Numbers which contain the digit d

Last Updated : 27 May, 2026

Given two integers n and d. Return an array containing all the numbers between 0 to n that contain the specific digit d.

Examples: 

Input: n = 20, d = 5
Output: [5, 15]
Explanation: For number till 20, 5 appears in 5 itself and 15.

Input: n = 50, d = 2
Output: [2, 12, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 32, 42]
Explanation: For number till 50, 2 appears in all these numbers.

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Table of Content

[Naive Approach] String Conversion - O(n⨯logn) Time and O(logn) Auxiliary Space

Convert every number i into a string and use built-in string search functions to check if the character 'd' exists inside it.

C++ 
#include <iostream>
#include <string>
#include <vector>

using namespace std;

// helper function to check if a number contains the digit d using string method
bool isDigitPresent(int num, int d)
{

    // convert number and digit to strings
    string numStr = to_string(num);
    string dStr = to_string(d);

    // check if the digit string is found inside the number string
    if (numStr.find(dStr) != string::npos)
    {
        return true;
    }

    return false;
}

// function to return an array of all valid numbers
vector<int> findNumbers(int n, int d)
{

    vector<int> res;

    // check all numbers one by one
    for (int i = 0; i <= n; i++)
    {

        // checking for digit
        if (isDigitPresent(i, d))
        {
            res.push_back(i);
        }
    }

    return res;
}

int main()
{

    int n = 47, d = 7;
    vector<int> v = findNumbers(n, d);
    if (!v.empty())
    {
        for (int it : v)
        {
            cout << it << " ";
        }
        cout << endl;
    }
    else
        cout << -1 << endl;

    return 0;
}
Java 
import java.util.*;

public class Solution {

    // helper function to check if a number contains the
    // digit d using string method
    public boolean isDigitPresent(int num, int d)
    {

        // convert number and digit to strings
        String numStr = String.valueOf(num);
        String dStr = String.valueOf(d);

        // check if number string contains the digit string
        return numStr.contains(dStr);
    }

    // function to return an array of all valid numbers
    public List<Integer> findNumbers(int n, int d)
    {

        List<Integer> res = new ArrayList<>();

        // check all numbers one by one
        for (int i = 0; i <= n; i++) {

            // checking for digit
            if (isDigitPresent(i, d)) {
                res.add(i);
            }
        }

        return res;
    }

    public static void main(String[] args)
    {

        int n = 47;
        int d = 7;

        Solution sol = new Solution();

        List<Integer> v = sol.findNumbers(n, d);

        if (!v.isEmpty()) {
            for (int it : v) {
                System.out.print(it + " ");
            }
            System.out.println();
        }
        else {
            System.out.println("-1");
        }
    }
}
Python 
class Solution:

    # helper function to check if a number contains the digit d using string method
    def is_digit_present(self, num, d):

        # convert number and digit to string and check inclusion
        return str(d) in str(num)

    # function to return a list of all valid numbers
    def findNumbers(self, n, d):

        res = []

        # check all numbers one by one
        for i in range(n + 1):

            # checking for digit
            if self.is_digit_present(i, d):
                res.append(i)

        return res


# driver code
if __name__ == '__main__':

    n = 47
    d = 7

    sol = Solution()

    v = sol.findNumbers(n, d)

    if v:
        print(" ".join(map(str, v)))
    else:
        print("-1")
C# 
using System;
using System.Collections.Generic;

class Solution {

    // helper function to check if a number contains the
    // digit d using string method
    public bool IsDigitPresent(int num, int d)
    {

        // convert number and digit to string and check
        // inclusion
        return num.ToString().Contains(d.ToString());
    }

    // function to return a list of all valid numbers
    public List<int> findNumbers(int n, int d)
    {

        List<int> res = new List<int>();

        // check all numbers one by one
        for (int i = 0; i <= n; i++) {

            // checking for digit
            if (IsDigitPresent(i, d)) {
                res.Add(i);
            }
        }

        return res;
    }

    // driver code
    static void Main()
    {

        int n = 47;
        int d = 7;

        Solution sol = new Solution();

        List<int> v = sol.findNumbers(n, d);
        if (v.Count > 0) {
            Console.WriteLine(string.Join(" ", v));
        }
        else {

            Console.WriteLine("-1");
        }
    }
}
JavaScript 
class Solution {

    // helper function to check if a number contains the
    // digit d using string method
    isDigitPresent(num, d)
    {

        // convert number and digit to string and check
        // inclusion
        return num.toString().includes(d.toString());
    }

    // function to return an array of all valid numbers
    findNumbers(n, d)
    {

        const res = [];

        // check all numbers one by one
        for (let i = 0; i <= n; i++) {

            // checking for digit
            if (this.isDigitPresent(i, d)) {
                res.push(i);
            }
        }

        return res;
    }
}

// small driver code
const n = 47;
const d = 7;

const sol = new Solution();

const v = sol.findNumbers(n, d);
if (v.length > 0) {
    console.log(v.join(" "));
}
else {
    console.log("-1");
}

Output
7 17 27 37 47

[Expected Approach] Digit Extraction - O(n⨯logn) Time and O(1) Auxiliary Space

We can iterate through all numbers from 0 up to n. For every number, we can repeatedly extract its last digit using the modulo operator (%10). If that extracted digit matches our target digit d, we immediately add the number to our result list. If it doesn't match, we strip that last digit away using integer division (/10) and check the next digit, continuing until the number becomes 0.

