Find the minimum positive integer such that it is divisible by A and sum of its digits is equal to B

Given two integers A and B, the task is to find the minimum positive integer N such that N is divisible by A and the sum of the digits of N is equal to B. If number is not found then print -1.

Examples:

Input: A = 20, B = 30
Output: 49980
49980 is divisible by 20 and sum of its digit = 4 + 9 + 9 + 8 + 0 = 30



Input: A = 5, B = 2
Output: 20

Approach:

  • Create empty queue q that stores the value of A and B and output number as a string and create integer type 2-D array visited[][] that stores the visited digit.
  • Insert Node into queue and check if queue is non-empty.
  • While the queue is non-empty, pop an element from the queue and for every digit from 1 to 9, concatenate the digit after the string num and check whether the number formed is the required number.
  • If the required number is found, print the number.
  • Else repeat the steps while the number is less than B and the queue is non-empty while pushing the non-visited number to the queue.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Array that stores visited digits
int visited[501][5001];
  
// Structure for queue Node.
struct Node {
    int a, b;
    string str;
};
  
// Function to return the minimum number such that it is
// divisible by 'a' and sum of its digits is equals to 'b'
int findNumber(int a, int b)
{
    // Create queue
    queue<Node> q;
  
    // Initially queue is empty
    Node temp = Node{ 0, 0, "" };
  
    // Initialize visited to 1
    visited[0][0] = 1;
  
    // Push temp in queue
    q.push(temp);
  
    // While queue is not empty
    while (!q.empty()) {
  
        // Get the front of the queue and pop it
        Node u = q.front();
        q.pop();
  
        // If popped element is the required number
        if (u.a == 0 && u.b == b)
  
            // Parse int from string and return it
            return std::stoi(u.str);
  
        // Loop for each digit and check the sum 
        // If not visited then push it to the queue
        for (int i = 0; i < 10; i++) {
            int dd = (u.a * 10 + i) % a;
            int ss = u.b + i;
            if (ss <= b && !visited[dd][ss]) {
                visited[dd][ss] = 1;
                q.push(Node{ dd, ss, u.str + char('0' + i) });
            }
        }
    }
  
    // Required number not found return -1.
    return -1;
}
  
// Driver code.
int main()
{
    int a = 25, b = 1;
    cout << findNumber(a, b);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.util.*;
class Solution
{
   
// Array that stores visited digits
static int visited[][]= new int[501][5001];
   
// Structure for queue Node.
static class Node {
    int a, b;
    String str;
    Node(int a1,int b1,String s)
    {
        a=a1;
        b=b1;
        str=s;
    }
}
   
// Function to return the minimum number such that it is
// divisible by 'a' and sum of its digits is equals to 'b'
static int findNumber(int a, int b)
{
    // Create queue
    Queue<Node> q= new LinkedList<Node>();
   
    // Initially queue is empty
    Node temp =new  Node( 0, 0, "" );
   
    // Initialize visited to 1
    visited[0][0] = 1;
   
    // Push temp in queue
    q.add(temp);
   
    // While queue is not empty
    while (q.size()!=0) {
   
        // Get the front of the queue and pop it
        Node u = q.peek();
        q.remove();
   
        // If popped element is the required number
        if (u.a == 0 && u.b == b)
   
            // Parse int from string and return it
            return Integer.parseInt(u.str);
   
        // Loop for each digit and check the sum 
        // If not visited then push it to the queue
        for (int i = 0; i < 10; i++) {
            int dd = (u.a * 10 + i) % a;
            int ss = u.b + i;
            if (ss <= b && visited[dd][ss]==0) {
                visited[dd][ss] = 1;
                q.add(new Node( dd, ss, u.str + (char)('0' + i) ));
            }
        }
    }
   
    // Required number not found return -1.
    return -1;
}
   
// Driver code.
public static void  main(String args[])
{
    //initilize visited
    for(int i=0;i<500;i++)
        for(int j=0;j<500;j++)
            visited[i][j]=0;
      
    int a = 25, b = 1;
    System.out.println(findNumber(a, b));
      
}
}
//contributed by Arnab Kundu

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
# Array that stores visited digits 
visited = [[0 for x in range(501)] 
              for y in range(5001)] 
  
# Structure for queue Node. 
class Node:
      
    def __init__(self, a, b, string):
        self.a = a
        self.b = b
        self.string = string
  
# Function to return the minimum number 
# such that it is divisible by 'a' and
# sum of its digits is equals to 'b' 
def findNumber(a, b): 
  
    # Use list as queue 
    q = []
  
    # Initially queue is empty 
    temp = Node(0, 0, "") 
  
    # Initialize visited to 1 
    visited[0][0] = 1
  
    # Push temp in queue 
    q.append(temp) 
  
    # While queue is not empty 
    while len(q) > 0
  
        # Get the front of the queue 
        # and pop it 
        u = q.pop(0
  
        # If popped element is the 
        # required number 
        if u.a == 0 and u.b == b: 
  
            # Parse int from string 
            # and return it 
            return int(u.string) 
  
        # Loop for each digit and check the sum 
        # If not visited then push it to the queue 
        for i in range(0, 10): 
            dd = (u.a * 10 + i) %
            ss = u.b +
              
            if ss <= b and visited[dd][ss] == False
                visited[dd][ss] = 1
                q.append(Node(dd, ss, u.string + str(i))) 
  
    # Required number not found return -1. 
    return -1
  
# Driver code. 
if __name__ == "__main__":
  
    a, b = 25, 1
    print(findNumber(a, b))
      
# This code is contributed by Rituraj Jain

chevron_right


Output:

100


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : andrew1234, rituraj_jain