Smallest multiple of a given number made of digits 0 and 9 only
Last Updated :
29 Aug, 2022
We are given an integer N. We need to write a program to find the least positive integer X made up of only digits 9's and 0's, such that, X is a multiple of N.
Note: It is assumed that the value of X will not exceed 106.
Examples:
Input : N = 5
Output : X = 90
Explanation: 90 is the smallest number made up
of 9's and 0's which is divisible by 5.
Input : N = 7
Output : X = 9009
Explanation: 9009 is smallest number made up
of 9's and 0's which is divisible by 7.
The idea to solve this problem is to generate and store all of the numbers which can be formed using digits 0 & 9. Then find the smallest number among these generated number which is divisible by N.
We will use the method of generating binary numbers to generate all numbers which can be formed by using digits 0 & 9.
Below is the implementation of above idea:
C++
// CPP program to find smallest multiple of a
// given number made of digits 0 and 9 only
#include <bits/stdc++.h>
using namespace std;
// Maximum number of numbers made of 0 and 9
#define MAX_COUNT 10000
// vector to store all numbers that can be formed
// using digits 0 and 9 and are less than 10^5
vector<string> vec;
/* Preprocessing function to generate all possible
numbers formed by 0 and 9 */
void generateNumbersUtil()
{
// Create an empty queue of strings
queue<string> q;
// enqueue the first number
q.push("9");
// This loops is like BFS of a tree with 9 as root
// 0 as left child and 9 as right child and so on
for (int count = MAX_COUNT; count > 0; count--)
{
string s1 = q.front();
q.pop();
// storing the front of queue in the vector
vec.push_back(s1);
string s2 = s1;
// Append "0" to s1 and enqueue it
q.push(s1.append("0"));
// Append "9" to s2 and enqueue it. Note that
// s2 contains the previous front
q.push(s2.append("9"));
}
}
// function to find smallest number made up of only
// digits 9’s and 0’s, which is a multiple of n.
string findSmallestMultiple(int n)
{
// traverse the vector to find the smallest
// multiple of n
for (int i = 0; i < vec.size(); i++)
// stoi() is used for string to int conversion
if (stoi(vec[i])%n == 0)
return vec[i];
}
// Driver Code
int main()
{
generateNumbersUtil();
int n = 7;
cout << findSmallestMultiple(n);
return 0;
}
// Java program to find smallest
// multiple of a given number
// made of digits 0 and 9 only
import java.util.*;
class GFG
{
// Maximum number of
// numbers made of 0 and 9
static int MAX_COUNT = 10000;
// vector to store all numbers
// that can be formed using
// digits 0 and 9 and are
// less than 10^5
static List<String> vec = new LinkedList<String>();
/* Preprocessing function
to generate all possible
numbers formed by 0 and 9 */
static void generateNumbersUtil()
{
// Create an empty
// queue of Strings
Queue<String> q = new LinkedList<String>();
// enqueue the
// first number
q.add("9");
// This loops is like BFS of
// a tree with 9 as root
// 0 as left child and 9 as
// right child and so on
for (int count = MAX_COUNT;
count > 0; count--)
{
String s1 = q.peek();
q.remove();
// storing the Peek of
// queue in the vector
vec.add(s1);
String s2 = s1;
// Append "0" to s1
// and enqueue it
q.add(s1 + "0");
// Append "9" to s2 and
// enqueue it. Note that
// s2 contains the previous Peek
q.add(s2 + "9");
}
}
// function to find smallest
// number made up of only
// digits 9's and 0's, which
// is a multiple of n.
static String findSmallestMultiple(int n)
{
// traverse the vector
// to find the smallest
// multiple of n
for (int i = 0; i < vec.size(); i++) // stoi() is used for
// String to int conversion
{
if (Integer.parseInt(vec.get(i)) % n == 0)
{
return vec.get(i);
}
}
return "";
}
// Driver Code
public static void main(String[] args)
{
generateNumbersUtil();
int n = 7;
System.out.println(findSmallestMultiple(n));
}
}
// This code is contributed by Rajput-Ji
# Python3 program to find smallest multiple of
# a given number made of digits 0 and 9 only
from queue import Queue
# Preprocessing function to generate
# all possible numbers formed by 0 and 9
def generateNumbersUtil():
global vec
# Create an empty queue of strings
q = Queue()
# enqueue the first number
q.put("9")
# This loops is like BFS of a tree
# with 9 as root, 0 as left child
# and 9 as right child and so on
for count in range(MAX_COUNT, -1, -1):
s1 = q.queue[0]
q.get()
# storing the front of queue
# in the vector
vec.append(s1)
s2 = s1
# Append "0" to s1 and enqueue it
s1 += "0"
q.put(s1)
# Append "9" to s2 and enqueue it. Note
# that s2 contains the previous front
s2 += "9"
q.put(s2)
