Leetcode #65: Valid Number
In this guide, we solve Leetcode #65 Valid Number in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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".
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: String
Intuition
We need to scan characters while tracking positions or counts.
A simple state machine keeps the logic precise.
Approach
Iterate through the string once and update the state for each character.
Use a map or array if you need fast lookups.
Steps:
- Iterate through characters.
- Maintain necessary state.
- Build or validate the output.
Python Solution
class Solution:
def isNumber(self, s: str) -> bool:
n = len(s)
i = 0
if s[i] in '+-':
i += 1
if i == n:
return False
if s[i] == '.' and (i + 1 == n or s[i + 1] in 'eE'):
return False
dot = e = 0
j = i
while j < n:
if s[j] == '.':
if e or dot:
return False
dot += 1
elif s[j] in 'eE':
if e or j == i or j == n - 1:
return False
e += 1
if s[j + 1] in '+-':
j += 1
if j == n - 1:
return False
elif not s[j].isnumeric():
return False
j += 1
return True
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.