Check if given string contains all the digits

Given a string str consisting of alphanumeric characters, the task is to check whether the string contains all the digits from 1 to 9.  The string can contain other characters, but it should contain all the digits from 1 to 9.

Examples:

Input: str = “Geeks12345for69708” 
Output: Yes 
Explanation: All the digits from 0 to 9 are present in the given string.

Input: str = “Amazing1234” 
Output: No 
Explanation: All the digits are not present in the string.

Approach: Create a frequency array to mark the frequency of each of the digits from 0 to 9. Finally, traverse the frequency array and if there is any digit that is not present in the given string then the answer will be “No” else the answer will be “Yes”.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

const int MAX = 10;

// Function that returns true
// if ch is a digit
bool isDigit(char ch)
{
    if (ch >= '0' && ch <= '9')
        return true;
    return false;
}

// Function that returns true
// if str contains all the
// digits from 0 to 9
bool allDigits(string str, int len)
{

    // To mark the present digits
    bool present[MAX] = { false };

    // For every character of the string
    for (int i = 0; i < len; i++) {

        // If the current character is a digit
        if (isDigit(str[i])) {

            // Mark the current digit as present
            int digit = str[i] - '0';
            present[digit] = true;
        }
    }

    // For every digit from 0 to 9
    for (int i = 0; i < MAX; i++) {

        // If the current digit is
        // not present in str
        if (!present[i])
            return false;
    }

    return true;
}

// Driver code
int main()
{
    string str = "Geeks12345for69708";
    int len = str.length();

    if (allDigits(str, len))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

static int MAX = 10;

// Function that returns true
// if ch is a digit
static boolean isDigit(char ch)
{
    if (ch >= '0' && ch <= '9')
        return true;
    return false;
}

// Function that returns true
// if str contains all the
// digits from 0 to 9
static boolean allDigits(String str, int len)
{

    // To mark the present digits
    boolean []present = new boolean[MAX];

    // For every character of the String
    for (int i = 0; i < len; i++)
    {

        // If the current character is a digit
        if (isDigit(str.charAt(i)))
        {

            // Mark the current digit as present
            int digit = str.charAt(i) - '0';
            present[digit] = true;
        }
    }

    // For every digit from 0 to 9
    for (int i = 0; i < MAX; i++)
    {

        // If the current digit is
        // not present in str
        if (!present[i])
            return false;
    }

    return true;
}

// Driver code
public static void main(String[] args)
{
    String str = "Geeks12345for69708";
    int len = str.length();

    if (allDigits(str, len))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
MAX = 10

# Function that returns true
# if ch is a digit
def isDigit(ch):
    ch = ord(ch)
    if (ch >= ord('0') and ch <= ord('9')):
        return True
    return False

# Function that returns true
# if st contains all the
# digits from 0 to 9
def allDigits(st, le):

    # To mark the present digits
    present = [False for i in range(MAX)]

    # For every character of the string
    for i in range(le):

        # If the current character is a digit
        if (isDigit(st[i])):

            # Mark the current digit as present
            digit = ord(st[i]) - ord('0')
            present[digit] = True

    # For every digit from 0 to 9
    for i in range(MAX):

        # If the current digit is
        # not present in st
        if (present[i] == False):
            return False

    return True

# Driver code
st = "Geeks12345for69708"
le = len(st)

if (allDigits(st, le)):
    print("Yes")
else:
    print("No")

# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;

class GFG
{

static int MAX = 10;

// Function that returns true
// if ch is a digit
static bool isDigit(char ch)
{
    if (ch >= '0' && ch <= '9')
        return true;
    return false;
}

// Function that returns true
// if str contains all the
// digits from 0 to 9
static bool allDigits(String str, int len)
{

    // To mark the present digits
    bool []present = new bool[MAX];

    // For every character of the String
    for (int i = 0; i < len; i++)
    {

        // If the current character is a digit
        if (isDigit(str[i]))
        {

            // Mark the current digit as present
            int digit = str[i] - '0';
            present[digit] = true;
        }
    }

    // For every digit from 0 to 9
    for (int i = 0; i < MAX; i++)
    {

        // If the current digit is
        // not present in str
        if (!present[i])
            return false;
    }

    return true;
}

// Driver code
public static void Main(String[] args)
{
    String str = "Geeks12345for69708";
    int len = str.Length;

    if (allDigits(str, len))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript implementation of the approach

let MAX = 10;

// Function that returns true
// if ch is a digit
function isDigit(ch)
{
    if (ch >= '0' && ch <= '9')
        return true;
    return false;
}

// Function that returns true
// if str contains all the
// digits from 0 to 9
function allDigits(str, len)
{

    // To mark the present digits
    let present = Array.from({length: MAX}, (_, i) => 0);

    // For every character of the String
    for (let i = 0; i < len; i++)
    {

        // If the current character is a digit
        if (isDigit(str[i]))
        {

            // Mark the current digit as present
            let digit = str[i] - '0';
            present[digit] = true;
        }
    }

    // For every digit from 0 to 9
    for (let i = 0; i < MAX; i++)
    {

        // If the current digit is
        // not present in str
        if (!present[i])
            return false;
    }

    return true;
}

// Driver code

      let str = "Geeks12345for69708";
    let len = str.length;

    if (allDigits(str, len))
        document.write("Yes");
    else
        document.write("No");

</script>

Output:

Yes

Time Complexity: O(n)
Auxiliary Space: O(1), no extra space is required, so it is a constant.