File: rearrange-spaces-between-words/solution.py

Date: 2026-06-06

Time: 18:40

rearrange-spaces-between-words/solution.py

Purpose

This file solves LeetCode 1592: Rearrange Spaces Between Words. Given a string with words separated by (possibly irregular) spaces, redistribute all spaces evenly between words, with any remainder appended to the end.

Key Components

reorderSpaces(text: str) -> str — The sole function. Contract:

• Input: A string containing lowercase letters and spaces. Leading/trailing/multiple consecutive spaces are all valid.
• Output: The same words in the same order, with spaces redistributed evenly between adjacent words. Any leftover spaces (when total spaces don't divide evenly by len(words) - 1) are appended after the last word.

Flow

1. Count total spaces — text.count(' ') counts every space in the original string, regardless of position.

2. Extract words — text.split() splits on any whitespace and discards empties, so " hello world " yields ["hello", "world"].

3. Single-word edge case — If only one word exists, there are zero gaps between words. All spaces go after the word: words[0] + ' ' * total_spaces.

4. Even distribution — divmod(total_spaces, len(words) - 1) computes between (spaces per gap) and extra (leftover spaces for the end).

5. Reassemble — Join words with between spaces, append extra spaces.

Patterns

• divmod for quotient+remainder — A Python idiom that avoids two separate // and % operations. Clean for distribution problems where you need both the "per-item share" and the "leftover".
• str.split() without arguments — Splits on all whitespace runs and strips leading/trailing whitespace. This is the right tool here because it extracts words regardless of how the original spacing was arranged.

Dependencies

• Imports: None — pure stdlib string operations.
• Imported by: rearrange-spaces-between-words/test_solution.py (directly). The large "Imported By" list in the prompt is noise from the test harness — those test files share a common test runner import pattern, not a direct dependency on this solution.

Invariants

• The total number of space characters in the output always equals total_spaces from the input. No spaces are created or destroyed.
• Word order is preserved — split() maintains left-to-right order.
• When len(words) == 1, all spaces trail the single word (the problem guarantees at least one word).

Error Handling

None. The function assumes valid input per the LeetCode contract (non-empty string with at least one word). No explicit validation, no exceptions raised. An empty string would cause words[0] to raise IndexError — but that's outside the problem's input constraints.

Beliefs

• rearrange-spaces-preserves-space-count — The output always contains exactly the same number of space characters as the input.
• rearrange-spaces-single-word-trailing — When the input contains exactly one word, all spaces are placed after that word (not before).
• rearrange-spaces-no-imports — The solution uses only Python builtins (str.count, str.split, divmod, str.join) with zero imports.
• rearrange-spaces-word-order-preserved — Words appear in the output in the same left-to-right order as the input; only spacing changes.