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

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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 a 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 1:

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)
Auxiliary Space: O(1)

Approach 2: Dynamic Programming

We can solve this problem using Dynamic Programming (DP). We can create a DP array of size N+1, where DP[i] stores whether it is possible to represent i as the sum of numbers having the digit D in them or not. We can initialize DP[0] as true since it is always possible to represent 0 as the sum of no numbers.

Then, for each i from 1 to N, we can check if i contains the digit D. If it does, then we can mark DP[i] as true. Otherwise, we can iterate over all possible j < i such that DP[j] is true, and check if i-j contains the digit D. If it does, then we can mark DP[i] as true.

At the end, we can check if DP[N] is true or not. If it is, then we can say that N can be represented as the sum of numbers having the digit D in them. Otherwise, we can say that it is not possible.

C++

#include <cstring>
#include <iostream>
 
using namespace std;
 
bool check(int N, int D)
{
    // Create DP array
    bool DP[N + 1];
 
    // Initialize DP[0]
    memset(DP, false, sizeof(DP));
    DP[0] = true;
 
    // Fill DP array
    for (int i = 1; i <= N; i++) {
        // Check if i contains digit D
        if (i % 10 == D || i / 10 == D) {
            DP[i] = true;
            continue;
        }
        // Iterate over all j < i such that DP[j] is true
        for (int j = 1; j < i; j++) {
            if (DP[j]
                && ((i - j) % 10 == D
                    || (i - j) / 10 == D)) {
                DP[i] = true;
                break;
            }
        }
    }
    // Return DP[N]
    return DP[N];
}
 
// Driver Code
int main()
{
    int N = 24;
    int D = 7;
    if (check(N, D)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
 
    return 0;
}

Java

import java.util.Arrays;
 
public class Main {
     
    static boolean check(int N, int D) {
        // Create DP array
        boolean[] DP = new boolean[N + 1];
 
        // Initialize DP[0]
        Arrays.fill(DP, false);
        DP[0] = true;
 
        // Fill DP array
        for (int i = 1; i <= N; i++) {
            // Check if i contains digit D
            if (i % 10 == D || i / 10 == D) {
                DP[i] = true;
                continue;
            }
            // Iterate over all j < i such that DP[j] is true
            for (int j = 1; j < i; j++) {
                if (DP[j]
                    && ((i - j) % 10 == D
                        || (i - j) / 10 == D)) {
                    DP[i] = true;
                    break;
                }
            }
        }
        // Return DP[N]
        return DP[N];
    }
 
    // Driver Code
    public static void main(String[] args) {
        int N = 24;
        int D = 7;
        if (check(N, D)) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }
}

Python3

def check(N, D):
    # Create DP array
    DP = [False] * (N + 1)
 
    # Initialize DP[0]
    DP[0] = True
 
    # Fill DP array
    for i in range(1, N+1):
        # Check if i contains digit D
        if i % 10 == D or i // 10 == D:
            DP[i] = True
            continue
        # Iterate over all j < i such that DP[j] is true
        for j in range(1, i):
            if DP[j] and ((i - j) % 10 == D or (i - j) // 10 == D):
                DP[i] = True
                break
 
    # Return DP[N]
    return DP[N]
 
# Driver code
N = 24
D = 7
if check(N, D):
    print("Yes")
else:
    print("No")

C#

using System;
 
namespace ConsoleApp {
class Program {
    static bool Check(int N, int D)
    {
        // Create DP array
        bool[] DP = new bool[N + 1];
        // Initialize DP[0]
        for (int i = 0; i < DP.Length; i++) {
            DP[i] = false;
        }
        DP[0] = true;
 
        // Fill DP array
        for (int i = 1; i <= N; i++) {
            // Check if i contains digit D
            if (i % 10 == D || i / 10 == D) {
                DP[i] = true;
                continue;
            }
 
            // Iterate over all j < i such that DP[j] is
            // true
            for (int j = 1; j < i; j++) {
                if (DP[j]
                    && ((i - j) % 10 == D
                        || (i - j) / 10 == D)) {
                    DP[i] = true;
                    break;
                }
            }
        }
        // Return DP[N]
        return DP[N];
    }
 
    // Driver Code
    static void Main(string[] args)
    {
        int N = 24;
        int D = 7;
        if (Check(N, D)) {
            Console.WriteLine("Yes");
        }
        else {
            Console.WriteLine("No");
        }
 
        Console.ReadKey();
    }
}
}
//This code is contributed by sarojmcy2e

Javascript

function check(N, D) {
  let DP = new Array(N + 1).fill(false);
  DP[0] = true;
 
  for (let i = 1; i <= N; i++) {
    if (i % 10 === D || Math.floor(i / 10) === D) {
      DP[i] = true;
      continue;
    }
 
    for (let j = 1; j < i; j++) {
      if (DP[j] && ((i - j) % 10 === D || Math.floor((i - j) / 10) === D)) {
        DP[i] = true;
        break;
      }
    }
  }
 
  return DP[N];
}
 
let N = 24;
let D = 7;
if (check(N, D)) {
  console.log("Yes");
} else {
  console.log("No");
}

Output

Yes

Time Complexity: O(N^(1/3)log(N)), where N is the input number.
Auxiliary Space: O(N^(1/3)), since we are storing all perfect cubes up to N^(1/3) in the map.

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