Complexity of algorithm (asymptotic) -

can confirm me complexity of algorithm o(n^2)?

a = 0 b = 0 c = n while (b <= c) {     (j = b; j<=c; j++)     {         = j * j -2 * j + 3     }     b = b + 3     c = c + 2 } 

the inner loop executes c - b + 1 times. each execution of inner loop body a = j * j -2 * j + 3 takes constant (bounded) time (assuming we're dealing fixed-width integer types, otherwise depend on multiplication algorithm used [and addition, that's hard implement in way multiplication faster]), execution of body of outer loop o(d) (Θ(d) even), d = c - b + 1.

the updates of variables controlling outer loop

b = b + 3 c = c + 2 

decrease difference c - b 1 in each execution of outer loop's body, hence outer loop executed n+1 times, , have total of o(n²), since

 n                   n+1  ∑ (n+2k - (3k) +1) = ∑ j = (n+1)(n+2)/2 k=0                  j=1 

it Θ(n²), unless compiler optimises , sets variables final values directly.

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answer original question typo:

the inner loop

for (j = b; j==c; j++) 

will execute either once - when b == c or not @ all, body of outer loop o(1). updates in outer loop

b = b + 3 c = c + 2 

mean difference c-b decreases 1 each time loop body executed, so

b = 0 c = n while (b <= c) 

will execute n+1 times - total: o(n).