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What is the output of the following code? double[] myList = {1, 5, 5, 5, 5, 1}; double max = myList[0]; int indexOfMax = 0; for (int i = 1; i < myList.length; i++) {   if (myList[i] > max) {     max = myList[i];     indexOfMax = i;   } } System.out.println(indexOfMax); Group of answer choices01254

Question

What is the output of the following code? double[] myList = {1, 5, 5, 5, 5, 1}; double max = myList[0]; int indexOfMax = 0; for (int i = 1; i < myList.length; i++) {   if (myList[i] > max) {     max = myList[i];     indexOfMax = i;   } } System.out.println(indexOfMax); Group of answer choices01254

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Solution 1

The output of the code will be 1.

Here's the step by step explanation:

  1. The code initializes an array myList with the elements {1, 5, 5, 5, 5, 1}.

  2. It then sets a variable max to the first element of the array (which is 1) and indexOfMax to 0 (the index of the first element).

  3. The code then enters a for loop that iterates over the array starting from the second element (index 1).

  4. Inside the loop, it checks if the current element is greater than max. If it is, it updates max to the current element and indexOfMax to the current index.

  5. In the first iteration, it finds that 5 is greater than 1, so it updates max to 5 and indexOfMax to 1.

  6. For the rest of the iterations, since all the other elements are either 5 or less, it does not update max or indexOfMax again.

  7. Finally, it prints indexOfMax, which is 1. So, the output of the code is 1.

This problem has been solved

Solution 2

The output of the code will be 1.

Here's the step by step explanation:

  1. The code initializes an array myList with the elements {1, 5, 5, 5, 5, 1}.

  2. It then sets a variable max to the first element of the array (which is 1) and indexOfMax to 0 (the index of the first element).

  3. The code then enters a for loop that iterates over the array starting from the second element (index 1).

  4. Inside the loop, it checks if the current element is greater than max. If it is, it updates max to the current element and indexOfMax to the current index.

  5. The first time it encounters a number greater than max is at index 1 (where the value is 5). It updates max to 5 and indexOfMax to 1.

  6. For the rest of the array, it doesn't find a number greater than 5, so max and indexOfMax don't change.

  7. Finally, it prints indexOfMax, which is 1.

This problem has been solved

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