Question

Consider this data sequence: "fish bird reptile reptile bird bird bird mammal fish". let's define a singleton to be a data element that is not repeated immediately before or after itself in the sequence. so, here there are four singletons (the first appearance of "fish", the first appearance of "bird", "mammal", and the second appearance of "fish"). write some code that uses a loop to read a sequence of words, terminated by the "xxxxx". the code assigns to the variable n the number of singletons that were read. (for example in the above data sequence it would assign 4 to n. assume that n has already been declared . but not initialized . assume that there will be at least one word before the terminating "xxxxx".

Answer 1

To determine the number of singletons in a sequence of words ending with 'xxxxx', you can write code containing a loop that reads each word, compares it with the previous word, and increments a counter for each singleton detected until the 'xxxxx' ending is encountered.

Step-by-step explanation:

To count the number of singletons in a given sequence of words ending with 'xxxxx', you can use the following pseudocode:

previous = None
n = 0
while True:
word = input('Enter word: ')
if word == 'xxxxx':
break
if previous != word and (previous is None or input('Enter next word: ') != word):
n += 1
previous = word
This code performs a loop that continues to accept words as input until the user enters 'xxxxx'. For each word, if it is not equal to the previous word and is not followed immediately by a duplicate, it is considered a singleton and n is incremented. The variable previous stores the last word processed to perform the comparison with the next word. Note that the second input within the loop is used to look ahead at the next word without moving past it in the sequence.The loop is used to read each word, the if statement helps in determining whether to increment the counter n, which is the final output representing the number of singletons in the sequence.

Answer 2

n = 0

string temp;
if wordset[0] != wordset[1] then {
n = n + 1
}
temp = wordset[0]
for i = 1 to wordset.length - 2 do {
if wordset[i] != temp AND wordset[i] != wordset[i + 1] then {
n = n + 1
}
temp = wordset[i]

}

}