Analysis of Selection Sort

Consider the following selection sort algorithm:

selection_sort(int s[], int n)
{
int i, j; // counters
int min; // index of minimum

for(i=0; i<n; i++) {
min = i;
for(j=i+1; j<n; j++)
if(s[j] < s[min]) min = j;
swap(&s[i], &s[min]);
}
}


performing the algorithmic analysis:

• for loop with i index operates O(n) times
• second for loop operates O(i) times, within the loop that runs n times, for an algorithmic complexity given below.

${\displaystyle O(i):\sum _{i=1}^{n}i={\dfrac {n(n+1)}{2}}\sim O(n^{2})}$

Analysis of Insertion Sort

Consider the following insertion sort algorithm:

for(i=1; i<n; i++) {
j = i;
while((j>0)&&(s[j]<s[j-1])) {
swap(&s[j],&s[j-1]);
j=j-1;
}
}


This algorithm requires focusing on the worst case scenario.

The outer for loop runs n times, and is O(n).

The inner for loop has two conditions that must be met. We can assume j>0 always, and focus on when the other condition would be true - when would a random index of S be less than its neighbor? Worst case assumption is, it will always be smaller, and so the while loop will run every single time. This gives us a while loop where j iterates from i to 0.

As before, the outer loop is O(n) and the inner loop is O(i), and therefore runs ${\displaystyle {\frac {n(n+1)}{2}}}$ times, or ${\displaystyle O(n^{2})}$.