Part $3$: Printing a Path

Part $3\mathrm{.}1$: Node findPredPath$($int i$)$

Node findPredPath$($int i$)$ should find the path to the predecessor of index $i($exactly the same as

Node findPred$($int i$))$ and return the predecessor Node. However, in this version you should store the path

somehow. You may look at ods. SkiplistSSet. java for inspiration $-$ it uses an array of Nodes called stack to save

part of the path on a call to add $($T x$).$ This may not be sufficient for your needs, but it is a good starting place.

Part $3\mathrm{.}2$: String toStringWithPath$()$

This method is currently exactly the same as String toStringWithNodes$().$ You are to modify it so that its out$-$

put highlights the path found on the previous call to Node findPredPath$($int i$).$ To highlight a path replace all

instances of$,>>>>,$ or $>>>>>>>>$along the path. You do NOT have to indicate downward steps. This method need work ONCE per call to

findPredPath$().$ If you call toStringWithPath$()$ twice in a row without a second call to findPredPath$(),$ it

does not need to highlight the path on the second call $($although it can if you like$).$ That means that you can destroy

the path that was stored as you highlight it if you wish.

IMPORTANT: Do not change how anything else is printed. The tests will check if the returned String is an

exact match to the correct version. If it is not exact, you get $0$ marks. For example, if you accidentally add a space

somewhere, you will get $0.$

An example of the expected output is provided below. It also highlight another function, toStringWithLength,

that provides the length of each next pointer $($which you may use for debugging$).$

Note that findPredPath$(4)$ finds a path to the predecessor of Node $4,$ which is Node $3($containing $\text{'}$ c $\text{'}).$