Given an array of numbers, the task is to print those numbers in ascending order separated by commas that have 1, 2, and 3 in their digits. If no number containing digits 1, 2, and 3 is present then print -1.
Examples:
Input : numbers[] = {123, 1232, 456, 234, 32145}
Output : 123, 1232, 32145Input : numbers[] = {9821, 627183, 12, 1234}
Output : 1234, 627183Input : numbers[] = {12, 232, 456, 234}
Output : -1
Try It Yourself
First finding all the number in form of array which contains 1, 2 & 3 then sort the number according to 1, 2 and 3 and then print it.
Follow the steps below to implement the above idea:
- Iterate over the given array numbers[] and check if the number is valid or not.
- For checking the validity of a number:
- Take three variables countOnes, countTwo, countThee, and initialize to 0.
- Increment the value for the above variables accordingly, if found.
- Finally, Check if all the above three variables contain some value, then return true.
- Otherwise, return false.
- For checking the validity of a number:
- If the number is valid then push the number into array oneTwoThree[].
- Sort the array and print the result.
Below is the implementation of the above approach:
// CPP program to print all number containing
// 1, 2 and 3 in any order.
#include <bits/stdc++.h>
using namespace std;
// convert the number to string and find
// if it contains 1, 2 & 3.
bool findContainsOneTwoThree(int number)
{
string str = to_string(number);
int countOnes = 0, countTwo = 0, countThree = 0;
for (int i = 0; i < str.length(); i++) {
if (str[i] == '1')
countOnes++;
else if (str[i] == '2')
countTwo++;
else if (str[i] == '3')
countThree++;
}
return (countOnes && countTwo && countThree);
}
// prints all the number containing 1, 2, 3
string printNumbers(int numbers[], int n)
{
vector<int> oneTwoThree;
for (int i = 0; i < n; i++) {
// check if the number contains 1,
// 2 & 3 in any order
if (findContainsOneTwoThree(numbers[i]))
oneTwoThree.push_back(numbers[i]);
}
// sort all the numbers
sort(oneTwoThree.begin(), oneTwoThree.end());
string result = "";
for (auto number : oneTwoThree) {
int value = number;
if (result.length() > 0)
result += ", ";
result += to_string(value);
}
return (result.length() > 0) ? result : "-1";
}
// Driver Code
int main()
{
int numbers[] = { 123, 1232, 456, 234, 32145 };
int n = sizeof(numbers) / sizeof(numbers[0]);
string result = printNumbers(numbers, n);
cout << result;
return 0;
}
// This code is contributed
// by Sirjan13
// Java program to print all number containing
// 1, 2 and 3 in any order.
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Iterator;
class GFG {
// prints all the number containing 1, 2, 3
// in any order
private static String printNumbers(int[] numbers)
{
ArrayList<Integer> array = new ArrayList<>();
for (int number : numbers) {
// check if the number contains 1,
// 2 & 3 in any order
if (findContainsOneTwoThree(number))
array.add(number);
}
// sort all the numbers
Collections.sort(array);
StringBuffer strbuf = new StringBuffer();
Iterator it = array.iterator();
while (it.hasNext()) {
int value = (int)it.next();
if (strbuf.length() > 0)
strbuf.append(", ");
strbuf.append(Integer.toString(value));
}
return (strbuf.length() > 0) ? strbuf.toString()
: "-1";
}
// convert the number to string and find
// if it contains 1, 2 & 3.
private static boolean
findContainsOneTwoThree(int number)
{
String str = Integer.toString(number);
return (str.contains("1") && str.contains("2")
&& str.contains("3"));
}
public static void main(String[] args)
{
int[] numbers = { 123, 1232, 456, 234, 32145 };
System.out.println(printNumbers(numbers));
}
}
# Python program for printing
# all numbers containing 1,2 and 3
def printNumbers(numbers):
# convert all numbers
# to strings
numbers = map(str, numbers)
result = []
for num in numbers:
# check if each number
# in the list has 1,2 and 3
if ('1' in num and
'2' in num and
'3' in num):
result.append(num)
# if there are no
# valid numbers
if not result:
result = ['-1']
return sorted(result)
# Driver Code
numbers = [123, 1232, 456,
234, 32145]
result = printNumbers(numbers)
print ', '.join(num for num in result)
# This code is contributed
# by IshitaTripathi
// C# program to print all number
// containing 1, 2 and 3 in any order.
using System;
using System.Collections.Generic;
using System.Text;
class GFG {
// prints all the number
// containing 1, 2, 3
// in any order
private static string printNumbers(int[] numbers)
{
List<int> array = new List<int>();
foreach(int number in numbers)
{
// check if the number contains 1,
// 2 & 3 in any order
if (findContainsOneTwoThree(number)) {
array.Add(number);
}
}
// sort all the numbers
array.Sort();
StringBuilder strbuf = new StringBuilder();
System.Collections.IEnumerator it
= array.GetEnumerator();
while (it.MoveNext()) {
int value = (int)it.Current;
if (strbuf.Length > 0) {
strbuf.Append(", ");
}
strbuf.Append(Convert.ToString(value));
}
return (strbuf.Length > 0) ? strbuf.ToString()
: "-1";
}
// convert the number
// to string and find
// if it contains 1, 2 & 3.
private static bool findContainsOneTwoThree(int number)
{
string str = Convert.ToString(number);
return (str.Contains("1") && str.Contains("2")
&& str.Contains("3"));
}
// Driver Code
public static void Main(string[] args)
{
int[] numbers
= new int[] { 123, 1232, 456, 234, 32145 };
Console.WriteLine(printNumbers(numbers));
}
}
// This code is contributed by Shrikant13
<script>
// JavaScript program to print all number containing
// 1, 2 and 3 in any order.
// convert the number to string and find
// if it contains 1, 2 & 3.
function findContainsOneTwoThree(number)
{
let str = number.toString();
let countOnes = 0, countTwo = 0, countThree = 0;
for(let i = 0; i < str.length; i++) {
if(str[i] == '1') countOnes++;
else if(str[i] == '2') countTwo++;
else if(str[i] == '3') countThree++;
}
return (countOnes && countTwo && countThree);
}
// prints all the number containing 1, 2, 3
function printNumbers(numbers, n)
{
let oneTwoThree = [];
for (let i = 0; i < n; i++)
{
// check if the number contains 1,
// 2 & 3 in any order
if (findContainsOneTwoThree(numbers[i]))
oneTwoThree.push(numbers[i]);
}
// sort all the numbers
oneTwoThree.sort();
let result = "";
for(let number of oneTwoThree)
{
let value = number;
if (result.length > 0)
result += ", ";
result += value.toString();
}
return (result.length > 0) ? result : "-1";
}
// Driver Code
let numbers = [ 123, 1232, 456, 234, 32145 ];
let n = numbers.length;
let result = printNumbers(numbers, n);
document.write(result);
// This code is contributed
// by shinjanpatra
</script>
Output
123, 1232, 32145
Time Complexity: O(N log(N)), where N is the length of the given number.
Auxiliary Space: O(N)
Please suggest if someone has a better solution that is more efficient in terms of space and time.