Answer :
Answer:
1) no of computations are :
sum =0 - (1 time )
i=0 ( 1 times)
i<n ( n times )
++sum(n times)
++i (n times)
Total = 3n+2
O(n) is the big O notation
2) no of computations are :
sum =0 - (1 time )
i=0 ( 1 times)
i<n ( n times )
++sum(n^2 times)
++i (n times)
j=0 ( n times)
j<n (n^2 times)
++j (n^2 times)
Total = 3n^2 + 3n+2
O(n^2) is the big O notation as we considered the maximum possible computation
3) no of computations are :
sum =0 - (1 time )
i=0 ( 1 times)
i<n ( n times )
++sum(n^3 times)
++i (n times)
j=0 ( n times)
j<n (n^3 times)
++j (n^3 times)
Total = 3n^3 + 3n+2
O(n^3) is the big O notation as we considered the maximum possible computation
4) no of computations are :
sum =0 - (1 time )
i=0 ( 1 times)
i<n ( n times )
++i (n times)
In the "j" th loop
j=0 ( n times)
j<i (this executes for max of n^2 times when i=n-1)
Similarly ++j (n^2 times when i=n-1)
Hence ++sum(n^2 times)
Total = 3n^2+ 3n+2
O(n^2) is the big O notation as we considered the maximum possible computation
6) no of computations are :
sum =0 - (1 time )
i=0 ( 1 times)
i<n ( n times )
++i (n times)
In the "j" th loop
j=0 ( n times)
j<i*i(this executes for max of n^3 times when i=n-1)
Similarly ++j (n^3 times when i=n-1)
In the "k" th loop
k=0 ( max of n^3 times when j=n^3)
k<j(max of n^4 times when j=n^3)
Similarly ++k(max of n^4 times when j=n^3)
Finally ++sum ( n^4 times when j=n^3)
Total = 3n^4+ 3n^3+3n+2
O(n^4) is the big O notation as we considered the maximum possible computation
5) no of computations are :
sum =0 - (1 time )
i=0 ( 1 times)
i<n ( n times )
++i (n times)
In the "j" th loop
j=0 ( n times)
j<i*i(this executes for max of n^3 times when i=n-1)
Similarly ++j (n^3 times when i=n-1)
In the "k" th loop
k=0 ( max of n^3 times when j=n^3)
k<j(max of n^4 times when j=n^3)
Similarly ++k(max of n^4 times when j=n^3)
Finally ++sum ( n^4 times when j=n^3)
Total = 3n^4+ 3n^3+3n+2
O(n^4) is the big O notation as we considered the maximum possible computation