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.
Explanation: when unused apps keep runningon the device, it will make the device to keep lagging, thereby reducing it efficiency. But once those apps are disabled or closed and the system is probably restarted, the device will regain it efficiency back.