In Java, what will be the content of the StringBuffer s after executing: StringBuffer s = new StringBuffer("Hello"); s.deleteCharAt(0);?

Introduction / Context:
This question tests your understanding of how StringBuffer operations manipulate character sequences in Java. While the code looks simple, it requires you to pay attention to zero based indexing and the behavior of the deleteCharAt method. Such questions are common in interviews for Java developer roles because they reveal whether a candidate can correctly reason about basic string and buffer operations without running the code.

Given Data / Assumptions:

- The class used is java.lang.StringBuffer.

- The initial buffer is created with the content "Hello".

- The method deleteCharAt is called with index 0.

- We use standard zero based indexing as in most Java collections and arrays.

Concept / Approach:
StringBuffer maintains a mutable sequence of characters, similar to a dynamic array. The characters in "Hello" are indexed as H at index 0, e at index 1, l at index 2, l at index 3, and o at index 4. The deleteCharAt method removes the character at the specified index and shifts subsequent characters to the left by one position. Therefore, calling deleteCharAt(0) on "Hello" removes the first character H, leaving "ello" in the buffer.

Step-by-Step Solution:
Step 1: Start with the initial content of the buffer: "Hello". Step 2: Write down the indices: H is at 0, e at 1, l at 2, l at 3, o at 4. Step 3: The call s.deleteCharAt(0) means remove the character at index position 0. Step 4: Removing H leaves the sequence e, l, l, o, which forms the string "ello". Step 5: Compare this result with the options and select "ello" as the correct one.

Verification / Alternative check:
An alternative check is to compare deleteCharAt with substring operations. For example, for a simple String, str.substring(1) on "Hello" would return "ello" by skipping the first character. Since deleteCharAt(0) in a mutable buffer has a similar effect of removing the first character, "ello" is again the expected result. This mental comparison confirms that the buffer content after the method call is "ello".

Why Other Options Are Wrong:
Option A, "llo", would be correct only if both H and e were removed, which does not happen here. Option B, "Hllo", suggests removal of e instead of H, which would occur with deleteCharAt(1), not with deleteCharAt(0). Option D, "H", would require removing all characters except the first one, which is not what deleteCharAt does when called with index 0. Therefore, only "ello" correctly reflects the result of removing the first character from "Hello".

Common Pitfalls:
A typical mistake is to forget that Java indexing starts at 0, not 1, which leads to confusion about which character is removed. Another pitfall is to assume that deleteCharAt returns a new buffer, when in fact it modifies the existing one in place. For interview purposes, always show that you can visualize the index positions and walk through the effect of each method call step by step, since this demonstrates clear understanding of fundamental string operations.

Final Answer:
After executing the code, the StringBuffer s contains the text "ello".