Question

What is the big O value for the following algorithm? Why is it that value?

algorithm A (val array <ptr to int>)
     1 n = 0
     2 loop ( n < array size ) 
    1 min = n;
    2 m = n;
    3 loop ( m < array size)
      1 if (array[m] < array[min])
            1  min = m;
    4 swap(array[min],array[n]);
     3 n = n + 1

I answered O(n^2) am I correct? As to how I arrived to this conclusion, the inner loops executes the n times where n = the array size and the outer loop executes n times where n is the array size n*n = n^2

Answer 1

Yes! you are correct!

This is selection sort algorithm.
Its Θ(n^2) to be more precise.

Edit : Why is it that value?

You take the first element. Compare it with all the other elements to find minimum in the array and place it in the first place. Iterations : n.
You take the second element. Compare it with rest of the array and find minimum in that part (second minimum in whole array) and place it in the second place. Iterations : n-1.
Continuing in this way for last element, Iterations : 1.

Total = n+n-1+ … +1 = n(n+1)/2. That is O(n^2).