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Check if a given string is a valid number (Integer or Floating Point) | SET 1(Basic approach)

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Validate if a given string is numeric. Examples:

Input : str = "11.5"
Output : true

Input : str = "abc"
Output : false

Input : str = "2e10"
Output : true

Input : 10e5.4
Output : false

The following cases need to be handled in the code.

  1. Ignore the leading and trailing white spaces.
  2. Ignore the ‘+’, ‘-‘ and’.’ at the start.
  3. Ensure that the characters in the string belong to {+, -, ., e, [0-9]}
  4. Ensure that no ‘.’ comes after ‘e’.
  5. A dot character ‘.’ should be followed by a digit.
  6. The character ‘e’ should be followed either by ‘+’, ‘-‘, or a digit.

Below is implementation of above steps. 

C++

// C++ program to check if input number
// is a valid number
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
 
int valid_number(string str)
{
    int i = 0, j = str.length() - 1;
 
    // Handling whitespaces
    while (i < str.length() && str[i] == ' ')
        i++;
    while (j >= 0 && str[j] == ' ')
        j--;
 
    if (i > j)
        return 0;
 
    // if string is of length 1 and the only
    // character is not a digit
    if (i == j && !(str[i] >= '0' && str[i] <= '9'))
        return 0;
 
    // If the 1st char is not '+', '-', '.' or digit
    if (str[i] != '.' && str[i] != '+'
        && str[i] != '-' && !(str[i] >= '0' && str[i] <= '9'))
        return 0;
 
    // To check if a '.' or 'e' is found in given
    // string. We use this flag to make sure that
    // either of them appear only once.
    bool flagDotOrE = false;
 
    for (i; i <= j; i++) {
        // If any of the char does not belong to
        // {digit, +, -, ., e}
        if (str[i] != 'e' && str[i] != '.'
            && str[i] != '+' && str[i] != '-'
            && !(str[i] >= '0' && str[i] <= '9'))
            return 0;
 
        if (str[i] == '.') {
            // checks if the char 'e' has already
            // occurred before '.' If yes, return 0.
            if (flagDotOrE == true)
                return 0;
 
            // If '.' is the last character.
            if (i + 1 > str.length())
                return 0;
 
            // if '.' is not followed by a digit.
            if (!(str[i + 1] >= '0' && str[i + 1] <= '9'))
                return 0;
        }
 
        else if (str[i] == 'e') {
            // set flagDotOrE = 1 when e is encountered.
            flagDotOrE = true;
 
            // if there is no digit before 'e'.
            if (!(str[i - 1] >= '0' && str[i - 1] <= '9'))
                return 0;
 
            // If 'e' is the last Character
            if (i + 1 > str.length())
                return 0;
 
            // if e is not followed either by
            // '+', '-' or a digit
            if (str[i + 1] != '+' && str[i + 1] != '-'
                && (str[i + 1] >= '0' && str[i] <= '9'))
                return 0;
        }
    }
 
    /* If the string skips all above cases, then
    it is numeric*/
    return 1;
}
 
// Driver code
int main()
{
    char str[] = "0.1e10";
    if (valid_number(str))
        cout << "true";
    else
        cout << "false";
    return 0;
}
 
// This code is contributed by rahulkumawat2107

                    

Java

// Java program to check if input number
// is a valid number
import java.io.*;
import java.util.*;
 
class GFG {
    public boolean isValidNumeric(String str)
    {
        str = str.trim(); // trims the white spaces.
 
        if (str.length() == 0)
            return false;
 
        // if string is of length 1 and the only
        // character is not a digit
        if (str.length() == 1 && !Character.isDigit(str.charAt(0)))
            return false;
 
        // If the 1st char is not '+', '-', '.' or digit
        if (str.charAt(0) != '+' && str.charAt(0) != '-'
            && !Character.isDigit(str.charAt(0))
            && str.charAt(0) != '.')
            return false;
 
        // To check if a '.' or 'e' is found in given
        // string. We use this flag to make sure that
        // either of them appear only once.
        boolean flagDotOrE = false;
 
        for (int i = 1; i < str.length(); i++) {
            // If any of the char does not belong to
            // {digit, +, -, ., e}
            if (!Character.isDigit(str.charAt(i))
                && str.charAt(i) != 'e' && str.charAt(i) != '.'
                && str.charAt(i) != '+' && str.charAt(i) != '-')
                return false;
 
            if (str.charAt(i) == '.') {
                // checks if the char 'e' has already
                // occurred before '.' If yes, return 0.
                if (flagDotOrE == true)
                    return false;
 
