As shown in the picture, we can use it to find an array's maximum and minimum elements.Assume that A(n) represents the algorithm's complexity function, which tests the number of element comparisons.1) Express and find A(n) for the best-case and worst-case.2) What about the average-case?Procedure MAXMIN(B, m, max, min) /* max for the maximum and min for the minimum of B(1:m) */begininteger i, m;max := B[1];min := B[1];for i := 2 to m doif B[i] > max thenmax := B[i]else if B[i] < min thenmin := B[