Base Case
  str.length() == 0
  str.length() == 1
  return true

  str.charAt(0) != str.charAt(str.length() - 1)
  return false

Rec  Case
  str.length() > 1

Rec  Step

  str.charAt(0) == str.charAt(str.length() - 1)
  (true for now)
  palindrome(str.substring(1, str.length() - 1))

  "kayak"
  "aya"
  "y"

public static boolean palindrome(String str) {
  boolean success = true;

  if (str.length() > 1) {
    if (str.charAt(0) == str.charAt(str.length() - 1)) {
      success = palindrome(str.substring(1, str.length() - 1));
    }
    else { // if (str.charAt(0) != str.charAt(str.length() - 1)) {
      success = false;
    }
  }

  return success;
}



       f     ?        l
       f              ?                      l   
Index: 0  1  2  3  4  5   6   7   8   9  10 11
List:  1  3  4  5  6  9  12  15  20  22  24



                                  f   ?      l
                      f           ?          l
       f              ?                      l   
Index: 0  1  2  3  4  5   6   7   8   9  10 11
List:  1  3  4  5  6  9  12  15  20  22  24

                          f   l
                          f   ?   l
                      f   ?       l
                      f           ?          l
       f              ?                      l   
Index: 0  1  2  3  4  5   6   7   8   9  10 11
List:  1  3  4  5  6  9  12  15  20  22  24

Base Case
  i have found the value i was looking for
    return the index where the value exists

  f ==  mid || l == mid
    i have searched the entire list and i have not found
    the element i was looking

    return false

Rec  Case
  f < mid || l < mid

Rec  Step
  binarySearch(int array[], int first, int last, int value)
  binarySearch(array, first, last, value)

  value < array[mid] // rec case A
    binarySearch(array, first, mid, value);
  value > array[mid] // rec case B
    binarySearch(array, mid + 1, last, value);