Smallest multiple of N formed using the given set of digits

Given a set of digits S and an integer N, the task is to find the smallest positive integer if exists which contains only the digits from S and is a multiple of N. Note that the digits from the set can be used multiple times. Examples:

Input: S[] = {5, 2, 3}, N = 12 Output: 252 We can observe that 252 is formed using {2, 5} and is a multiple of 12 Input: S[] = {1, 3, 5, 7, 9}, N = 2 Output: -1 Multiple of 2 would always be an even number but from the given set of digits even number can’t be formed.

A simple approach is to sort the set of digits and then move from smallest to the largest number formed using the given digits. We check each number whether it satisfies the given condition. Implementing it this way would result in exponential time complexity. A better approach is to use Modular Arithmetic. So, we maintain a queue in which we will store the modulus of numbers formed using set of digits with the given number N. Initially in the queue, there would be (single digit number) % N but we can calculate (double digit number) % N by using,

New_mod = (Old_mod * 10 + digit) % N

By using the above expression we can calculate modulus values of multiple digit numbers. This is an application of dynamic programming as we are building our solution from smaller state to larger state. We maintain an another vector to ensure that element to be inserted in queue is already present in the queue or not. It uses hashing to ensure O(1) time complexity. We used another vector of pair which also uses hashing to store the values and its structure is

result[new_mod] = { current_element_of_queue, digit}

This vector would be used to construct the solution:

1. Sort the set of digits.
2. Initialize two vectors dp and result, with INT_MAX and {-1, 0} respectively.
3. Initialize a queue and insert digit % N.
4. Do while queue is not empty.
5. Remove front value from queue and for each digit in the set, find (formed number) % N using above written expression.
6. If we didn’t get 0 as a modulus value before queue is empty then smallest positive number does not exist else trace the result vector from the 0th index until we get -1 at any index.
7. Put all these values in another vector and reverse it.
8. This is our required solution.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the required number
int findSmallestNumber(vector<int>& arr, int n)
{

    // Initialize both vectors with their initial values
    vector<int> dp(n, numeric_limits<int>::max() - 5);
    vector<pair<int, int> > result(n, make_pair(-1, 0));

    // Sort the vector of digits
    sort(arr.begin(), arr.end());

    // Initialize the queue
    queue<int> q;
    for (auto i : arr) {
        if (i != 0) {

            // If modulus value is not present
            // in the queue
            if (dp[i % n] > 1e9) {

                // Compute digits modulus given number and
                // update the queue and vectors
                q.push(i % n);
                dp[i % n] = 1;
                result[i % n] = { -1, i };
            }
        }
    }

    // While queue is not empty
    while (!q.empty()) {

        // Remove the first element of the queue
        int u = q.front();
        q.pop();
        for (auto i : arr) {

            // Compute new modulus values by using old queue
            // values and each digit of the set
            int v = (u * 10 + i) % n;

            // If value is not present in the queue
            if (dp[u] + 1 < dp[v]) {
                dp[v] = dp[u] + 1;
                q.push(v);
                result[v] = { u, i };
            }
        }
    }

    // If required condition can't be satisfied
    if (dp[0] > 1e9)
        return -1;
    else {

        // Initialize new vector
        vector<int> ans;
        int u = 0;
        while (u != -1) {

            // Constructing the solution by backtracking
            ans.push_back(result[u].second);
            u = result[u].first;
        }

        // Reverse the vector
        reverse(ans.begin(), ans.end());
        for (auto i : ans) {

            // Return the required number
            return i;
        }
    }
}

// Driver code
int main()
{
    vector<int> arr = { 5, 2, 3 };
    int n = 12;

    cout << findSmallestNumber(arr, n);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
class GFG
{

  // Function to return the required number
  static int findSmallestNumber(ArrayList<Integer> arr, int n)
  {

    // Initialize both vectors with their initial values
    ArrayList<Integer> dp = new ArrayList<Integer>();
    for (int i = 0; i < n; i++)
      dp.add(Integer.MAX_VALUE - 5);

