How would I compute the time complexity of this algorithm? The outer for loop runs n times. The inside for loop runs n-1, n-2, n-3, n-4, ... nn times after each iteration.
/*
* isUniqueBrute - This algorithm uses the brute force approach by comparing each character
* in the string with another. Only a single comparison is made per pair of characters.
*/
bool isUniqueBrute(string str)
{
char buffer;
for (int i = 0; i < str.length(); i++)
{
for (int j = i+1; j < str.length(); j++)
{
if (str[i] == str[j])
return false;
}
}
return true;
}
You perform the calculation by doing the math:
outer loop runs n times: O(n)
inner loop runs n-1, n-2, n-3, ... 0 times.
So you break down the iterations:
1 + 1 + 1 + 1 + ... n times = outer loop
n-1 + n-2 + n-3 + ... n times = inner loop
Rearranging:
1+n-1 = n
1+n-2 = n-1
1+n-3 = n-2
... n times
Add those up, and you get the sum of all numbers from 1..n. I'm sure you know the formula by now, but:
n * (n+1) / 2
Which, if you expand it, includes n² as the dominant element. So this function is O(n²).
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.