So now we'll take a look at the time complexity of the insertion sort algorithm. For that, let us take an array like this. Now you would see that I would again begin from this point and compare four with five. I would find that I am doing one comparison here and swapping. So if I say that TN is my number of comparisons, I am doing exactly one comparison here when I'm comparing four with five. So let me write one here and I would swap them. So in my next iteration my array will look something like this.

Now I would begin at this point and compare three with five. So I will swap them since three is less than five and I would once again compare three with four, I would see that three is less than four so I would again swap them. So how many times did I compare three with these? I compared two times. So let me write two here. Now in the next iteration my array looks like this.

I start from this point and compare all the elements which are to the left of two. I would see that five is exchanged with two. Similarly, since two is less than four I'll swap them and since two is also less than three, I'll swap them again. So what is the total number of comparisons I'm doing? Two is getting compared to five, then four, then three. So I'm doing a total number of three comparisons. Now my array looks like this.

When I move to the last of my elements I would be again needing to compare it with all the elements to its left. So one would get compared to five and since one is less than five we would swap them. So this is one comparison. Now one will get compared to four and since one is less than four I would swap them. So that was my second comparison. Now my one will get compared to three and since one is less than three, again I will swap them. This was my third comparison and lastly my one will get compared to two and since one is less than two I'll swap them again. This was my fourth comparison. So in this particular iteration I did a total of four comparisons. So now if you observe carefully for any given array of length n here my n is equal to five. Since there are five elements in my array, my first iteration will have one comparison, my second iteration will have two number of comparisons, my third will have three and fourth will have four. So now for an array of length n, these number of

comparisons can be generalized and written in terms of n. Like this here I have four. So this basically reduces to n minus one because my n was five here, this three can be written as n minus two. Similarly I'll go on from n minus one, n minus two, n minus three and my number of comparisons will keep on decreasing till I reach a point where I'm doing two and one number of comparisons. Now let me just find the summation of all these total number of comparisons to calculate my time complexity. So here t of n will be written as one plus two plus three till n minus one. So you have seen this expression before. This is nothing but the sum of all natural numbers starting from one and going till n minus one. So you have already seen that this summation is equal to n minus one into n divided by two. This can be alternatively written as half of N square minus N. Now again, looking at this expression, you would see that I can ignore the half here since it is a constant. And I can also ignore the

N here since N square will always be greater than N. So basically my T of N gets reduced to theta of N square. So in conclusion, we can say that the time complexity of insertion sort is of the order of n square.

Video Recap

time complexity of the insertion sort algorithm
The speaker uses an example array of 5 elements to illustrate the number of comparisons needed in each iteration of the algorithm
The total number of comparisons for an array of length n can be generalized and written as $n - 1, n - 2, n - 3$ and so on until 1
The time complexity of the algorithm can be calculated by finding the summation of all these comparisons
The final time complexity of insertion sort is of the order of n square $(O (n^{2}))$

Erratum: At timestamp 04:58, the tome complexity should be O $(n^{2})$ and not theta $(n^{2})$

So, insertion sort again comes out to have a time complexity of O(N $^{2}$ ). However, it follows O(N) in the best case. So, if the array is almost sorted, i.e. it is nearer to the best case than the worst case, the insertion sort will have a time complexity somewhere between O(N $^{2}$ ) and O(N). It will be nearer to O(N) if the array is almost sorted, and nearer to O(N $^{2}$ ) if the array is arranged in the opposite order of our preference. For example, consider an array where all elements are in descending order, and you want to sort it in ascending order. In such a case, it will be closer to O(N $^{2}$ ). Watch the video below to understand this better.

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That was interesting, wasn’t it? It’s always a good practice to verify what you learn with something in real time Here as well insertion sort comes out to have time complexity of O(N $^{2}$ ) in the worst case, and O(N) in the best case. Remember that you only need to do comparisons; no swaps are done when the cards are already in ascending order.