    ArrayList <ArrayList<Integer>> result = new ArrayList <ArrayList<Integer>>();
    for (int i = 0; i < n; i++)
      result.add(new ArrayList<Integer>(Arrays.asList(-1, 0)));

    // Sort the vector of digits
    Collections.sort(arr);

    // Initialize the queue
    ArrayList<Integer> q = new ArrayList<Integer>();
    for (int i : arr) {
      if (i != 0) {

        // If modulus value is not present
        // in the queue
        if (dp.get(i % n) > 1000000000) {

          // Compute digits modulus given number and
          // update the queue and vectors
          q.add(i % n);
          dp.set(i % n, 1);
          result.set(i % n, new ArrayList<Integer>(Arrays.asList(-1, i)));
        }
      }
    }

    // While queue is not empty
    while (q.size() != 0) {

      // Remove the first element of the queue
      int u = q.get(0);
      q.remove(0);
      for (int i : arr) {

        // Compute new modulus values by using old queue
        // values and each digit of the set
        int v = (u * 10 + i) % n;

        // If value is not present in the queue
        if (dp.get(u) + 1 < dp.get(v)) {
          dp.set(v, dp.get(u) + 1);
          q.add(v);
          result.set(v, new ArrayList<Integer>(Arrays.asList(u, i)));
        }
      }
    }

    // If required condition can't be satisfied
    if (dp.get(0) > 1000000000)
      return -1;
    else {

      // Initialize new vector
      ArrayList<Integer> ans = new ArrayList<Integer>();
      int u = 0;
      while (u != -1) {

        // Constructing the solution by backtracking
        ans.add(result.get(u).get(1));
        u = result.get(u).get(0);
      }

      // Reverse the vector
      Collections.reverse(ans);

      for (int i : ans) {

        // Return the required number
        return i;
      }
    }

    // This line should never be reached
    return -1;
  }

  // Driver code
  public static void main(String[] args)
  {
    ArrayList<Integer> arr = new ArrayList<Integer>(Arrays.asList(5, 2, 3));
    int n = 12;

    System.out.println(findSmallestNumber(arr, n));
  }
}

// This code is contributed by phasing17

Python3

# Python3 implementation of the approach

# Function to return the required number
def findSmallestNumber(arr, n):

    # Initialize both vectors with their initial values
    dp = [float('inf')] * n
    result = [(-1, 0)] * n

    # Sort the vector of digits
    arr.sort()

    # Initialize the queue
    q = []
    for i in arr:
        if i != 0:

            # If modulus value is not
            # present in the queue
            if dp[i % n] > 10 ** 9:

                # Compute digits modulus given number
                # and update the queue and vectors
                q.append(i % n)
                dp[i % n] = 1
                result[i % n] = -1, i

    # While queue is not empty
    while len(q) > 0:

        # Remove the first element of the queue
        u = q.pop(0)
        for i in arr:

            # Compute new modulus values by using old
            # queue values and each digit of the set
            v = (u * 10 + i) % n

            # If value is not present in the queue
            if dp[u] + 1 < dp[v]:
                dp[v] = dp[u] + 1
                q.append(v)
                result[v] = u, i

    # If required condition can't be satisfied
    if dp[0] > 10 ** 9:
        return -1
    else:

        # Initialize new vector
        ans = []
        u = 0
        while u != -1:

            # Constructing the solution by backtracking
            ans.append(result[u][1])
            u = result[u][0]

        # Reverse the vector
        ans = ans[::-1]
        for i in ans:

            # Return the required number
            return i

# Driver code
if __name__ == "__main__":

    arr = [5, 2, 3]
    n = 12

    print(findSmallestNumber(arr, n))

# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{

  // Function to return the required number
  static int findSmallestNumber(List<int> arr, int n)
  {

    // Initialize both vectors with their initial values
    List<int> dp = new List<int>();
    for (int i = 0; i < n; i++)
      dp.Add(Int32.MaxValue - 5);

