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Ans. int duplicateArray[] = { 1, 2, 2, 3, 4, 5, 6, 8, 9}
Set unique = new HashSet();
for (int i = 0; i < duplicateArray.length; i) {
if (unique.contains(duplicateArray[i])) {
System.out.println(duplicateArray[i]);
} else {
unique.add(duplicateArray[i]);
}
}
Complexity O(n) = nHashSet contains and add has O(n) = 1
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