Numbers having Unique (or Distinct) digits
Last Updated :
23 Jul, 2025
Given a range, print all numbers having unique digits.
Examples :
Input : 10 20
Output : 10 12 13 14 15 16 17 18 19 20 (Except 11)
Input : 1 10
Output : 1 2 3 4 5 6 7 8 9 10
Approach:
As the problem is pretty simple, the only thing to be done is :-
1- Find the digits one by one and keep marking visited digits.
2- If all digits occurs one time only then print that number.
3- Else not.
Implementation:
C++14
#include <array>
#include <iostream>
// Method to print unique digit numbers
// in the range from l to r.
void printUnique(int l, int r)
{
// Start traversing the numbers
for (int i = l; i <= r; ++i) {
int num = i;
std::array<bool, 10> visited = { false };
// Find digits and maintain its hash
while (num != 0) {
// If a digit occurs more than 1 time
// then break
if (visited[num % 10])
break;
visited[num % 10] = true;
num = num / 10;
}
// num will be 0 only when the above loop
// doesn't get broken, that means the
// number is unique, so print it.
if (num == 0)
std::cout << i << " ";
}
}
// Driver method
int main()
{
int l = 1, r = 20;
printUnique(l, r);
return 0;
}
// C++ implementation to find unique digit
// numbers in a range
#include<bits/stdc++.h>
using namespace std;
// Function to print unique digit numbers
// in range from l to r.
void printUnique(int l, int r)
{
// Start traversing the numbers
for (int i=l ; i<=r ; i++)
{
int num = i;
bool visited[10] = {false};
// Find digits and maintain its hash
while (num)
{
// if a digit occurs more than 1 time
// then break
if (visited[num % 10])
break;
visited[num%10] = true;
num = num/10;
}
// num will be 0 only when above loop
// doesn't get break that means the
// number is unique so print it.
if (num == 0)
cout << i << " ";
}
}
// Driver code
int main()
{
int l = 1, r = 20;
printUnique(l, r);
return 0;
}
public class UniqueDigitNumbers {
// Method to print unique digit numbers
// in the range from l to r.
static void printUnique(int l, int r) {
// Start traversing the numbers
for (int i = l; i <= r; ++i) {
int num = i;
boolean[] visited = new boolean[10];
// Find digits and maintain its hash
while (num != 0) {
// If a digit occurs more than 1 time
// then break
if (visited[num % 10]) {
break;
}
visited[num % 10] = true;
num = num / 10;
}
// num will be 0 only when the above loop
// doesn't get broken, that means the
// number is unique, so print it.
if (num == 0) {
System.out.print(i + " ");
}
}
}
// Driver method
public static void main(String[] args) {
int l = 1, r = 20;
printUnique(l, r);
}
}
# Python3 implementation
# to find unique digit
# numbers in a range
# Function to print
# unique digit numbers
# in range from l to r.
def printUnique(l,r):
# Start traversing
# the numbers
for i in range (l, r + 1):
num = i;
visited = [0,0,0,0,0,0,0,0,0,0];
# Find digits and
# maintain its hash
while (num):
# if a digit occurs
# more than 1 time
# then break
if visited[num % 10] == 1:
break;
visited[num % 10] = 1;
num = (int)(num / 10);
# num will be 0 only when
# above loop doesn't get
# break that means the
# number is unique so
# print it.
if num == 0:
print(i, end = " ");
# Driver code
l = 1;
r = 20;
printUnique(l, r);
# This code is
# contributed by mits
// C# implementation to find unique digit
// numbers in a range
using System;
public class GFG {
// Method to print unique digit numbers
// in range from l to r.
static void printUnique(int l, int r)
{
// Start traversing the numbers
for (int i = l ; i <= r ; i++)
{
int num = i;
bool []visited = new bool[10];
// Find digits and maintain
// its hash
while (num != 0)
{
// if a digit occurs more
// than 1 time then break
if (visited[num % 10])
break;
visited[num % 10] = true;
num = num / 10;
}
// num will be 0 only when
// above loop doesn't get
// break that means the number
// is unique so print it.
if (num == 0)
Console.Write(i + " ");
}
}
// Driver method
public static void Main()
{
int l = 1, r = 20;
printUnique(l, r);
}
}
// This code is contributed by Sam007.
<script>
// Javascript implementation to find unique digit
// numbers in a range
// Function to print unique digit numbers
// in range from l to r.
function printUnique(l, r)
{
// Start traversing the numbers
for (let i=l ; i<=r ; i++)
{
let num = i;
let visited = new Array(10);
// Find digits and maintain its hash
while (num)
{
// if a digit occurs more than 1 time
// then break
if (visited[num % 10])
break;
visited[num%10] = true;
num = Math.floor(num/10);
}
// num will be 0 only when above loop
// doesn't get break that means the
// number is unique so print it.
if (num == 0)
document.write(i + " ");
}
}
// Driver code
let l = 1, r = 20;
printUnique(l, r);
// This code is contributed by Mayank Tyagi
</script>
<?php
// PHP implementation to find unique
// digit numbers in a range
// Function to print unique digit
// numbers in range from l to r.
