I am pretty sure that the answer for this is n^3, because the 3rd for loop is executed n times and the middle is executed n+4 times and the top one is executed n times. I understand how to use sigma notation for the bottom and top for loop because is uses n, but what do I do different for the middle one because it uses j instead of n

``````for (int j = 4; j < n; ++j) {
cin >> val; }
for (int i = 0; i < j; ++i) {
b = b * val;}
for (int k = 0; k < n; ++k)
c = b + c;}
}``````

if these are nested loops, you summation should be:

sum(j=4..n){sum(i=0..j){sum(k=0..n)ijk}}
=> sum(j=4..n){sum(i=0..j)ij[n(n+1)/2]} (because sum(k=0..n)k = [n(n+1)/2] = n^2}

and then you can solve further using the sum operations.

## All 2 Replies

if these are nested loops, you summation should be:

sum(j=4..n){sum(i=0..j){sum(k=0..n)ijk}}
=> sum(j=4..n){sum(i=0..j)ij[n(n+1)/2]} (because sum(k=0..n)k = [n(n+1)/2] = n^2}

and then you can solve further using the sum operations.

``````//O(n)
for (int j = 4; j < n; ++j) {
cin >> val;
}

//O(n)
for (int i = 0; i < j; ++i) {
b = b * val;
}

//misformatted?
for (int k = 0; k < n; ++k)
c = b + c;}
}``````
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