Check if N can be represented as sum of positive integers containing digit D at least once

Given a positive integer N and a digit D, the task is to check if N can be represented as a sum of positive integers containing the digit D at least once. If it is possible to represent N in such format, then print “Yes”. Otherwise, print “No”.

Examples:

Input: N = 24, D = 7
Output: Yes
Explanation: The value 24 can be represented as 17 + 7, both containing the digit 7.

Input: N = 27 D = 2
Output: Yes

Approach: Follow the steps to solve the problem:

• Check if the given N contains digit D or not. If found to be true, then print “Yes”.
• Otherwise, iterate until N is greater than 0 and perform the following steps:
  • Subtracting the value D from the value of N.
  • Check if the updated value of N contains digit D or not. If found to be true, then print “Yes” and break out of the loop.
• After completing the above steps, if none of the above conditions satisfies, then print “No”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <iostream>
using namespace std;

// Function to check if N contains
// digit D in it
bool findDigit(int N, int D)
{
    // Iterate until N is positive
    while (N > 0) {

        // Find the last digit
        int a = N % 10;

        // If the last digit is the
        // same as digit D
        if (a == D) {
            return true;
        }

        N /= 10;
    }

    // Return false
    return false;
}

// Function to check if the value of
// N can be represented as sum of
// integers having digit d in it
bool check(int N, int D)
{
    // Iterate until N is positive
    while (N > 0) {

        // Check if N contains digit
        // D or not
        if (findDigit(N, D) == true) {
            return true;
        }

        // Subtracting D from N
        N -= D;
    }

    // Return false
    return false;
}

// Driver Code
int main()
{
    int N = 24;
    int D = 7;
    if (check(N, D)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }

    return 0;
}

Java

// Java approach for the above approach
import java.util.*;

class GFG{

// Function to check if N contains
// digit D in it
static boolean findDigit(int N, int D)
{

    // Iterate until N is positive
    while (N > 0)
    {

        // Find the last digit
        int a = N % 10;

        // If the last digit is the
        // same as digit D
        if (a == D)
        {
            return true;
        }
        N /= 10;
    }

    // Return false
    return false;
}

// Function to check if the value of
// N can be represented as sum of
// integers having digit d in it
static boolean check(int N, int D)
{

    // Iterate until N is positive
    while (N > 0)
    {

        // Check if N contains digit
        // D or not
        if (findDigit(N, D) == true)
        {
            return true;
        }

        // Subtracting D from N
        N -= D;
    }

    // Return false
    return false;
}

// Driver Code
public static void main(String[] args)
{
    int N = 24;
    int D = 7;

    if (check(N, D))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}

// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach

# Function to check if N contains
# digit D in it
def findDigit(N, D):

    # Iterate until N is positive
    while (N > 0):

        # Find the last digit
        a = N % 10

        # If the last digit is the
        # same as digit D
        if (a == D):
            return True

        N /= 10

    # Return false
    return False

# Function to check if the value of
# N can be represented as sum of
# integers having digit d in it
def check(N, D):

    # Iterate until N is positive
    while (N > 0):

        # Check if N contains digit
        # D or not
        if (findDigit(N, D) == True):
            return True

        # Subtracting D from N
        N -= D

    # Return false
    return False

# Driver Code
if __name__ == '__main__':

    N = 24
    D = 7

    if (check(N, D)):
        print("Yes")
    else:
        print("No")

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;

class GFG{

// Function to check if N contains
// digit D in it
static bool findDigit(int N, int D)
{

    // Iterate until N is positive
    while (N > 0)
    {

        // Find the last digit
        int a = N % 10;

        // If the last digit is the
        // same as digit D
        if (a == D)
        {
            return true;
        }
        N /= 10;
    }

    // Return false
    return false;
}

// Function to check if the value of
// N can be represented as sum of
// integers having digit d in it
static bool check(int N, int D)
{

    // Iterate until N is positive
    while (N > 0)
    {

        // Check if N contains digit
        // D or not
        if (findDigit(N, D) == true)
        {
            return true;
        }

        // Subtracting D from N
        N -= D;
    }

    // Return false
    return false;
}


// Driver Code
public static void Main()
{
    int N = 24;
    int D = 7;

    if (check(N, D))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }

}
}

// This code is contributed by code_hunt.

Javascript

<script>

// javascript program for the above approach

// Function to check if N contains
// digit D in it
function findDigit(N, D)
{

    // Iterate until N is positive
    while (N > 0)
    {

        // Find the last digit
        let a = N % 10;

        // If the last digit is the
        // same as digit D
        if (a == D)
        {
            return true;
        }
        N = Math.floor(N / 10);
    }

    // Return false
    return false;
}

// Function to check if the value of
// N can be represented as sum of
// integers having digit d in it
function check(N, D)
{

    // Iterate until N is positive
    while (N > 0)
    {

        // Check if N contains digit
        // D or not
        if (findDigit(N, D) == true)
        {
            return true;
        }

        // Subtracting D from N
        N -= D;
    }

    // Return false
    return false;
}

// Driver Code

    let N = 24;
    let D = 7;

    if (check(N, D))
    {
        document.write("Yes");
    }
    else
    {
        document.write("No");
    }
</script>

Output:

Yes

Time Complexity: O(N)
Space Complexity: O(1)