# 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++

 `// CPP program to print the number which ` `// contain the digit d from 0 to n ` `#include ` `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; ` `} `

## Java

 `// 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); ` `    ``} ` `} `

## Python 3

 `# 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 `

## C#

 `// 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. `

## PHP

 ` 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. ` `?> `

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

## Java

 `// 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); ` `    ``} ` `} `

## Python3

# 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

# 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

Output:

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

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