    List <int[]> result = new List<int[]>();
    for (int i = 0; i < n; i++)
      result.Add(new [] {-1, 0});

    // Sort the vector of digits
    arr.Sort();

    // Initialize the queue
    List<int> q = new List<int>();
    foreach (int i in arr) {
      if (i != 0) {

        // If modulus value is not present
        // in the queue
        if (dp[i % n] > 1000000000) {

          // Compute digits modulus given number and
          // update the queue and vectors
          q.Add(i % n);
          dp[i % n] = 1;
          result[i % n] = new [] {-1, i};
        }
      }
    }

    // While queue is not empty
    while (q.Count != 0) {

      // Remove the first element of the queue
      int u = q[0];
      q.RemoveAt(0);
      foreach (int i in arr) {

        // Compute new modulus values by using old queue
        // values and each digit of the set
        int v = (u * 10 + i) % n;

        // If value is not present in the queue
        if (dp[u] + 1 < dp[v]) {
          dp[v] = dp[u] + 1;
          q.Add(v);
          result[v] = new[] { u, i };
        }
      }
    }

    // If required condition can't be satisfied
    if (dp[0] > 1000000000)
      return -1;
    else {

      // Initialize new vector
      List<int> ans = new List<int>();
      int u = 0;
      while (u != -1) {

        // Constructing the solution by backtracking
        ans.Add(result[u][1]);
        u = result[u][0];
      }

      // Reverse the vector
      ans.Reverse();

      foreach (var i in ans) {

        // Return the required number
        return i;
      }
    }

    // This line should never be reached
    return -1;
  }

  // Driver code
  public static void Main(string[] args)
  {
    List<int> arr = new List<int>(new[] { 5, 2, 3 });
    int n = 12;

    Console.WriteLine(findSmallestNumber(arr, n));
  }
}

// This code is contributed by phasing17

Javascript

// JavaScript implementation of the approach
let inf = 100000000000

// Function to return the required number
function findSmallestNumber(arr, n){

    // Initialize both vectors with their initial values
    let dp = new Array(n).fill(inf)
    let result = new Array(n)
    for (var i = 0; i < n; i++)
        result[i] = [-1, 0]


    // Sort the vector of digits
    arr.sort()

    // Initialize the queue
    let q = []
    for (var i of arr)
    {
        if (i != 0)
        {
            // If modulus value is not
            // present in the queue
            if (dp[i % n] > 1000000000)
            {

                // Compute digits modulus given number
                // and update the queue and vectors
                q.push(i % n)
                dp[i % n] = 1
                result[i % n] = [-1, i]
            }
        }
    }
    // While queue is not empty
    while (q.length > 0)
    {
        // Remove the first element of the queue
        let u = q.shift()
        for (var i of arr)
        {
            // Compute new modulus values by using old
            // queue values and each digit of the set
            let v = (u * 10 + i) % n

            // If value is not present in the queue
            if (dp[u] + 1 < dp[v])
            {
                dp[v] = dp[u] + 1
                q.push(v)
                result[v] = [u, i]
            }
        }
    }

    // If required condition can't be satisfied
    if (dp[0] > 1000000000)
        return -1

    else
    {

        // Initialize new vector
        let ans = []
        let u = 0
        while (u != -1)
        {

            // Constructing the solution by backtracking
            ans.push(result[u][1])
            u = result[u][0]
        }

        // Reverse the vector
        ans.reverse()
        for (var i of ans)
        {

            // Return the required number
            return i
        }
    }
}

// Driver code
let arr = [5, 2, 3]
let n = 12

console.log(findSmallestNumber(arr, n))

// This code is contributed by phasing17

Output

2

Time Complexity: O(N*M), where N is the size of the input ArrayList arr and M is the number 12 (used in the modulo operation and loop iterations)
Auxiliary Space: O(N), as we are using extra space for dp and queue. Where N is the number of elements in the array.