Find the Number which contain the digit d

Given two integer number n and d. The task is to find the number between 0 to n which contain the specific digit d.

Examples:

Input : n = 20
        d = 5
Output : 5 15

Input : n = 50
        d = 2
Output : 2 12 20 21 22 23 24 25 26 27 28 29 32 42 

Approach 1:
Take a loop from 0 to n and check each number one by one, if the number contains digit d then print it otherwise increase the number. Continue this process until loop ended.

C++

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// CPP program to print the number which
// contain the digit d from 0 to n
#include <bits/stdc++.h>
using namespace std;
  
// Returns true if d is present as digit
// in number x.
bool isDigitPresent(int x, int d)
{
    // Breal loop if d is present as digit
    while (x > 0)
    {
        if (x % 10 == d)
            break;
  
        x = x / 10;
    }
  
    // If loop broke
    return (x > 0);
}
  
// function to display the values
void printNumbers(int n, int d)
{
    // Check all numbers one by one
    for (int i = 0; i <= n; i++)
  
        // checking for digit
        if (i == d || isDigitPresent(i, d))
            cout << i << " ";
}
  
// Driver code
int main()
{
    int n = 47, d = 7;
    printNumbers(n, d);
    return 0;
}

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Java

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// Java program to print the number which
// contain the digit d from 0 to n
  
class GFG
{
    // Returns true if d is present as digit
    // in number x.
    static boolean isDigitPresent(int x, int d)
    {
        // Breal loop if d is present as digit
        while (x > 0)
        {
            if (x % 10 == d)
                break;
  
            x = x / 10;
        }
  
        // If loop broke
        return (x > 0);
    }
  
    // function to display the values
    static void printNumbers(int n, int d)
    {
        // Check all numbers one by one
        for (int i = 0; i <= n; i++)
  
            // checking for digit
            if (i == d || isDigitPresent(i, d))
                System.out.print(i + " ");
    }
  
  
    // Driver code
    public static void main(String[] args)
    {
        int n = 47, d = 7;
        printNumbers(n, d);
    }
}

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Python 3

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# Python3 program to print the number which
# contain the digit d from 0 to n
  
# Returns true if d is present as digit
# in number x.
def isDigitPresent(x, d):
  
    # Breal loop if d is present as digit
    while (x > 0):
      
        if (x % 10 == d):
            break
  
        x = x / 10
      
  
    # If loop broke
    return (x > 0)
  
  
# function to display the values
def printNumbers(n, d):
  
    # Check all numbers one by one
    for i in range(0, n+1):
  
        # checking for digit
        if (i == d or isDigitPresent(i, d)):
            print(i,end=" ")
  
# Driver code
n = 47
d = 7
print("The number of values are")
printNumbers(n, d)
#This code is contributed by
#Smitha Dinesh Semwal

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C#

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// C# program to print the number which
// contain the digit d from 0 to n
using System; 
  
class GFG {
      
    // Returns true if d is present as digit
    // in number x.
    static bool isDigitPresent(int x, int d)
    {
          
        // Breal loop if d is present as digit
        while (x > 0)
        {
            if (x % 10 == d)
                break;
  
            x = x / 10;
        }
  
        // If loop broke
        return (x > 0);
    }
  
    // function to display the values
    static void printNumbers(int n, int d)
    {
          
        // Check all numbers one by one
        for (int i = 0; i <= n; i++)
  
            // checking for digit
            if (i == d || isDigitPresent(i, d))
                Console.Write(i + " ");
    }
  
  
    // Driver code
    public static void Main()
    {
        int n = 47, d = 7;
          
        printNumbers(n, d);
    }
}
  
// This code contribute by parashar.

