Need help understanding how the second loop is executed, I know for the second loop you would multiply by the first loop. First I count the number of instructions and then apply big-oh notation. This is my approach: would j=2*n be executed 2n times, j>=1 n+1, and j– n times? New to big-oh notation and can’t seem to find a list of ‘for loops’ examples anywhere, would love the feedback and information on where to practice more.
for(i=5;i<=n/2;i++) 1+n/2+n = 1+3n/2
for(j=2*n; j>=1; j–) (1+3n/2)(2n+n+1+n)
acc -= j + acc++; \1+1+1 (2n+n+1+n)