b</mi><mn\; id="EditorElement\_1889851">1</mn>$:

return TRUE

else $ifb1<mtext\; id="EditorElement\_1889884"></mtext><mi\; id="EditorElement\_1889885">a</mi><mn\; id="EditorElement\_1889886">1</mn>$:

return FALSE

else $ifa1==b1$:

return LTRec$(a2....an,b2....bn)$

$We$ can write a recurrence for the runtime $of$ this algorithm

$asT(n)={?}_{(}{)}_{{?}_{?}}$Blank $1_{{?}_{?}}{?}_{{?}_{?}}{?}_{{?}_{?}}$

$\{Blank1$ Options:

$T(n-1)+c,T(1)=d$

$T(n-1)+cno$ base case

$T(n-1)+T(n-1)+c,T(1)=d$

$T(\frac{n}{2})+c*n^2,T(1)=d$

$T(n-1)+cn,T(1)=1\}$

From this, $we$ can conclude a tight Big$-O$ bound for this

algorithm's runtime is $\_\_\_\_\_$Blank $2\_\_\_\_$

$\{$Blank $2$ Options:

O$($n$)$

O$($log n$)$

O$(1)$

O$($n$^2)$

O$($n log n$)\}$