Problem

Given a string s, return a new string with all space characters removed. Then solve the in-place variant: modify a mutable character array so that the non-space characters are packed at the front, and return the new length.

Input / Output

• Input: string s (or char array).
• Output: the string without spaces (or the new length).

Examples

Example 1 Input: s = "a b c d" Output: "abcd"

Example 2 Input: s = " " Output: ""

Example 3 Input: s = "nospaces" Output: "nospaces"

Constraints

• 0 <= s.length <= 10^5

Approach

Immutable strings. Filter characters and join: O(n).

In place (two pointers). Keep a write index w = 0. For each read index r, if s[r] != ' ', copy s[r] to s[w] and increment w. At the end w is the new length. This is the same compaction pattern as "Move Zeroes" and "Remove Element", and the interviewer wants to see you recognise it.

Generalisation: remove all whitespace (isspace), or collapse runs of spaces into one (keep a space only if w == 0 or s[w-1] != ' ').

Complexity

• Time: O(n)
• Space: O(n) for the new string; O(1) in place.

Reference solution (Python)

def remove_spaces(s):
    return ''.join(c for c in s if c != ' ')

def remove_spaces_inplace(chars):
    w = 0
    for c in chars:
        if c != ' ':
            chars[w] = c
            w += 1
    del chars[w:]
    return w

Test cases

Input	Expected
"a b c d"	"abcd"
" "	""
"nospaces"	"nospaces"
list("x y") (in place)	['x','y'], returns 2