2

I have the following simple algorithm to find duplicate characters in a string:

for i = 1 -> n
    for j = i + 1 -> n
        if A[i] == A[j] return true
return false

Why is the running time of this algorithm $\mathcal{O}(n^2)$? If the first iteration is $n$ steps then, $n-1, n-2,n-3,..,1$ it seems to me that adding all these would never be $n^2$ or am I wrong?

Raphael
  • 73,212
  • 30
  • 182
  • 400
Armen Babakanian
  • 123
  • 4

3 Answers3

3

When we sum all numbers we get $n(n-1)/2$, which is in the class of functions $O(n^2)$. Make sure that you understand what big O notation is about.

Another way of seeing that this sum is $O(n^2)$ since each summand is at most $n$ and there are at most $n$ summands, so the sum is at most $n^2$.

Yuval Filmus
  • 280,205
  • 27
  • 317
  • 514
1

Don't confuse the $n$ in $O(n)$ with the $n$ in your algorithm! If you consider the number of times the line if A[i] == A[j] return true is evaluated, this will in the worst case indeed be $(n-1) + (n-2) + (n-3) + \dots + 1$. There is a closed formula for this sum:

$\sum _{i=1}^n i = \frac{n(n+1)}{2} = \frac{n^2 + n }{2}$, (see Wiki)

so asymptotically you get the runtime of $O(n^2)$.

Jasper
  • 194
  • 1
  • 9
1

The inner loop is executed $\sum_{1\le i\le n}\sum_{i\lt j\le n} 1$ times. $$\sum_{1\le i\le n}\sum_{i\lt j\le n} 1=\sum_{1\le i\lt j\le n} 1=\sum_{1\le j\le n}\sum_{1\le i\le j-1}1=\sum_{1\le j\le n}(j-1)$$ $$=0+\dots+n-1=(n-1)n/2.$$

This is $\cal O(n^2)$.

mrk
  • 3,748
  • 23
  • 35