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May 15, 2021 09:15 am GMT

Solution: Valid Number

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode's forums.

Leetcode Problem #65 (Hard): Valid Number

Description:


(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)

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

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

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

  • (Optional) A sign character (either '+' or '-').
  • One of the following formats:
    • At least one digit, followed by a dot '.'.
    • At least one digit, followed by a dot '.', followed by at least one digit.
    • A dot '.', followed by at least one digit.

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

  • (Optional) A sign character (either '+' or '-').
  • At least one digit.

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.

Examples:

Example 1:
Input:s = "0"
Output:true
Example 2:
Input:s = "e"
Output:false
Example 3:
Input:s = "."
Output:false
Example 4:
Input:s = ".1"
Output:true

Constraints:

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

Idea:


(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)

To solve this problem, we should should just make a list of the possible error conditions and then check for each one.

The error conditions are:

  • More than one exponent character ('e'/'E'), or seeing an 'e'/'E' when a number has not yet been seen.
  • More than one sign, or a sign appearing after a decimal or number have been seen. This gets reset when passing an 'e'/'E'.
  • More than one decimal, or a decimal appearing after an 'e'/'E' has been seen.
  • Any other non-number character appearing.
  • Reaching the end of S without an active number.

To help with this process, we can set up some boolean flags for the different things of which we're keeping track (num, exp, sign, dec). We'll also need to remember to reset all flags except exp when we find an 'e'/'E', as we're starting a new integer expression.

  • Time Complexity: O(N) where N is the number of characters in S
  • Space Complexity: O(1)

Javascript Code:


(Jump to: Problem Description || Solution Idea)

var isNumber = function(S) {    let exp = false, sign = false, num = false, dec = false    for (let c of S)        if (c >= '0' && c <= '9') num = true             else if (c === 'e' || c === 'E')            if (exp || !num) return false            else exp = true, sign = false, num = false, dec = false        else if (c === '+' || c === '-')            if (sign || num || dec) return false            else sign = true        else if (c === '.')            if (dec || exp) return false            else dec = true        else return false    return num};

Python Code:


(Jump to: Problem Description || Solution Idea)

class Solution:    def isNumber(self, S: str) -> bool:            num, exp, sign, dec = False, False, False, False        for c in S:            if c >= '0' and c <= '9': num = True                 elif c == 'e' or c == 'E':                if exp or not num: return False                else: exp, num, sign, dec = True, False, False, False            elif c == '+' or c == '-':                if sign or num or dec: return False                else: sign = True            elif c == '.':                if dec or exp: return False                else: dec = True            else: return False        return num

Java Code:


(Jump to: Problem Description || Solution Idea)

class Solution {    public boolean isNumber(String S) {        boolean num = false, exp = false, sign = false, dec = false;        for (int i = 0; i < S.length(); i++) {            char c = S.charAt(i);            if (c >= '0' && c <= '9') num = true ;                else if (c == 'e' || c == 'E')                if (exp || !num) return false;                else {                    exp = true;                    sign = false;                    num = false;                    dec = false;                }            else if (c == '+' || c == '-')                if (sign || num || dec) return false;                else sign = true;            else if (c == '.')                if (dec || exp) return false;                else dec = true;            else return false;        }        return num;    }}

C++ Code:


(Jump to: Problem Description || Solution Idea)

class Solution {public:    bool isNumber(string S) {        bool num = false, exp = false, sign = false, dec = false;        for (auto c : S)            if (c >= '0' && c <= '9') num = true ;                else if (c == 'e' || c == 'E')                if (exp || !num) return false;                else exp = true, sign = false, num = false, dec = false;            else if (c == '+' || c == '-')                if (sign || num || dec) return false;                else sign = true;            else if (c == '.')                if (dec || exp) return false;                else dec = true;            else return false;        return num;    }};

Original Link: https://dev.to/seanpgallivan/solution-valid-number-3gf5

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