Consider sorting n numbers stored in array A[1:n] by first finding the smallest element of A[1:n] and exchanging it with the element in A[1]. Then find the smallest element of A[2:n], and exchange it with A[2]. Then find the smallest element of A[3:n], and exchange it with A[3]. Continue in this manner for the first n-1 elements of A. Write pseudocode for this algorithm, which is known as selection sort. What loop invariant does this algorithm maintain? Why does it need to run for only the first n-1 elements, rather than for all n elements? Give the worst-case running time of selection sort in big theta notation. Is the best-case running time any better?

it's not a programming question.

Two words (spelt in uppercase) are similar if and only if all letters in the 1st word are in the 2nd word and all letters in the 2nd word are in the 1st word; including the same number of duplications. For example, ABBA and BABA are similar; however, BABA and BAAA are not similar. To check two words which are similar, we can use two integer arrays, each has 26 elements (as we have 26 letters), you can assume that 1st array element stores the number of letter 'A', 2nd array element stores the number of letter 'B', and so on. What you need to do now is to check whether the two integer arrays are equal to decide whether the two words are similar. We propose following UML for this question: O TwoWords - String word1 -String word2 O Exam_Q4 +TwoWords0 - static Scanner input +TwoWords(String word1, String word2) O+TwoWords(TwoWords tw) O+ String getWord10 O+ static void main(String args) O+ String getWord20 O+void setinfo(String word1, String word2) O -intj countinfo (String word) O+boolean…

Given a sorted array of size N and an integer K, find the position at which K is present in the array using binary search. Example 1: Input: N = 5 arr[] = {1 2 3 4 5} K = 4 Output: 3 Explanation: 4 appears at index 3.