This code is intended to check whether an array of...

This code is intended to check whether an array of integers contains any repeated elements. However, the inner for loop's initialization and condition are missing (as indicated by /* missing code */). Based on the logic, the missing part should presumably iterate over all elements in the array except the current one being checked in the outer loop. Here's what it does step by step, assuming the missing part is filled correctly:

What the code is intended to do:

1. Iterate through each element of the array (arr) using an outer for loop with the index k.
2. Inside the outer loop, another for loop (the missing part) will iterate through the array with a different index (j). This inner loop checks for matching elements, ensuring it doesn't compare the same index.
3. For each pair of indices (k and j), the code compares the elements of arr at these indices. If any two elements are equal (arr[j] == arr[k]), the method immediately returns true.
4. If the loops complete without finding any duplicate elements, the method returns false.

The missing part:

The inner loop should look like this:

for (int j = k + 1; j < arr.length; j++)

This setup ensures the loop compares arr[k] only with the elements after index k, avoiding redundant comparisons or comparing an element with itself.

Fixed Code:

public static boolean hasRepeats(int[] arr)
{
  for (int k = 0; k < arr.length; k++)
  {
    for (int j = k + 1; j < arr.length; j++) // Compare with elements after k
    {
      if (arr[j] == arr[k])
      {
        return true; // A repeated element is found
      }
    }
  }
  return false; // No repeats detected
}

Example Usage:

If the input array is {1, 2, 3, 4, 2}, the method will return true because the number 2 is repeated. If the input is {1, 2, 3, 4}, the method will return false as there are no duplicates.

Complexity:

• Time complexity: O(n²), because there are nested loops iterating over the array.
• Space complexity: O(1), as no additional data structures are used.