function printUnique($l, $r)
{
// Start traversing the numbers
for ($i = $l ; $i <= $r; $i++)
{
$num = $i;
$visited = (false);
// Find digits and
// maintain its hash
while ($num)
{
// if a digit occurs more
// than 1 time then break
if ($visited[$num % 10])
$visited[$num % 10] = true;
$num = (int)$num / 10;
}
// num will be 0 only when above
// loop doesn't get break that
// means the number is unique
// so print it.
if ($num == 0)
echo $i , " ";
}
}
// Driver code
$l = 1; $r = 20;
printUnique($l, $r);
// This code is contributed by aj_36
?>
Output1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20
Time Complexity : O(nlogn)
Auxiliary Space: O(1)
Another Approach: Use set STL for C++ and Java Collections for Java in order to check if a number has only unique digits. Then we can compare the size of string s formed from a given number and newly created set. For example, let us consider the number 1987, then we can convert the number into the string,
Implementation:
C++
int n;
cin>>n;
string s = to_string(n);
//creating scanner class object for taking user input
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
//converting the number to string using String.valueof(number);
string s = String.valueof(n);
n = int(input())
s = str(n)
int n = Convert.ToInt32(Console.ReadLine());
// converting the number to string using
// String.valueof(number);
string s = Convert.ToString(n);
After that, initialize a set with the contents of string s.
C++
set<int> uniDigits(s.begin(), s.end());
HashSet<Integer> uniDigits = new HashSet<Integer>();
for(int i:s.tocharArray())
{
uniDigits.add(i);
}
var uniDigits = new HashSet<char>(s.ToCharArray());
let uniDigits = new Set();
Then we can compare the size of string s and newly created set uniDigits.
Here is the code for the above approach:
C++14
#include <iostream>
#include <unordered_set>
// Function to print unique numbers
void printUnique(int l, int r) {
// Iterate from l to r
for (int i = l; i <= r; ++i) {
// Convert the number to a string
std::string s = std::to_string(i);
// Convert String to set using C++ unordered_set
std::unordered_set<char> uniDigits(s.begin(), s.end());
// Output if condition satisfies
if (s.length() == uniDigits.size()) {
std::cout << i << " ";
}
}
}
// Driver Code
int main() {
// Input of the lower and higher limits
int l = 1, r = 20;
// Function Call
printUnique(l, r);
return 0;
}
// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print unique
// numbers
void printUnique(int l, int r){
// Iterate from l to r
for (int i = l; i <= r; i++) {
// Convert the no. to
// string
string s = to_string(i);
// Convert string to set using stl
set<int> uniDigits(s.begin(), s.end());
// Output if condition satisfies
if (s.size() == uniDigits.size()) {
cout << i << " ";
}
}
}
// Driver Code
int main()
{
// Input of the lower and
// higher limits
int l = 1, r = 20;
// Function Call
printUnique(l, r);
return 0;
}
// Java code for the above approach
import java.io.*;
class GFG{
// Function to print unique
// numbers
static void printUnique(int l, int r)
{
// Iterate from l to r
for(int i = l; i <= r; i++)
{
// Convert the no. to
// String
String s = String.valueOf(i);
// Convert String to set using Java Collections
HashSet<Integer> uniDigits = new HashSet<Integer>();
for(int c : s.toCharArray())
uniDigits.add(c);
// Output if condition satisfies
if (s.length() == uniDigits.size())
{
System.out.print(i+ " ");
}
}
}
// Driver Code
public static void main(String[] args)
{
// Input of the lower and
// higher limits
int l = 1, r = 20;
// Function Call
printUnique(l, r);
}
}
// This code is contributed by Princi Singh
# Function to print unique
# numbers
def print_unique(l, r):
# Iterate from l to r
for i in range(l, r + 1):
# Convert the no. to
# string
s = str(i)
# Convert string to set
uni_digits = set(s)
# Output if condition satisfies
if len(s) == len(uni_digits):
print(i, end=" ")
# Driver Code
# Input of the lower and
# higher limits
l = 1
r = 20
# Function Call
print_unique(l, r)
// C# code for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to print unique
// numbers
static void printUnique(int l, int r)
{
// Iterate from l to r
for(int i = l; i <= r; i++)
{
// Convert the no. to
// String
String s = String.Join("", i);
// Convert String to set using stl
HashSet<int> uniDigits = new HashSet<int>();
foreach(int c in s.ToCharArray())
uniDigits.Add(c);
// Output if condition satisfies
if (s.Length == uniDigits.Count)
{
Console.Write(i + " ");
}
}
}
// Driver Code
public static void Main(String[] args)
{
// Input of the lower and
// higher limits
int l = 1, r = 20;
// Function Call
printUnique(l, r);
}
}
// This code is contributed by Princi Singh
<script>
// JavaScript code for the above approach
// Function to print unique
// numbers
function printUnique(l, r) {
// Iterate from l to r
for (let i = l; i <= r; i++) {
// Convert the no. to
// string
let s = String(i);
// Convert string to set using stl
let uniDigits = new Set();
for(let c of s.split(""))
uniDigits.add(c);
// Output if condition satisfies
if (s.length == uniDigits.size) {
document.write(i + " ");
}
}
}
// Driver Code
// Input of the lower and
// higher limits
let l = 1, r = 20;
// Function Call
printUnique(l, r);
</script>
Output1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20
Time Complexity: O(nlogn)
Auxiliary Space: O(n)
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