This is true. Recording relative positions in most cases will reduce total programming time and make it over all more efficient.
Answer:
Option a. int max = aList.get(0); for (int count = 1; count < aList.size(); count++) { if (aList.get(count) > max) { max = aList.get(count); } }
is the correct code snippet.
Explanation:
Following is given the explanation for the code snippet to find largest value in an integer array list aList.
- From the array list aList, very first element having index 0 will be stored in the variable max (having data type int).
- By using for starting from count =1 to count = size of array (aList), we will compare each element of the array with first element of the array.
- If any of the checked element get greater from the first element, it gets replaced in the variable max and the count is increased by 1 so that the next element may be checked.
- When the loop will end, the variable max will have the greatest value from the array aList.
i hope it will help you!
Answer:
A
Explanation:
The answer is A mate, if you need any help let me know please
Answer:
C. Optimization
Explanation:
Optimization entails improving your ads campaign making use of the available tools.
Optimization mostly makes ad to target the right audience and engage the right people.
