unSortMap.put (7, "e"); unSortMap.put (50, "m"); Map<Integer, String> treeMap = new TreeMap <Integer, String> (new Comparator<Integer> () { @Override public int compare (Integer o1, Integer o2)…

The correct answer is C. e z b d. Option C is correct because the custom Comparator uses o2.compareTo(o1) instead of the natural o1.compareTo(o2), which reverses the sort order - TreeMap stores entries by descending key rather than ascending. When iterated, the values are printed from the highest key to the…

Question

unSortMap.put (7, "e");
unSortMap.put (50, "m");
Map<Integer, String> treeMap = new TreeMap <Integer, String> (new Comparator<Integer> () {
@Override public int compare (Integer o1, Integer o2) {return
o2.compareTo(o1); } });
treeMap.putAll (unSortMap);
for (Map.Entry<Integer, String> entry : treeMap.entrySet ()) {
System.out.print (entry.getValue () + " ");
}
}
What is the result?

Options

AA compilation error occurs.
Bd b e j
Ce z b d
Dz b d e j

How the community answered

(34 responses)

A
3% (1)
B
6% (2)
C
85% (29)
D
6% (2)

Explanation

Option C is correct because the custom Comparator uses o2.compareTo(o1) instead of the natural o1.compareTo(o2), which reverses the sort order - TreeMap stores entries by descending key rather than ascending. When iterated, the values are printed from the highest key to the lowest, yielding e z b d.

Why the distractors are wrong:

A is wrong because the code is syntactically valid Java - anonymous Comparator classes with generics compile without issue.
B (d b e j) reflects ascending key order (natural TreeMap behavior without the custom Comparator), not the reversed order the code actually applies.
D (z b d e j) contains five values, suggesting a misread of the map's contents or confusion about which entries are included.

Memory tip: When you see o2.compareTo(o1) in a Comparator, mentally swap the letters - o2 before o1 means the second argument wins, which flips to descending order. Think: "backwards arguments, backwards sort."

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