                // If '.' is the last character.
                if (i + 1 >= str.length())
                    return false;
 
                // if '.' is not followed by a digit.
                if (!Character.isDigit(str.charAt(i + 1)))
                    return false;
            }
 
            else if (str.charAt(i) == 'e') {
                // set flagDotOrE = 1 when e is encountered.
                flagDotOrE = true;
 
                // if there is no digit before 'e'.
                if (!Character.isDigit(str.charAt(i - 1)))
                    return false;
 
                // If 'e' is the last Character
                if (i + 1 >= str.length())
                    return false;
 
                // if e is not followed either by
                // '+', '-' or a digit
                if (!Character.isDigit(str.charAt(i + 1))
                    && str.charAt(i + 1) != '+'
                    && str.charAt(i + 1) != '-')
                    return false;
            }
        }
 
        /* If the string skips all above cases, then
           it is numeric*/
        return true;
    }
 
    /* Driver Function to test isValidNumeric function */
    public static void main(String[] args)
    {
        String input = "0.1e10";
        GFG gfg = new GFG();
        System.out.println(gfg.isValidNumeric(input));
    }
}

                    

Python3

# Python3 program to check if input number
# is a valid number
def valid_number(str):
    i = 0
    j = len(str) - 1
 
    # Handling whitespaces
    while i<len(str) and str[i] == ' ':
        i += 1
    while j >= 0 and str[j] == ' ':
        j -= 1
 
    if i > j:
        return False
 
    # if string is of length 1 and the only
    # character is not a digit
    if (i == j and not(str[i] >= '0' and
                       str[i] <= '9')):
        return False
 
    # If the 1st char is not '+', '-', '.' or digit
    if (str[i] != '.' and str[i] != '+' and
        str[i] != '-' and not(str[i] >= '0' and
        str[i] <= '9')):
        return False
 
    # To check if a '.' or 'e' is found in given
    # string.We use this flag to make sure that
    # either of them appear only once.
    flagDotOrE = False
 
    for i in range(j + 1):
         
        # If any of the char does not belong to
        # {digit, +, -,., e}
        if (str[i] != 'e' and str[i] != '.' and
            str[i] != '+' and str[i] != '-' and not
           (str[i] >= '0' and str[i] <= '9')):
            return False
 
        if str[i] == '.':
             
            # check if the char e has already
            # occurred before '.' If yes, return 0
            if flagDotOrE:
                return False
            if i + 1 > len(str):
                return False
            if (not(str[i + 1] >= '0' and
                    str[i + 1] <= '9')):
                return False
        elif str[i] == 'e':
             
            # set flagDotOrE = 1 when e is encountered.
            flagDotOrE = True
 
            # if there is no digit before e
            if (not(str[i - 1] >= '0' and
                    str[i - 1] <= '9')):
                return False
                 
            # if e is the last character
            if i + 1 > len(str):
                return False
                 
            # if e is not followed by
            # '+', '-' or a digit
            if (str[i + 1] != '+' and str[i + 1] != '-' and
               (str[i + 1] >= '0' and str[i] <= '9')):
                return False
                 
    # If the string skips all the
    # above cases, it must be a numeric string
    return True
 
# Driver Code
if __name__ == '__main__':
    str = "0.1e10"
    if valid_number(str):
        print('true')
    else:
        print('false')
 
# This code is contributed by
# chaudhary_19 (Mayank Chaudhary)

                    

C#

// C# program to check if input number
// is a valid number
using System;
 
class GFG {
    public Boolean isValidNumeric(String str)
    {
        str = str.Trim(); // trims the white spaces.
 
        if (str.Length == 0)
            return false;
 
        // if string is of length 1 and the only
        // character is not a digit
        if (str.Length == 1 && !char.IsDigit(str[0]))
            return false;
 
        // If the 1st char is not '+', '-', '.' or digit
        if (str[0] != '+' && str[0] != '-'
            && !char.IsDigit(str[0]) && str[0] != '.')
            return false;
 
