Problem

You are given a string s made up of words separated by single spaces. Each word contains only lowercase English letters. Return a new string in which, for every word, the first and the last character are converted to uppercase while every other character stays unchanged. Words of length 1 have the same first and last character, so that single character becomes uppercase.

Input / Output

• Input: a string s with 1 <= s.length <= 10^5, words separated by exactly one space, no leading or trailing spaces.
• Output: the transformed string.

Examples

Example 1 Input: s = "hello world" Output: "HellO WorlD" Explanation: In hello the first letter h and the last letter o become uppercase. Same for world.

Example 2 Input: s = "a quick brown fox" Output: "A QuicK BrowN FoX" Explanation: The single-letter word a becomes A.

Example 3 Input: s = "abc" Output: "AbC"

Constraints

• 1 <= s.length <= 10^5
• s contains only lowercase letters and spaces.
• There is exactly one space between consecutive words.

Approach

This is a straightforward string-scanning problem, but interviewers use it to check whether you can handle boundaries cleanly without splitting the string into a list (which costs extra memory).

1. Convert the string to a mutable character array.
2. Walk through the array with an index i. A character at index i is the first letter of a word if i == 0 or s[i-1] == ' '. It is the last letter of a word if i == n-1 or s[i+1] == ' '.
3. If either condition holds and the character is a letter, replace it with its uppercase form.
4. Join the characters back into a string.

Because each position is inspected once and the checks are constant time, the algorithm is linear.

Alternative: split on spaces, transform each word (w[0].upper() + w[1:-1] + w[-1].upper() with a guard for length 1) and join with a single space. This is simpler to write but allocates a list of words.

Complexity

• Time: O(n), where n is the length of the string.
• Space: O(n) for the output (O(1) extra if the language allows in-place mutation).

Follow-ups often asked

• Handle multiple consecutive spaces or leading/trailing spaces without collapsing them.
• Do it in place in a language with mutable strings (C/C++).
• Extend to Unicode words where "letter" is not just a–z.

Reference solution (Python)

def capitalize_ends(s: str) -> str:
    chars = list(s)
    n = len(chars)
    for i, c in enumerate(chars):
        if c == ' ':
            continue
        first = i == 0 or chars[i - 1] == ' '
        last = i == n - 1 or chars[i + 1] == ' '
        if first or last:
            chars[i] = c.upper()
    return ''.join(chars)

Test cases

Input	Expected
"hello world"	"HellO WorlD"
"a quick brown fox"	"A QuicK BrowN FoX"
"x"	"X"
"ab"	"AB"