Given the following Java code: public static boolean f(int [] arr) { for (int i = 0; i < arr.length - 1; i++) { for (int j = i + 1; j < arr.length; j++) { if (arr[i] < arr[j]) // *HERE* return false;} return true;} (i) Find the number of times the comparison marked by *HERE* will be evaluated for each input: {1, 2} {10, 20, 30} {30, 20, 10} {-4, 7, 1} (ii) For an array of size ????, what is the big-oh runtime of this code in the worst case?

Answer:See explaination Explanation:{1, 2}Runs only once for 1 < 2{10, 20, 30}Runs 1 for 10 < 20Runs 1 for 10 < 30Runs 1 for 20 < 30Total 3 times{30, 20, 10}Runs 1 for 30 < 20Runs 1 for 30 < 10Runs 1 for 20 < 10Total 3 times{-4, 7, 1}Runs 1 for -4 < 7Runs 1 for -4 < 1Runs 1 for -7 < 1​​​​​​​Total 3 timesGeneralizing it will run for n*(n-1)/2 so for 2 element it will run for 2*1/2 = 1For 3 element it will run for 3*2/2 = 3 times(ii) (3 points) For an array of size , what is the big-oh runtime of this code in the worst case?Time complexity in Worst case is O(n^2). Reason being it uses two for loop where first loop runs for n time and the second loop runs for n*n time in worst case hence its O(n^2)