Question
a)
public int findPopular(int[] a)
{
if (a == null || a.length == 0)
return 0;
Arrays.sort(a);
int previous = a[0];
int popular = a[0];
int count = 1;
int maxCount = 1;
for (int i = 1; i < a.length; i++)
{
if (a[i] == previous)
count++;
else
{
if (count > maxCount)
{
popular = a[i-1];
maxCount = count;
}
previous = a[i];
count = 1;
}
}
return count > maxCount ? a[a.length-1] : popular;
}
b)
public int findPopular(int[] a)
{
if (a == null || a.length == 0)
return 0;
Arrays.sort(a);
int previous = a[0];
int popular = a[0];
int count = 1;
int maxCount = 1;
for (int i = 1; i < a.length; i++)
{
if (a[i] == previous)
count++;
else
{
if (count > maxCount)
{
popular = a[i];
maxCount = count;
}
previous = a[i];
count = 1;
}
}
return count > maxCount ? a[a.length-1] : popular;
}
c)
public int findPopular(int[] a)
{
if (a == null || a.length == 0)
return 0;
Arrays.sort(a);
int previous = a[0];
int popular = a[0];
int count = 1;
int maxCount = 1;
for (int i = 1; i < a.length; i++)
{
if (a[i+1] == previous)
count++;
else
{
if (count > maxCount)
{
popular = a[i-1];
maxCount = count;
}
previous = a[i];
count = 1;
}
}
return count > maxCount ? a[a.length-1] : popular;
}
d) None of the mentioneda.
a
b.
b
c.
c
d.
d
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Q. Select the code snippet which prints the element with maximum frequency.
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