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:
- Sort the set of digits.
- Initialize two vectors dp and result, with INT_MAX and {-1, 0} respectively.
- Initialize a queue and insert digit % N.
- Do while queue is not empty.
- Remove front value from queue and for each digit in the set, find (formed number) % N using above written expression.
- 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.
- Put all these values in another vector and reverse it.
- This is our required solution.
Below is the implementation of the above approach:Â
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;Â
// Function to return the required numberint findSmallestNumber(vector<int>& arr, int n){Â
    // 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));Â
    // Sort the vector of digits    sort(arr.begin(), arr.end());Â
    // Initialize the queue    queue<int> q;    for (auto i : arr) {        if (i != 0) {Â
            // If modulus value is not present            // in the queue            if (dp[i % n] > 1e9) {Â
                // Compute digits modulus given number and                // update the queue and vectors                q.push(i % n);                dp[i % n] = 1;                result[i % n] = { -1, i };            }        }    }Â
    // While queue is not empty    while (!q.empty()) {Â
        // Remove the first element of the queue        int u = q.front();        q.pop();        for (auto i : arr) {Â
            // Compute new modulus values by using old queue            // values and each digit of the set            int v = (u * 10 + i) % n;Â
            // 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 };            }        }    }Â
    // If required condition can't be satisfied    if (dp[0] > 1e9)        return -1;    else {Â
        // Initialize new vector        vector<int> ans;        int u = 0;        while (u != -1) {Â
            // Constructing the solution by backtracking            ans.push_back(result[u].second);            u = result[u].first;        }Â
        // Reverse the vector        reverse(ans.begin(), ans.end());        for (auto i : ans) {Â
            // Return the required number            return i;        }    }}Â
// Driver codeint main(){Â Â Â Â vector<int> arr = { 5, 2, 3 };Â Â Â Â int n = 12;Â
    cout << findSmallestNumber(arr, n);Â
    return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{Â
  // Function to return the required number  static int findSmallestNumber(ArrayList<Integer> arr, int n)  {Â
    // 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);Â
    ArrayList <ArrayList<Integer>> result = new ArrayList <ArrayList<Integer>>();    for (int i = 0; i < n; i++)      result.add(new ArrayList<Integer>(Arrays.asList(-1, 0)));Â
    // Sort the vector of digits    Collections.sort(arr);Â
    // Initialize the queue    ArrayList<Integer> q = new ArrayList<Integer>();    for (int i : arr) {      if (i != 0) {Â
        // If modulus value is not present        // in the queue        if (dp.get(i % n) > 1000000000) {Â
          // 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)));        }      }    }Â
    // While queue is not empty    while (q.size() != 0) {Â
      // Remove the first element of the queue      int u = q.get(0);      q.remove(0);      for (int i : arr) {Â
        // Compute new modulus values by using old queue        // values and each digit of the set        int v = (u * 10 + i) % n;Â
        // 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)));        }      }    }Â
    // If required condition can't be satisfied    if (dp.get(0) > 1000000000)      return -1;    else {Â
      // Initialize new vector      ArrayList<Integer> ans = new ArrayList<Integer>();      int u = 0;      while (u != -1) {Â
        // Constructing the solution by backtracking        ans.add(result.get(u).get(1));        u = result.get(u).get(0);      }Â
      // Reverse the vector      Collections.reverse(ans);Â
      for (int i : ans) {Â
        // Return the required number        return i;      }    }Â
    // This line should never be reached    return -1;  }Â
  // Driver code  public static void main(String[] args)  {    ArrayList<Integer> arr = new ArrayList<Integer>(Arrays.asList(5, 2, 3));    int n = 12;Â
    System.out.println(findSmallestNumber(arr, n));  }}Â
// This code is contributed by phasing17 |
Python3
# Python3 implementation of the approach Â
# 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 Â
    # Sort the vector of digits     arr.sort() Â
    # Initialize the queue     q = []     for i in arr:         if i != 0: Â
            # If modulus value is not             # present in the queue             if dp[i % n] > 10 ** 9: Â
                # 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:Â
        # Remove the first element of the queue         u = q.pop(0)         for i in arr: Â
            # Compute new modulus values by using old             # queue values and each digit of the set             v = (u * 10 + i) % n Â
            # 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 Â
    # If required condition can't be satisfied     if dp[0] > 10 ** 9:        return -1    else: Â
        # Initialize new vector         ans = []         u = 0        while u != -1: Â
            # Constructing the solution by backtracking             ans.append(result[u][1])             u = result[u][0]                  # Reverse the vector         ans = ans[::-1]         for i in ans: Â
            # Return the required number             return i           # Driver code if __name__ == "__main__":       arr = [5, 2, 3]     n = 12Â
    print(findSmallestNumber(arr, n)) Â
# This code is contributed by Rituraj Jain |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;Â
class GFG{Â
  // Function to return the required number  static int findSmallestNumber(List<int> arr, int n)  {Â
    // 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);Â
    List <int[]> result = new List<int[]>();    for (int i = 0; i < n; i++)      result.Add(new [] {-1, 0});Â
    // Sort the vector of digits    arr.Sort();Â
    // Initialize the queue    List<int> q = new List<int>();    foreach (int i in arr) {      if (i != 0) {Â
        // If modulus value is not present        // in the queue        if (dp[i % n] > 1000000000) {Â
          // 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};        }      }    }Â
    // While queue is not empty    while (q.Count != 0) {Â
      // Remove the first element of the queue      int u = q[0];      q.RemoveAt(0);      foreach (int i in arr) {Â
        // Compute new modulus values by using old queue        // values and each digit of the set        int v = (u * 10 + i) % n;Â
        // 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 };        }      }    }Â
    // If required condition can't be satisfied    if (dp[0] > 1000000000)      return -1;    else {Â
      // Initialize new vector      List<int> ans = new List<int>();      int u = 0;      while (u != -1) {Â
        // Constructing the solution by backtracking        ans.Add(result[u][1]);        u = result[u][0];      }Â
      // Reverse the vector      ans.Reverse();Â
      foreach (var i in ans) {Â
        // Return the required number        return i;      }    }Â
    // This line should never be reached    return -1;  }Â
  // Driver code  public static void Main(string[] args)  {    List<int> arr = new List<int>(new[] { 5, 2, 3 });    int n = 12;Â
    Console.WriteLine(findSmallestNumber(arr, n));  }}Â
// This code is contributed by phasing17 |
Javascript
// JavaScript implementation of the approach let inf = 100000000000Â
// 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() Â
    // 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)            { Â
                // 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 Â
            // 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]            }        }    }Â
    // If required condition can't be satisfied     if (dp[0] > 1000000000)        return -1              else    { Â
        // Initialize new vector         let ans = []         let u = 0         while (u != -1)        { Â
            // Constructing the solution by backtracking             ans.push(result[u][1])             u = result[u][0]         }                 // Reverse the vector         ans.reverse()        for (var i of ans)        { Â
            // Return the required number             return i         }    }}          // Driver code let arr = [5, 2, 3] let n = 12 Â
console.log(findSmallestNumber(arr, n)) Â
// This code is contributed by phasing17 |
Output:
2
Time Complexity: O(N*N), as we are using nested loops to traverse N*N times. Where N is the number of elements in the array.
Auxiliary Space: O(N), as we are using extra space for dp and queue. Where N is the number of elements in the array.
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