This problem assesses the your ability to work with string manipulation and correctly use the stack data structure to solve a problem.

Below is a Python implementation of the is_valid_parentheses function to determine if the input string contains valid parentheses:

def is_valid_parentheses(s: str) -> bool:
    """
    Determine if the input string contains valid parentheses.

    Parameters:
    - s: Input string.

    Returns:
    - bool: True if the parentheses are valid, False otherwise.
    """
    stack = []  # Stack to keep track of open parentheses
    parentheses_map = {')': '(', '}': '{', ']': '['}  # Mapping of closing to opening parentheses

    for char in s:
        if char in parentheses_map.values():
            # If the character is an open parenthesis, push it onto the stack
            stack.append(char)
        elif char in parentheses_map:
            # If the character is a closing parenthesis
            # Check if the stack is not empty and the top of the stack matches the corresponding open parenthesis
            if not stack or stack.pop() != parentheses_map[char]:
                return False
        else:
            # If the character is neither open nor closing parenthesis, return False
            return False

    # The parentheses are valid if the stack is empty
    return not stack

# Example usage:
s1 = "()"
result1 = is_valid_parentheses(s1)
print(result1)  # Output: True

s2 = "()[]{}"
result2 = is_valid_parentheses(s2)
print(result2)  # Output: True

s3 = "(]"
result3 = is_valid_parentheses(s3)
print(result3)  # Output: False

s4 = "([)]"
result4 = is_valid_parentheses(s4)
print(result4)  # Output: False

s5 = "{[]}"
result5 = is_valid_parentheses(s5)
print(result5)  # Output: True

This solution uses a stack to keep track of open parentheses as they are encountered in the string. It checks if the closing parentheses match the corresponding open parentheses.

Time Complexity:

The time complexity is O(n), where n is the length of the input string.

Space Complexity:

The space complexity is O(n) due to the stack.