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Check if a number can be represented as the sum of numbers with at least one digit equal to K

  • Difficulty Level : Medium
  • Last Updated : 26 Oct, 2021

Given integers N and K, the task is to check if a number can be represented as the sum of numbers that have at least one digit equal to K.

Example:

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Input: N = 68, K = 7
Output: YES
Explanation: 68 = (27 + 17 + 17 + 7). Each number has atleast one digit equal to 7.

Input: N = 23, K = 3
Output: YES
Explanation: 23 itself contains a digit equal to 3.



 

Approach: The given problem can be solved by using simple concepts of math. Follow the steps below to solve the problem:

  • Initialize a variable temp to k, and also take a counter say count, assign it with 0
  • Iterate till the last digits of temp and N are not equal and at each iteration
    • Increment the value of temp by k
    • Keep a count of iterations and break the loop if the count becomes greater than 10
  • Check if the last digits of temp and N are equal, and if the value of temp <= N:
    • If the above condition is satisfied return true
    • Else return false
  • Also if k * 10 <= N, return true
  • Else return false

Below is the implementation of the above approach:

C++




// C++ implementation for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to Check if a number can
// be equal to sum of numbers having
// at least one digit equal to k
bool checkEqualtoSum(int N, int k)
{
    // Temporary variable to
    // store k
    int temp = k;
     
    // Variable for count
    int count = 0;
     
    // Iterating till count is less or
    // equal to 10 and N % 10 is not
    // equal to temp % 10
    while(count <= 10 &&
                  N % 10 != temp % 10) {
       
        temp += k;
        count++;
    }
     
    // If N % 10 is equal to temp % 10
    // and temp is less or equal to N,
    // return true
    if(N % 10 == temp % 10 && temp <= N)
        return true;
     
    // If k * 10 <= N, return true
    if(k * 10 <= N)
        return true;
     
    // Else return false
    return false;
}
 
// Driver Code
int main()
{
    int N = 68;
    int K = 7;
 
      // Call the function
    if(checkEqualtoSum(N, K))
          cout << "YES";
    else cout << "NO";
     
    return 0;
}


Java




// Java program for the above approach
import java.io.*;
class GFG {
 
    // Function to Check if a number can
    // be equal to sum of numbers having
    // at least one digit equal to k
    static boolean checkEqualtoSum(int N, int k)
    {
        // Temporary variable to
        // store k
        int temp = k;
 
        // Variable for count
        int count = 0;
 
        // Iterating till count is less or
        // equal to 10 and N % 10 is not
        // equal to temp % 10
        while (count <= 10 && N % 10 != temp % 10) {
 
            temp += k;
            count++;
        }
 
        // If N % 10 is equal to temp % 10
        // and temp is less or equal to N,
        // return true
        if (N % 10 == temp % 10 && temp <= N)
            return true;
 
        // If k * 10 <= N, return true
        if (k * 10 <= N)
            return true;
 
        // Else return false
        return false;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given Input
        int N = 68;
        int K = 7;
 
        // Call the function
        if (checkEqualtoSum(N, K))
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}
 
// This code is contributed by dwivediyash


Python3




# python implementation for the above approach
 
# Function to Check if a number can
# be equal to sum of numbers having
# at least one digit equal to k
 
 
def checkEqualtoSum(N, k):
 
    # Temporary variable to
    # store k
    temp = k
 
    # Variable for count
    count = 0
 
    # Iterating till count is less or
    # equal to 10 and N % 10 is not
    # equal to temp % 10
    while(count <= 10 and N % 10 != temp % 10):
 
        temp += k
        count += 1
 
    # If N % 10 is equal to temp % 10
    # and temp is less or equal to N,
    # return true
    if(N % 10 == temp % 10 and temp <= N):
        return True
 
    # If k * 10 <= N, return true
    if(k * 10 <= N):
        return True
 
    # Else return false
    return False
 
 
# Driver Code
if __name__ == "__main__":
 
    N = 68
    K = 7
 
    # Call the function
    if(checkEqualtoSum(N, K)):
        print("YES")
    else:
        print("NO")
 
    # This code is contributed by rakeshsahni


Javascript




<script>
    // JavaScript implementation for the above approach
 
    // Function to Check if a number can
    // be equal to sum of numbers having
    // at least one digit equal to k
    const checkEqualtoSum = (N, k) => {
        // Temporary variable to
        // store k
        let temp = k;
 
        // Variable for count
        let count = 0;
 
        // Iterating till count is less or
        // equal to 10 and N % 10 is not
        // equal to temp % 10
        while (count <= 10 &&
            N % 10 != temp % 10) {
 
            temp += k;
            count++;
        }
 
        // If N % 10 is equal to temp % 10
        // and temp is less or equal to N,
        // return true
        if (N % 10 == temp % 10 && temp <= N)
            return true;
 
        // If k * 10 <= N, return true
        if (k * 10 <= N)
            return true;
 
        // Else return false
        return false;
    }
 
    // Driver Code
    let N = 68;
    let K = 7;
 
    // Call the function
    if (checkEqualtoSum(N, K))
        document.write("YES");
    else document.write("NO");
 
    // This code is contributed by rakeshsahni
 
</script>


C#




// C# implementation for the above approach
using System;
class gFG
{
   
    // Function to Check if a number can
    // be equal to sum of numbers having
    // at least one digit equal to k
    static bool checkEqualtoSum(int N, int k)
    {
        // Temporary variable to
        // store k
        int temp = k;
 
        // Variable for count
        int count = 0;
 
        // Iterating till count is less or
        // equal to 10 and N % 10 is not
        // equal to temp % 10
        while (count <= 10 && N % 10 != temp % 10) {
 
            temp += k;
            count++;
        }
 
        // If N % 10 is equal to temp % 10
        // and temp is less or equal to N,
        // return true
        if (N % 10 == temp % 10 && temp <= N)
            return true;
 
        // If k * 10 <= N, return true
        if (k * 10 <= N)
            return true;
 
        // Else return false
        return false;
    }
 
    // Driver Code
    public static void Main()
    {
        int N = 68;
        int K = 7;
 
        // Call the function
        if (checkEqualtoSum(N, K))
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }
}
 
// This code is contributed by ukasp.


 
 

Output

YES

 

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

 




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