HardString

Valid Number

1 approach, code in every language

Given a string, determine whether it represents a valid number. A valid number is an optional sign followed by either an integer or a decimal, optionally followed by an exponent part. A decimal must have digits on at least one side of the dot, an exponent begins with 'e' or 'E' followed by an optional sign and one or more digits, and any leading exponent must be preceded by a valid number.

No surrounding whitespace is allowed and no stray characters are permitted. Your job is to accept only strings that follow these rules and reject everything else.

Example 1

Input: s = "-3.14e2"

Output: true

A leading sign, a decimal part, and a valid exponent together form a well-formed number.

Example 2

Input: s = "1e"

Output: false

The exponent marker 'e' is not followed by any digits, so the string is incomplete.

Constraints

1 ≤ s.length ≤ 20
s consists of English letters, digits, and the characters '+', '-', '.'
The answer is a single boolean

Single pass with state flags

Scan the string once while tracking whether a digit, a dot, and an exponent have been seen, rejecting the moment any rule is violated.

Loading animation…

Intuition

Instead of writing an intricate regular expression, notice that each character type has simple local rules: a sign is only legal at the very start or right after an exponent marker, a dot cannot repeat and cannot appear after an exponent, and an exponent needs a digit before it and a digit after it. By resetting the digit flag when an exponent appears we force the mantissa-and-exponent structure, and a final check that a digit was seen guarantees the number is complete.

Algorithm walkthrough

1Keep three booleans: seenDigit, seenDot, and seenExp, all starting false.
2For a digit, mark seenDigit true.
3For a sign, allow it only at index 0 or immediately after an 'e'/'E'; otherwise reject.
4For a dot, reject if a dot or an exponent was already seen, else mark seenDot; for 'e'/'E', reject if an exponent was already seen or no digit came before, then mark seenExp and clear seenDigit to demand digits after it.
5Reject any other character, and after the loop return seenDigit so that trailing 'e' or a bare sign fails.

Implementation

class Solution:
    def isNumber(self, s: str) -> bool:
        seen_digit = seen_dot = seen_exp = False
        for i, c in enumerate(s):
            if c.isdigit():
                seen_digit = True
            elif c in '+-':
                if i > 0 and s[i - 1] not in 'eE':
                    return False
            elif c == '.':
                if seen_dot or seen_exp:
                    return False
                seen_dot = True
            elif c in 'eE':
                if seen_exp or not seen_digit:
                    return False
                seen_exp = True
                seen_digit = False
            else:
                return False
        return seen_digit

Time

O(n)

Space

O(1)

One linear scan with a fixed number of boolean flags; no extra structures grow with the input.

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After the pattern

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