# function to find smallest number made
# up of only digits 9’s and 0’s, which
# is a multiple of n.
def findSmallestMultiple(n):
global vec
# traverse the vector to find
# the smallest multiple of n
for i in range(len(vec)):
# int is used for string to
# conversion
if (int(vec[i]) % n == 0):
return vec[i]
# Driver Code
# Maximum number of numbers
# made of 0 and 9
MAX_COUNT = 10000
# stack to store all numbers that
# can be formed using digits 0 and
# 9 and are less than 10^5
vec = []
generateNumbersUtil()
n = 7
print(findSmallestMultiple(n))
# This code is contributed by PranchalK
// C# program to find smallest
// multiple of a given number
// made of digits 0 and 9 only
using System;
using System.Collections.Generic;
class GFG
{
// Maximum number of
// numbers made of 0 and 9
static int MAX_COUNT = 10000;
// vector to store all numbers
// that can be formed using
// digits 0 and 9 and are
// less than 10^5
static List<string> vec = new List<string>();
/* Preprocessing function
to generate all possible
numbers formed by 0 and 9 */
static void generateNumbersUtil()
{
// Create an empty
// queue of strings
Queue<string> q = new Queue<string>();
// enqueue the
// first number
q.Enqueue("9");
// This loops is like BFS of
// a tree with 9 as root
// 0 as left child and 9 as
// right child and so on
for (int count = MAX_COUNT;
count > 0; count--)
{
string s1 = q.Peek();
q.Dequeue();
// storing the Peek of
// queue in the vector
vec.Add(s1);
string s2 = s1;
// Append "0" to s1
// and enqueue it
q.Enqueue(s1 + "0");
// Append "9" to s2 and
// enqueue it. Note that
// s2 contains the previous Peek
q.Enqueue(s2 + "9");
}
}
// function to find smallest
// number made up of only
// digits 9’s and 0’s, which
// is a multiple of n.
static string findSmallestMultiple(int n)
{
// traverse the vector
// to find the smallest
// multiple of n
for (int i = 0; i < vec.Count; i++)
// stoi() is used for
// string to int conversion
if (int.Parse(vec[i]) % n == 0)
return vec[i];
return "";
}
// Driver Code
static void Main()
{
generateNumbersUtil();
int n = 7;
Console.Write(findSmallestMultiple(n));
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
// JavaScript program to find smallest multiple of
// a given number made of digits 0 and 9 only
let vec = [];
// Preprocessing function to generate
// all possible numbers formed by 0 and 9
function generateNumbersUtil(){
// Create an empty queue of strings
let q = []
// enqueue the first number
q.push("9")
// This loops is like BFS of a tree
// with 9 as root, 0 as left child
// and 9 as right child and so on
for (var count = MAX_COUNT; count > -1; count--){
s1 = q[0]
q.shift()
// storing the front of queue
// in the vector
vec.push(s1)
s2 = s1
// Append "0" to s1 and enqueue it
s1 += "0"
q.push(s1)
// Append "9" to s2 and enqueue it. Note
// that s2 contains the previous front
s2 += "9"
q.push(s2)
}
}
// function to find smallest number made
// up of only digits 9’s and 0’s, which
// is a multiple of n.
function findSmallestMultiple(n){
// traverse the vector to find
// the smallest multiple of n
for (var i = 0; i < vec.length; i++)
// int is used for string to
// conversion
if (parseInt(vec[i]) % n == 0)
return vec[i]
}
// Driver Code
// Maximum number of numbers
// made of 0 and 9
let MAX_COUNT = 10000
// stack to store all numbers that
// can be formed using digits 0 and
// 9 and are less than 10^5
vec = []
generateNumbersUtil()
let n = 7
console.log(findSmallestMultiple(n))
// This code is contributed by phasing17
Time Complexity: O(n), where n represents the given integer.
Auxiliary Space: O(10000), no extra space is required, so it is a constant.
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