Time complexity of this code (from Cracking the Coding Interview “Group Anagrams”)
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0
The author claims the following code to be O(n * log(n)). This seems to imply that the Comparator is doing constant-time work, but there's a sort embedded in the Comparator's compare() function. Doesn't this increase the time complexity?
class AnagramComparator implements Comparator<String> {
public String sortChars(String s) {
char content = s.toCharArray();
Array.sort(content);
return new String(content);
}
public int compare(String s1, String s2) {
return sortChars(s1).compareTo(sortChars(s2));
}
}
Arrays.sort(array, new AnagramComparator());
big-o
asked Dec 28 '18 at 3:03
UnknownBeef
306
add a comment |
0
The author claims the following code to be O(n * log(n)). This seems to imply that the Comparator is doing constant-time work, but there's a sort embedded in the Comparator's compare() function. Doesn't this increase the time complexity?
class AnagramComparator implements Comparator<String> {
public String sortChars(String s) {
char content = s.toCharArray();
Array.sort(content);
return new String(content);
}
public int compare(String s1, String s2) {
return sortChars(s1).compareTo(sortChars(s2));
}
}
Arrays.sort(array, new AnagramComparator());
big-o
asked Dec 28 '18 at 3:03
UnknownBeef
306
1
The embedded sort call's time complexity depends on the lengths of the strings, not the number of strings (n). The author has probably assumed this to be small.
– meowgoesthedog
Dec 28 '18 at 9:23
So, if we call the size of the biggest strings, and assumes << n, we can disregard the embedded sort's time complexity ofO(n * s * log(s))becauseO(n * log(n))is the dominant term?
– UnknownBeef
Dec 28 '18 at 18:41
Are you referring to line 4? If so, since we are sorting an array of sizes, isn't that anO(s * log(s))operation each time?
– UnknownBeef
Dec 29 '18 at 6:07
I meant thatO(n * log(n) * s * log(s)) = O(n * s * log(n)), notO(n * s * log(s)).
– meowgoesthedog
Dec 29 '18 at 9:37
add a comment |
0
0
0
The author claims the following code to be O(n * log(n)). This seems to imply that the Comparator is doing constant-time work, but there's a sort embedded in the Comparator's compare() function. Doesn't this increase the time complexity?
class AnagramComparator implements Comparator<String> {
public String sortChars(String s) {
char content = s.toCharArray();
Array.sort(content);
return new String(content);
}
public int compare(String s1, String s2) {
return sortChars(s1).compareTo(sortChars(s2));
}
}
Arrays.sort(array, new AnagramComparator());
big-o
asked Dec 28 '18 at 3:03
UnknownBeef
306
The author claims the following code to be O(n * log(n)). This seems to imply that the Comparator is doing constant-time work, but there's a sort embedded in the Comparator's compare() function. Doesn't this increase the time complexity?
class AnagramComparator implements Comparator<String> {
public String sortChars(String s) {
char content = s.toCharArray();
Array.sort(content);
return new String(content);
}
public int compare(String s1, String s2) {
return sortChars(s1).compareTo(sortChars(s2));
}
}
Arrays.sort(array, new AnagramComparator());
big-o
big-o
asked Dec 28 '18 at 3:03
UnknownBeef
306
asked Dec 28 '18 at 3:03
UnknownBeef
306
asked Dec 28 '18 at 3:03
UnknownBeef
306
asked Dec 28 '18 at 3:03
UnknownBeef
306
asked Dec 28 '18 at 3:03
UnknownBeef
306
306
1
The embedded sort call's time complexity depends on the lengths of the strings, not the number of strings (n). The author has probably assumed this to be small.
– meowgoesthedog
Dec 28 '18 at 9:23
So, if we call the size of the biggest strings, and assumes << n, we can disregard the embedded sort's time complexity ofO(n * s * log(s))becauseO(n * log(n))is the dominant term?
– UnknownBeef
Dec 28 '18 at 18:41
Are you referring to line 4? If so, since we are sorting an array of sizes, isn't that anO(s * log(s))operation each time?
– UnknownBeef
Dec 29 '18 at 6:07
I meant thatO(n * log(n) * s * log(s)) = O(n * s * log(n)), notO(n * s * log(s)).
– meowgoesthedog
Dec 29 '18 at 9:37
add a comment |
1
The embedded sort call's time complexity depends on the lengths of the strings, not the number of strings (n). The author has probably assumed this to be small.
– meowgoesthedog
Dec 28 '18 at 9:23
So, if we call the size of the biggest strings, and assumes << n, we can disregard the embedded sort's time complexity ofO(n * s * log(s))becauseO(n * log(n))is the dominant term?
– UnknownBeef
Dec 28 '18 at 18:41
Are you referring to line 4? If so, since we are sorting an array of sizes, isn't that anO(s * log(s))operation each time?
– UnknownBeef
Dec 29 '18 at 6:07
I meant thatO(n * log(n) * s * log(s)) = O(n * s * log(n)), notO(n * s * log(s)).
– meowgoesthedog
Dec 29 '18 at 9:37
1
1
The embedded sort call's time complexity depends on the lengths of the strings, not the number of strings (
n). The author has probably assumed this to be small.– meowgoesthedog
Dec 28 '18 at 9:23
The embedded sort call's time complexity depends on the lengths of the strings, not the number of strings (
n). The author has probably assumed this to be small.– meowgoesthedog
Dec 28 '18 at 9:23
So, if we call the size of the biggest string
s, and assume s << n, we can disregard the embedded sort's time complexity of O(n * s * log(s)) because O(n * log(n)) is the dominant term?– UnknownBeef
Dec 28 '18 at 18:41
So, if we call the size of the biggest string
s, and assume s << n, we can disregard the embedded sort's time complexity of O(n * s * log(s)) because O(n * log(n)) is the dominant term?– UnknownBeef
Dec 28 '18 at 18:41
Are you referring to line 4? If so, since we are sorting an array of size
s, isn't that an O(s * log(s)) operation each time?– UnknownBeef
Dec 29 '18 at 6:07
Are you referring to line 4? If so, since we are sorting an array of size
s, isn't that an O(s * log(s)) operation each time?– UnknownBeef
Dec 29 '18 at 6:07
I meant that
O(n * log(n) * s * log(s)) = O(n * s * log(n)), not O(n * s * log(s)).– meowgoesthedog
Dec 29 '18 at 9:37
I meant that
O(n * log(n) * s * log(s)) = O(n * s * log(n)), not O(n * s * log(s)).– meowgoesthedog
Dec 29 '18 at 9:37
add a comment |
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1
The embedded sort call's time complexity depends on the lengths of the strings, not the number of strings (
n). The author has probably assumed this to be small.– meowgoesthedog
Dec 28 '18 at 9:23
So, if we call the size of the biggest string
s, and assumes << n, we can disregard the embedded sort's time complexity ofO(n * s * log(s))becauseO(n * log(n))is the dominant term?– UnknownBeef
Dec 28 '18 at 18:41
Are you referring to line 4? If so, since we are sorting an array of size
s, isn't that anO(s * log(s))operation each time?– UnknownBeef
Dec 29 '18 at 6:07
I meant that
O(n * log(n) * s * log(s)) = O(n * s * log(n)), notO(n * s * log(s)).– meowgoesthedog
Dec 29 '18 at 9:37