Dada una string str que consta de caracteres alfanuméricos, la tarea es verificar si la string contiene todos los dígitos del 1 al 9. La string puede contener otros caracteres, pero debe contener todos los dígitos del 1 al 9.
Ejemplos:
Entrada: str = “Geeks12345for69708”
Salida: Sí
Explicación: Todos los dígitos del 0 al 9 están presentes en la string dada.Entrada: str = “Amazing1234”
Salida: No
Explicación: No todos los dígitos están presentes en la string.
Enfoque: Cree una array de frecuencias para marcar la frecuencia de cada uno de los dígitos del 0 al 9. Finalmente, recorra la array de frecuencias y si hay algún dígito que no está presente en la string dada, la respuesta será «No», de lo contrario, la la respuesta será “Sí”.
A continuación se muestra la implementación del enfoque anterior:
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>
Producción:
Yes
Time Complexity - O(n)