For each of the algorithms uniquel and unique2, which solves the element uniqueness problem, perform an experimental study on the same computer to determine: 1) What are these two algorithms doing here? Are they doing the same thing or not? 2) Which algorithm is faster and why? 3) Can you come up with an even faster algorithm? If so, what is it ***Returns true if there are no duplicate lements in the array.*/ ublic static boolean uniquel (int[] data) { int n = data.length; for (int i=0; j

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For each of the algorithms unique 1 and unique2, which solves the
element uniqueness problem, perform an experimental study on
the same computer to determine:
1) What are these two algorithms doing here? Are they doing the
same thing or not?
2) Which algorithm is faster and why?
3) Can you come up with an even faster algorithm? If so, what is it?
/** Returns true if there are no duplicate
elements in the array.*/
public static boolean uniquel (int[] data) {
int n = data.length;
for (int j 0; j<n-1; j++)
for (int k-j+1; k<n; k++)
if (data[j] == data[k])
return false;
return true;
/** Returns true if there are no duplicate
elements in the array.*/
public static boolean unique2(int[] data) {
int n = data.length;
int[] temp = Arrays.copyOf(data, n);
Arrays.sort(temp);
for (int j 0; j<n-1; j++)
if (temp[j] temp[j+1])
return false;
return true;
[12]
Transcribed Image Text:For each of the algorithms unique 1 and unique2, which solves the element uniqueness problem, perform an experimental study on the same computer to determine: 1) What are these two algorithms doing here? Are they doing the same thing or not? 2) Which algorithm is faster and why? 3) Can you come up with an even faster algorithm? If so, what is it? /** Returns true if there are no duplicate elements in the array.*/ public static boolean uniquel (int[] data) { int n = data.length; for (int j 0; j<n-1; j++) for (int k-j+1; k<n; k++) if (data[j] == data[k]) return false; return true; /** Returns true if there are no duplicate elements in the array.*/ public static boolean unique2(int[] data) { int n = data.length; int[] temp = Arrays.copyOf(data, n); Arrays.sort(temp); for (int j 0; j<n-1; j++) if (temp[j] temp[j+1]) return false; return true; [12]
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