Question

5. What is the Big Oh method m2? public static void m2(int[] arr, int n) for (int í = 1; í <= n- 1; i++) pM2(arr [i], arr, 0, i - 1); // end m2 private static void pM2(int entry, int[l arr, int begin, int end) int i- end; for(; (i 〉= begin) && (entry 〈 arr [i]); i--) arr [1 + 1] = arr L1] arr[i + 1] - entry; return // end pM2

Answer 1

5)

Answer: Big Oh : O(n^2)

explanation:

in method m2

for loop runs from i=1 to n-1 // means n-1 times

and each time pM2 is called

in method pM2

for loop runs i=end to begin in worst case //means (end-begin)+1 times

//total complexity is

//total number of times

1+2+3+4+.......+n = n(n+1)/2 times

hence .....O(n^2)

4)

Answer: Big Oh : O(n^2)

explanation:

in method m1

for loop runs from i=0 to n-1 // means n-1 times

and each time pM1 is called //every time last will be n-1

in method pM1

for loop runs i=first+1 to last in worst case //means (last-(first+1))+1 times

//total complexity is

//total number of times

n-1+n-2+n-3+.......+3+2+1 = n(n-1)/2 times

hence .....O(n^2)