Problem

Given a string s, return the number of words it contains. A word is a maximal sequence of non-space characters. The string may contain leading, trailing or multiple consecutive spaces, and may be empty.

Input / Output

• Input: string s.
• Output: integer word count.

Examples

Example 1 Input: s = "the quick brown fox" Output: 4

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

Example 3 Input: s = "" or s = " " Output: 0

Constraints

• 0 <= s.length <= 10^5

Approach

A word starts at every index i where s[i] is not a space and (i == 0 or s[i-1] is a space). Count those positions in one pass:

count = 0
for i in range(n):
    if s[i] != ' ' and (i == 0 or s[i-1] == ' '):
        count += 1

This handles all the whitespace edge cases without allocating a list, unlike len(s.split()), which is fine in practice but is not what a "do it manually" interviewer wants.

Extensions: treat tabs and newlines as separators (isspace()), or count only words made of letters.

Complexity

• Time: O(n)
• Space: O(1)

Reference solution (Python)

def count_words(s):
    count = 0
    for i, c in enumerate(s):
        if not c.isspace() and (i == 0 or s[i - 1].isspace()):
            count += 1
    return count

Test cases

Input	Expected
"the quick brown fox"	4
" hello world "	2
""	0
"a"	1