Suppose that you want to count the number of duplicates in an unsorted array of n elements. A duplicate is an element that appears multiple times; if a given element appears x times, x - 1 of them are considered duplicates. For example, consider the following array:{10, 6, 2, 5, 6, 6, 8, 10, 5}It includes four duplicates: one extra 10, two extra 6s, and one extra 5.Below are two algorithms for counting duplicates in an array of integers:Algorithm A:public static int numDuplicatesA(int[] arr) { int numDups = 0; for (int i = 0; i < arr. length - 1; i++) { for (int j = i + 1; j < arr. length; j++) { if (arr[j] == arr[i]) { numDups++; break; } } } return numDups;}Algorithm B:public static int numDuplicatesB(int[] arr) { Sort. mergesort(arr); int numDups = 0; for (int i = 1; i < arr. length; i++) { if (arr[i] == arr[i - 1]) { numDups+}} return numDups;}What is the worst-case time efficiency of algorithm A in terms of the length n of the array?What is the worst-case time efficiency of algorithm B?Make use of big-O notation, and explain briefly how you came up with the big-O expressions that you use.