LeetCode #65 — HARD

Valid Number

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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".

Formally, a valid number is defined using one of the following definitions:

An integer number followed by an optional exponent.
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:

Digits followed by a dot '.'.
Digits followed by a dot '.' followed by digits.
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.

Example 1:

Input: s = "0"

Output: true

Example 2:

Input: s = "e"

Output: false

Example 3:

Input: s = "."

Output: false

Constraints:

1 <= s.length <= 20
s consists of only English letters (both uppercase and lowercase), digits (0-9), plus '+', minus '-', or dot '.'.

Step 01

Brute Force Baseline

Problem summary: 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". Formally, a valid number is defined using one of the following definitions: An integer number followed by an optional exponent. 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: Digits followed by a dot '.'. Digits followed by a dot '.' followed by digits. A dot '.' followed by digits. An exponent is defined with an exponent notation 'e' or 'E' followed by an integer number. The digits are

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"0"

Example 2

"e"

Example 3

"."

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

class Solution {
    public boolean isNumber(String s) {
        s = s.trim();
        if (s.isEmpty()) return false;

        boolean seenDigit = false;
        boolean seenDot = false;
        boolean seenExp = false;
        boolean digitAfterExp = true;

        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);

            if (Character.isDigit(ch)) {
                seenDigit = true;
                if (seenExp) digitAfterExp = true;
            } else if (ch == '+' || ch == '-') {
                // Sign is valid only at start or right after exponent.
                if (i > 0 && s.charAt(i - 1) != 'e' && s.charAt(i - 1) != 'E') return false;
            } else if (ch == '.') {
                // Dot allowed only once and only before exponent.
                if (seenDot || seenExp) return false;
                seenDot = true;
            } else if (ch == 'e' || ch == 'E') {
                // Exponent allowed once and only after at least one digit.
                if (seenExp || !seenDigit) return false;
                seenExp = true;
                digitAfterExp = false;
            } else {
                return false;
            }
        }

        return seenDigit && digitAfterExp;
    }
}

func isNumber(s string) bool {
    i, j := 0, len(s)-1
    for i <= j && s[i] == ' ' {
        i++
    }
    for j >= i && s[j] == ' ' {
        j--
    }
    if i > j {
        return false
    }
    s = s[i : j+1]

    seenDigit := false
    seenDot := false
    seenExp := false
    digitAfterExp := true

    for idx := 0; idx < len(s); idx++ {
        ch := s[idx]
        if ch >= '0' && ch <= '9' {
            seenDigit = true
            if seenExp {
                digitAfterExp = true
            }
        } else if ch == '+' || ch == '-' {
            if idx > 0 && s[idx-1] != 'e' && s[idx-1] != 'E' {
                return false
            }
        } else if ch == '.' {
            if seenDot || seenExp {
                return false
            }
            seenDot = true
        } else if ch == 'e' || ch == 'E' {
            if seenExp || !seenDigit {
                return false
            }
            seenExp = true
            digitAfterExp = false
        } else {
            return false
        }
    }

    return seenDigit && digitAfterExp
}

class Solution:
    def isNumber(self, s: str) -> bool:
        s = s.strip()
        if not s:
            return False

        seen_digit = False
        seen_dot = False
        seen_exp = False
        digit_after_exp = True

        for i, ch in enumerate(s):
            if ch.isdigit():
                seen_digit = True
                if seen_exp:
                    digit_after_exp = True
            elif ch in '+-':
                if i > 0 and s[i - 1] not in 'eE':
                    return False
            elif ch == '.':
                if seen_dot or seen_exp:
                    return False
                seen_dot = True
            elif ch in 'eE':
                if seen_exp or not seen_digit:
                    return False
                seen_exp = True
                digit_after_exp = False
            else:
                return False

        return seen_digit and digit_after_exp

impl Solution {
    pub fn is_number(s: String) -> bool {
        let s = s.trim();
        if s.is_empty() {
            return false;
        }

        let bytes = s.as_bytes();
        let mut seen_digit = false;
        let mut seen_dot = false;
        let mut seen_exp = false;
        let mut digit_after_exp = true;

        for i in 0..bytes.len() {
            let ch = bytes[i];
            if ch.is_ascii_digit() {
                seen_digit = true;
                if seen_exp {
                    digit_after_exp = true;
                }
            } else if ch == b'+' || ch == b'-' {
                if i > 0 && bytes[i - 1] != b'e' && bytes[i - 1] != b'E' {
                    return false;
                }
            } else if ch == b'.' {
                if seen_dot || seen_exp {
                    return false;
                }
                seen_dot = true;
            } else if ch == b'e' || ch == b'E' {
                if seen_exp || !seen_digit {
                    return false;
                }
                seen_exp = true;
                digit_after_exp = false;
            } else {
                return false;
            }
        }

        seen_digit && digit_after_exp
    }
}

function isNumber(s: string): boolean {
  s = s.trim();
  if (s.length === 0) return false;

  let seenDigit = false;
  let seenDot = false;
  let seenExp = false;
  let digitAfterExp = true;

  for (let i = 0; i < s.length; i++) {
    const ch = s[i];
    if (ch >= '0' && ch <= '9') {
      seenDigit = true;
      if (seenExp) digitAfterExp = true;
    } else if (ch === '+' || ch === '-') {
      if (i > 0 && s[i - 1] !== 'e' && s[i - 1] !== 'E') return false;
    } else if (ch === '.') {
      if (seenDot || seenExp) return false;
      seenDot = true;
    } else if (ch === 'e' || ch === 'E') {
      if (seenExp || !seenDigit) return false;
      seenExp = true;
      digitAfterExp = false;
    } else {
      return false;
    }
  }

  return seenDigit && digitAfterExp;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.