  • Create an empty array to store the valid numbers.
  • Loop through every integer i from 0 to n.
  • For each integer i, create a temporary variable num = i.
  • Run a while (num > 0) loop.
  • Check if the last digit matches d (num % 10 == d). If it does, add i to the result list and break the loop .
  • Otherwise, remove the last digit by setting num = num / 10.
C++ 
#include <bits/stdc++.h>
using namespace std;

// returns true if d is present as digit
// helper function to check if a number contains the digit d
bool isDigitPresent(int num, int d) {

    // if both the number and the target digit are 0
    if (num == 0 && d == 0) {
        return true;
    }

    // extract digits one by one
    while (num > 0) {
        if (num % 10 == d) {
            return true;
        }
        num /= 10;
    }

    return false;
}
    
// function to display the values
vector<int> findNumbers(int n, int d)
{
    vector<int> res;

    // check all numbers one by one
    for (int i = 0; i <= n; i++) {

        // checking for digit
        if (isDigitPresent(i, d))
            res.push_back(i);
    }

    return res;
}

// Driver code
int main()
{
    int n = 47, d = 7;
    vector<int> v = findNumbers(n, d);
    if(v.size()) for(auto it:v) cout<<it<<" ";
    else cout<<-1<<endl;
    return 0;
}
Java 
import java.util.*;
public class Solution {
    
    // helper function to check if a number contains the
    // digit d
    public boolean isDigitPresent(int num, int d)
    {

        // if both the number and the target digit are 0
        if (num == 0 && d == 0) {
            return true;
        }

        // extract digits one by one
        while (num > 0) {
            if (num % 10 == d) {
                return true;
            }
            num /= 10;
        }

        return false;
    }

    // function to display the values
    public List<Integer> findNumbers(int n, int d)
    {
        List<Integer> res = new ArrayList<>();

        // check all numbers one by one
        for (int i = 0; i <= n; i++) {

            // checking for digit
            if (isDigitPresent(i, d))
                res.add(i);
        }

        return res;
    }

    // Driver code  
    public static void main(String[] args)
    {
        int n = 47, d = 7; 
        Solution sol = new Solution();
        List<Integer> v = sol.findNumbers(n, d);

        if (v.size() > 0) {
            for (int it : v) {
                System.out.print(it + " ");
            }
            System.out.println();
        }
        else {
            System.out.println(-1);
        }
    }
}
Python 
class Solution:

    # returns true if d is present as digit
    # helper function to check if a number contains the digit d
    def isDigitPresent(self, num, d):

        # if both the number and the target digit are 0
        if num == 0 and d == 0:
            return True

        # extract digits one by one
        while num > 0:
            if num % 10 == d:
                return True
            num //= 10  
            
        return False

    # function to display the values
    def findNumbers(self, n, d):
        res = []

        # check all numbers one by one
        for i in range(n + 1):

            # checking for digit
            if self.isDigitPresent(i, d):
                res.append(i)

        return res


# Driver code
if __name__ == '__main__':
    n = 47
    d = 7
    sol = Solution()
    v = sol.findNumbers(n, d)

    if len(v) > 0:
        for it in v:
            print(it, end=" ")
        print()
    else:
        print("-1")
C# 
using System;
using System.Collections.Generic;
class Solution {

    // helper function to check if a number contains the
    // digit d
    public bool isDigitPresent(int num, int d)
    {

        // if both the number and the target digit are 0
        if (num == 0 && d == 0) {
            return true;
        }

        // extract digits one by one
        while (num > 0) {
            if (num % 10 == d) {
                return true;
            }
            num /= 10;
        }

        return false;
    }

    // function to display the values
    public List<int> findNumbers(int n, int d)
    {
        List<int> res = new List<int>();

        // check all numbers one by one
        for (int i = 0; i <= n; i++) {

            // checking for digit
            if (isDigitPresent(i, d))
                res.Add(i);
        }

        return res;
    }

    // Driver code placed directly inside the Solution class
    static void Main()
    {
        int n = 47, d = 7;

        Solution sol = new Solution();
        List<int> v = sol.findNumbers(n, d);

        if (v.Count > 0) {
            foreach(var it in v)
            {
                Console.Write(it + " ");
            }
            Console.WriteLine();
        }
        else {
            Console.WriteLine(-1);
        }
    }
}
JavaScript 
class Solution {

    // helper function to check if a number contains the
    // digit d
    isDigitPresent(num, d)
    {

        // if both the number and the target digit are 0
        if (num === 0 && d === 0) {
            return true;
        }

        // extract digits one by one
        while (num > 0) {
            if (num % 10 === d) {
                return true;
            }
            num = Math.floor(num / 10);
        }

        return false;
    }

    // function to display the values
    findNumbers(n, d)
    {
        let res = [];

        // check all numbers one by one
        for (let i = 0; i <= n; i++) {

            // checking for digit
            if (this.isDigitPresent(i, d))
                res.push(i);
        }

        return res;
    }
}

// Driver code
let n = 47, d = 7;
let sol = new Solution();
let v = sol.findNumbers(n, d);

if (v.length > 0) {
    let output = "";
    for (let it of v) {
        output += it + " ";
    }
    console.log(output);
}
else {
    console.log(-1);
}

Output
7 17 27 37 47


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