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PHP

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<?php
// PHP program to print the number which
// contain the digit d from 0 to n
  
// Returns true if d is present as digit
// in number x.
function isDigitPresent($x, $d)
{
      
    // Breal loop if d is 
    // present as digit
    while ($x > 0)
    {
        if ($x % 10 == $d)
            break;
  
        $x = $x / 10;
    }
  
    // If loop broke
    return ($x > 0);
}
  
// function to display the values
function printNumbers($n, $d)
{
      
    // Check all numbers one by one
    for ($i = 0; $i <= $n; $i++)
  
        // checking for digit
        if ($i == $d || isDigitPresent($i, $d))
            echo $i , " ";
}
  
    // Driver Code
    $n = 47; 
    $d = 7;
    printNumbers($n, $d);
      
// This code contributed by ajit.
?>

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Output:

The number of values are
7 17 27 37 47

Approach 2:
This approach uses every number as a String and checks digit is present or not. This approach use of String.indexOf() function to check if the character is present in the string or not.

String.indexOf() >= 0 means chaaracter is present

and String.indexOf() = -1 means character is not present

C++

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// CPP program to print the number which 
// contain the digit d from 0 to n
#include<bits/stdc++.h>
using namespace std;
  
// function to display the values
void printNumbers(int n, int d)
{
          
    // Converting d to character
    string st = "";
    st += to_string(d);
    char ch = st[0];
          
    string p = "";
    p += ch;
          
    // Loop to check each digit one by one.
    for (int i = 0; i <= n; i++)
    {
              
        // initialize the string
        st = "";
        st = st + to_string(i);
        int idx = st.find(p);
              
        // checking for digit
        if (i == d || idx!=-1)       
            cout << (i) << " ";
    }
}
  
// Driver code
int main()
{
    int n = 100, d = 5;
    printNumbers(n, d);
}
      
// This code is contributed by
// Surendra_Gangwar

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Java

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// Java program to print the number which 
// contain the digit d from 0 to n
  
public class GFG {
  
    // function to display the values
    static void printNumbers(int n, int d)
    {
          
        // Converting d to character
        String st = "" + d;
        char ch = st.charAt(0);
          
        // Loop to check each digit one by one.
        for (int i = 0; i <= n; i++) {
              
            // initialize the string
            st = "";
            st = st + i;
  
            // checking for digit
            if (i == d || st.indexOf(ch) >= 0)                
                System.out.print(i + " ");
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int n = 100, d = 5;
        printNumbers(n, d);
    }
}

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Python3

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# Python 3 program to print the number 
# which contain the digit d from 0 to n
  
def index(st, ch):
    for i in range(len(st)):
        if(st[i] == ch):
            return i;
    return -1
      
# function to display the values
def printNumbers(n, d):
      
# Converting d to character
    st = "" + str(d)
    ch = st[0]
  
    # Loop to check each digit one by one.
    for i in range(0, n + 1, 1):
          
        # initialize the string
        st = ""
        st = st + str(i)
          
        # checking for digit
        if (i == d or index(st, ch) >= 0):
            print(i, end = " ")
  
# Driver code
if __name__ == '__main__':
    n = 100
    d = 5
    printNumbers(n, d)
  
# This code is contributed by
# Shashank_Sharma

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C#

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// C# program to print the number which 
// contain the digit d from 0 to n 
using System;
  
class GFG 
  
    // function to display the values 
    static void printNumbers(int n, int d) 
    
          
        // Converting d to character 
        String st = "" + d; 
        char ch = st[0]; 
          
        // Loop to check each digit one by one. 
        for (int i = 0; i < n; i++) 
        
              
            // initialize the string 
            st = ""
            st = st + i; 
  
            // checking for digit 
            if (i == d || st.IndexOf(ch) >= 0)             
                Console.Write(i + " "); 
        
    
  
    // Driver code 
    public static void Main() 
    
        int n = 100, d = 5; 
        printNumbers(n, d); 
    
  
/* This code contributed by PrinciRaj1992 */

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Output:

5 15 25 35 45 50 51 52 53 54 55 56 57 58 59 65 75 85 95


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