Valid Number LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Valid Number in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – Valid Number– LeetCode Problem

Valid Number– LeetCode Problem

Problem:

A valid number can be split up into these components (in order):

1. A decimal number or an integer.
2. (Optional) An 'e' or 'E', followed by an integer.

A decimal number can be split up into these components (in order):

1. (Optional) A sign character (either '+' or '-').
2. One of the following formats:
   1. One or more digits, followed by a dot '.'.
   2. One or more digits, followed by a dot '.', followed by one or more digits.
   3. A dot '.', followed by one or more digits.

An integer can be split up into these components (in order):

1. (Optional) A sign character (either '+' or '-').
2. One or more digits.

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"].

Given a string s, return true if s is a valid number.

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 '.'.

Valid Number– LeetCode Solutions

Valid Number in C++:

class Solution {
 public:
  bool isNumber(string s) {
    trim(s);
    if (s.empty())
      return false;

    bool seenNum = false;
    bool seenDot = false;
    bool seenE = false;

    for (int i = 0; i < s.length(); ++i) {
      switch (s[i]) {
        case '.':
          if (seenDot || seenE)
            return false;
          seenDot = true;
          break;
        case 'e':
          if (seenE || !seenNum)
            return false;
          seenE = true;
          seenNum = false;
          break;
        case '+':
        case '-':
          if (i > 0 && s[i - 1] != 'e')
            return false;
          seenNum = false;
          break;
        default:
          if (!isdigit(s[i]))
            return false;
          seenNum = true;
      }
    }

    return seenNum;
  }

 private:
  void trim(string& s) {
    s.erase(0, s.find_first_not_of(' '));
    s.erase(s.find_last_not_of(' ') + 1);
  }
};

Valid Number in Java:

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

    boolean seenNum = false;
    boolean seenDot = false;
    boolean seenE = false;

    for (int i = 0; i < s.length(); ++i) {
      switch (s.charAt(i)) {
        case '.':
          if (seenDot || seenE)
            return false;
          seenDot = true;
          break;
        case 'e':
          if (seenE || !seenNum)
            return false;
          seenE = true;
          seenNum = false;
          break;
        case '+':
        case '-':
          if (i > 0 && s.charAt(i - 1) != 'e')
            return false;
          seenNum = false;
          break;
        default:
          if (!Character.isDigit(s.charAt(i)))
            return false;
          seenNum = true;
      }
    }

    return seenNum;
  }
}

Valid Number in Python:

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

    seenNum = False
    seenDot = False
    seenE = False

    for i, c in enumerate(s):
      if c == '.':
        if seenDot or seenE:
          return False
        seenDot = True
      elif c == 'e':
        if seenE or not seenNum:
          return False
        seenE = True
        seenNum = False
      elif c in '+-':
        if i > 0 and s[i - 1] != 'e':
          return False
        seenNum = False
      else:
        if not c.isdigit():
          return False
        seenNum = True

    return seenNum