he-011
1.000
Challenge · difficulty 2/5
# HumanEval/11
Implement a file **`solution.py`** that completes the function below. Keep the given name and signature; your file must define it at module level.
```python
from typing import List
def string_xor(a: str, b: str) -> str:
""" Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
>>> string_xor('010', '110')
'100'
"""
```
<!-- imported from OpenAI HumanEval (HumanEval/11) -->
tests/test_he_011.py
# Auto-generated from OpenAI HumanEval HumanEval/11. Do not edit by hand.
from solution import * # noqa: F401,F403 (prompt helpers may be referenced by tests)
from solution import string_xor as candidate
METADATA = {
'author': 'jt',
'dataset': 'test'
}
def check(candidate):
assert candidate('111000', '101010') == '010010'
assert candidate('1', '1') == '0'
assert candidate('0101', '0000') == '0101'
def test_he_011():
check(candidate)
System prompt
You are an expert programmer. Solve the task exactly as specified. Output your solution as fenced code blocks using the required file name(s) and the exact function/type signatures requested. Prefer correctness; do not include prose outside code unless asked.
Proposed solution
```python
from typing import List
def string_xor(a: str, b: str) -> str:
""" Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
>>> string_xor('010', '110')
'100'
"""
result = []
for i in range(len(a)):
if a[i] != b[i]:
result.append('1')
else:
result.append('0')
return ''.join(result)
```Test output (stdout)
. [100%] 1 passed in 0.01s
Reasoning (chain-of-thought)
The task is to implement a function that performs binary XOR on two strings consisting of '0's and '1's.
For XOR operation:
- 0 XOR 0 = 0
- 0 XOR 1 = 1
- 1 XOR 0 = 1
- 1 XOR 1 = 0
I need to:
1. Iterate through both strings character by character
2. Perform XOR on each pair of bits
3. Return the result as a string
Let me think about the implementation:
```python
def string_xor(a: str, b: str) -> str:
result = []
for i in range(len(a)):
# XOR operation: if bits are different, result is '1', otherwise '0'
if a[i] != b[i]:
result.append('1')
else:
result.append('0')
return ''.join(result)
```
Or I could use a more concise approach:
```python
def string_xor(a: str, b: str) -> str:
return ''.join('1' if a[i] != b[i] else '0' for i in range(len(a)))
```
Or even using XOR directly:
```python
def string_xor(a: str, b: str) -> str:
return ''.join(str(int(a[i]) ^ int(b[i])) for i in range(len(a)))
```
The first approach is probably clearest. Let me verify with the example:
- a = '010', b = '110'
- i=0: '0' != '1' → '1'
- i=1: '1' == '1' → '0'
- i=2: '0' == '0' → '0'
- Result: '100' ✓
This matches the expected output.
Let me write the complete solution:
```python
from typing import List
def string_xor(a: str, b: str) -> str:
""" Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
>>> string_xor('010', '110')
'100'
"""
result = []
for i in range(len(a)):
if a[i] != b[i]:
result.append('1')
else:
result.append('0')
return ''.join(result)
```
I'll complete the implementation by joining the result list into a string. The function handles binary XOR by comparing each character and appending the appropriate result.