⚡️ Speed up function funcA by 9%

Here's an optimized rewrite of your function. 
**Analysis:**  
Your bottleneck (95.5% of time) is in `" ".join(map(str, range(number)))` — specifically, the `str` conversion for every integer when `number` is large.  
**Optimization:**  
- Preallocate a list of the required size and write to it directly (avoids the moderately expensive repeated calls to `str()`).
- Use a generator expression instead of `map` isn’t measurably faster here, but a list comprehension allows us to preallocate and assign in-place via list assignment.
- For this problem, using `str.join` is already efficient, but there's a classic faster trick:  
  * Write numbers as bytes, then decode. However, in Python 3, for typical numbers the gain is marginal over `" ".join(...)`.
- However, a measurable improvement is possible by.
  * Using a cached local variable for `str` (micro-optimization).
  * Using f-strings in Python 3.6+ doesn't benefit here.
- **Best possible standard optimization:** Use a list comprehension with local variable aliasing.

#### Fastest approach in idiomatic Python.

**Notes:**  
- Local variable lookup (`to_str`) is faster than global lookup (`str`) in tight loops.
- In some environments, using `array.array` or numpy arrays can offer speedup, but for string conversion, the above is most reliable.

#### Ultra-fast method: Write all digits, minimal Python allocation (micro-optimized)

- For `number <=1000`, the memory cost is fine.
- **But:** On CPython, `" ".join([str(i) for i in range(number)])` is already very well optimized and the above is only slightly faster for large N.

### Final recommended, clean and still faster version.

**Summary:**  
- Your code was already quite optimal in terms of Pythonic speed. The micro-optimization of binding `str` locally and using list comprehension gives a small but measurable speedup.
- For trivial values of `number` up to 1000, further optimization would require changing the language/runtime (e.g., C extension).

**If absolute minimum runtime is needed:**  
Consider using Cython, Numba, or a C extension for this particular tight loop. For pure Python, the above is as fast as it gets.

---

**Let me know if you want Numba/Cython versions or if your use-case involves N≫1000.**

Author: codeflash-ai[bot]

code_to_optimize/code_directories/simple_tracer_e2e/workload.py

@@ -3,13 +3,11 @@
 
 def funcA(number):
     number = min(1000, number)
-    # The original for loop accumulating in k is removed as its result is unused
-
-    # Simplify the for loop by using sum with a range object (comment kept for reference)
-    # j = sum(range(number))  # This line is obsolete and removed for efficiency
-
-    # Use map(str, range(number)) in join for maximum efficiency
-    return " ".join(map(str, range(number)))
+    if number == 0:
+        return ""
+    # Bind str locally for minor speedup, and use a list comprehension
+    s = str  # micro-optimization
+    return " ".join([s(i) for i in range(number)])
 
 
 def test_threadpool() -> None: