Valid Number

Given a string s, return whether s is a valid number. For example, all the following are valid numbers: "2", "0089", "-0.1", "+3.14", "4.", "-.9", "2e10", "-90E3", "3e+7", "+6e-1", "53.5e93", "-123.456e789", while the following are not valid numbers: "abc", "1a", "1e", "e3", "99e2.5", "--6", "-+3", "95a54e53".

A valid number is defined using one of the following definitions:

1. An integer number followed by an optional exponent. 2. A decimal number followed by an optional exponent.

An integer number is defined with an optional sign '-' or '+' followed by digits.

A decimal number is defined with an optional sign '-' or '+' followed by one of the following definitions:

1. Digits followed by a dot '.'. 2. Digits followed by a dot '.' followed by digits. 3. A dot '.' followed by digits.

An exponent is defined with an exponent notation 'e' or 'E' followed by an integer number.

The digits are defined as one or more digits.

## Valid Number

This problem asks us to determine whether a given string represents a valid number, considering various formats such as integers, decimals, and exponents. We need to account for optional signs, digits, dots, and exponent notations ('e' or 'E').

### Brute Force Solution

A brute-force approach would involve manually checking all possible combinations of characters and their positions according to the rules. This can be done using a series of if-else statements or regular expressions.  However, this is not efficient and prone to errors due to the complexity of the rules.

```python
def isNumber_bruteforce(s: str) -> bool:
    try:
        float(s)
        # Additional checks to rule out cases like 'inf' and 'nan'
        if s.lower() == 'inf' or s.lower() == '-inf' or s.lower() == 'nan':
            return False
        return True
    except ValueError:
        return False

Optimal Solution: State Machine

A more robust and elegant solution involves using a state machine to parse the input string. We can define different states representing different parts of a valid number (e.g., integer part, decimal part, exponent part) and transitions between these states based on the input characters.

def isNumber(s: str) -> bool:
    s = s.strip()  # Remove leading/trailing whitespace
    n = len(s)
    state = 0
    # 0: initial, 1: digit (before dot), 2: dot, 3: digit (after dot),
    # 4: e, 5: sign after e, 6: digit after e, 7: trailing spaces
    states = [
        {" ": 0, "s": 1, "d": 1, ".": 2},
        {"d": 1, ".": 2, "e": 4, " ": 7},
        {"d": 3, "e": 4, " ": 7},
        {"d": 3, "e": 4, " ": 7},
        {"s": 5, "d": 6},
        {"d": 6},
        {"d": 6, " ": 7},
        {" ": 7}
    ]
    
    def to_state_key(char):
        if char.isdigit():
            return "d"
        elif char in ["+", "-"]:
            return "s"
        else:
            return char

    for char in s:
        key = to_state_key(char)
        if key in states[state]:
            state = states[state][key]
        else:
            return False

    return state in [1, 3, 6, 7]

Big(O) Time Complexity Analysis

The time complexity of the state machine solution is O(n), where n is the length of the input string. This is because we iterate through the string once, and the state transitions take constant time.

Big(O) Space Complexity Analysis

The space complexity is O(1), as we use a fixed number of states and variables regardless of the input string length.

Edge Cases

• Empty string: The code handles empty strings gracefully by returning False due to the s.strip() and subsequent checks.
• Leading/trailing whitespace: The s.strip() method removes leading and trailing whitespace, ensuring correct parsing.
• Multiple dots or 'e's: The state machine prevents multiple dots or 'e's by only allowing transitions to appropriate states.
• Sign after 'e': The state machine correctly handles signs (+/-) immediately following an exponent ('e' or 'E').
• Invalid characters: Any character not in the set of allowed characters (digits, '+', '-', '.', 'e', 'E', ' ') will cause the state machine to return False.
• 'inf' and 'nan': The brute force approach has a special check to filter out inf and nan as they are valid floats but the prompt suggests they are invalid inputs