# 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 using namespace std; const int MAX = 10; // Function that returns true// if ch is a digitbool 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 9bool 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 codeint main(){    string str = "Geeks12345for69708";    int len = str.length();     if (allDigits(str, len))        cout << "Yes";    else        cout << "No";     return 0;}

## Java

 // Java implementation of the approachclass GFG{ static int MAX = 10; // Function that returns true// if ch is a digitstatic 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 9static 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 codepublic 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 approachMAX = 10 # Function that returns true# if ch is a digitdef 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 9def 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 codest = "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 approachusing System; class GFG{ static int MAX = 10; // Function that returns true// if ch is a digitstatic 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 9static 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 codepublic 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



Output:
Yes

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

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