Time Complexity

Question

What will be TC here?

Ans given $O(n^{2})$ , while I am getting $O(n)$

Comments

 Shubhgupta  yes corrected ,thank you

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ok, but 1st case j is running less time right?

that means in 1st case actual running time will be $1+2+3+....+n=\text{almost equal to}\frac{n(n+1)}{2}=O(n^{2})$

And in 2nd case actual running time will be $n+n+.....+n=n(1+1+1+1....+1)=n^{2}=\text{exact equal to}O(n^{2})$

right now?

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thats true,  srestha mam $x^{2} > \frac{x^{2} + x}{2}$

Answer 1

yes it will be o(n²)

when i=0   j=0 so zero times 

when i=1  j=1,0 but only execute for j=1 so one time

when i=2 j=2,1,0 but j will run for j=2,1 so 2 times

.........................................similarly when i=n-1 j will execute n-1 times

so total =0+1+2+3+...................n-1

=O(n²)

 

Answer 2

in this problem  find how many times for loop execute

 

i=1, j=1

i=2,j=2
,,,,,,,,

i=n-1, j=n-1

Total Complexity = 0 + 1 + 2............. + n-1 = n(n-1)/2 = O(n^2)  is correct answer