        // To check if a '.' or 'e' is found in given
        // string. We use this flag to make sure that
        // either of them appear only once.
        Boolean flagDotOrE = false;
 
        for (int i = 1; i < str.Length; i++) {
            // If any of the char does not belong to
            // {digit, +, -, ., e}
            if (!char.IsDigit(str[i]) && str[i] != 'e'
                && str[i] != '.' && str[i] != '+'
                && str[i] != '-')
                return false;
 
            if (str[i] == '.') {
 
                // checks if the char 'e' has already
                // occurred before '.' If yes, return 0.
                if (flagDotOrE == true)
                    return false;
 
                // If '.' is the last character.
                if (i + 1 >= str.Length)
                    return false;
 
                // if '.' is not followed by a digit.
                if (!char.IsDigit(str[i + 1]))
                    return false;
            }
 
            else if (str[i] == 'e') {
                // set flagDotOrE = 1 when e is encountered.
                flagDotOrE = true;
 
                // if there is no digit before 'e'.
                if (!char.IsDigit(str[i - 1]))
                    return false;
 
                // If 'e' is the last Character
                if (i + 1 >= str.Length)
                    return false;
 
                // if e is not followed either by
                // '+', '-' or a digit
                if (!char.IsDigit(str[i + 1]) && str[i + 1] != '+'
                    && str[i + 1] != '-')
                    return false;
            }
        }
 
        /* If the string skips all above cases,
        then it is numeric*/
        return true;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        String input = "0.1e10";
        GFG gfg = new GFG();
        Console.WriteLine(gfg.isValidNumeric(input));
    }
}
 
// This code is contributed by Rajput-Ji

                    

Javascript

// Javascript code for the above approach
 
function valid_number(str) {
  let i = 0;
  let j = str.length - 1;
 
  // Handling whitespaces
  while (i < str.length && str[i] == ' ') {
    i++;
  }
  while (j >= 0 && str[j] == ' ') {
    j--;
  }
 
  if (i > j) {
    return false;
  }
 
  // if string is of length 1 and the only character is not a digit
  if (i == j && !(str[i] >= '0' && str[i] <= '9')) {
    return false;
  }
 
  // If the 1st char is not '+', '-', '.' or digit
  if (
    str[i] != '.' &&
    str[i] != '+' &&
    str[i] != '-' &&
    !(str[i] >= '0' && str[i] <= '9')
  ) {
    return false;
  }
 
  // To check if a '.' or 'e' is found in given string.
  // We use this flag to make sure that either of them appear only once.
  let flagDotOrE = false;
 
  for (let k = 0; k <= j; k++) {
    // If any of the char does not belong to {digit, +, -,., e}
    if (
      str[k] != 'e' &&
      str[k] != '.' &&
      str[k] != '+' &&
      str[k] != '-' &&
      !(str[k] >= '0' && str[k] <= '9')
    ) {
      return false;
    }
 
    if (str[k] == '.') {
      // check if the char e has already occurred before '.'
      // If yes, return false
      if (flagDotOrE) {
        return false;
      }
      if (k + 1 > str.length) {
        return false;
      }
      if (!(str[k + 1] >= '0' && str[k + 1] <= '9')) {
        return false;
      }
    } else if (str[k] == 'e') {
      // set flagDotOrE = true when 'e' is encountered.
      flagDotOrE = true;
 
      // if there is no digit before 'e'
      if (!(str[k - 1] >= '0' && str[k - 1] <= '9')) {
        return false;
      }
 
      // if 'e' is the last character
      if (k + 1 > str.length) {
        return false;
      }
 
      // if 'e' is not followed by '+', '-' or a digit
      if (
        str[k + 1] != '+' &&
        str[k + 1] != '-' &&
        !(str[k + 1] >= '0' && str[k + 1] <= '9')
      ) {
        return false;
      }
    }
  }
 
  // If the string skips all the above cases, it must be a numeric string
  return true;
}
 
// Driver Code
let str = "0.1e10";
if (valid_number(str)) {
  console.log("true");
} else {
  console.log("false");
}
 
// This code is contributed by adityashatmfh

                    

Output:

true

Time Complexity: O(n)  



Last Updated : 09 Mar, 2023
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