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Program to check if a given number is Lucky (all digits are different)

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  • Difficulty Level : Basic
  • Last Updated : 05 Aug, 2022
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A number is lucky if all digits of the number are different. How to check if a given number is lucky or not.
Examples: 
 

Input: n = 983
Output: true
All digits are different

Input: n = 9838
Output: false
8 appears twice

We strongly recommend you to minimize your browser and try this yourself first.
The idea is to traverse through every digit of given number and mark the traversed digit as visited. Since the total number of digits is 10, we need a boolean array of size only 10 to mark visited digits.
Below is the implementation of above idea. 
 

C++




// C++ program to check if a given number is lucky
#include<iostream>
using namespace std;
 
// This function returns true if n is lucky
bool isLucky(int n)
{
    // Create an array of size 10 and initialize all
    // elements as false. This array is used to check
    // if a digit is already seen or not.
    bool arr[10];
    for (int i=0; i<10; i++)
        arr[i] = false;
 
    // Traverse through all digits of given number
    while (n > 0)
    {
        // Find the last digit
        int digit = n%10;
 
        // If digit is already seen, return false
        if (arr[digit])
           return false;
 
        // Mark this digit as seen
        arr[digit] = true;
 
        // REmove the last digit from number
        n = n/10;
    }
    return true;
}
 
// Driver program to test above function.
int main()
{
    int arr[] = {1291, 897, 4566, 1232, 80, 700};
    int n = sizeof(arr)/sizeof(arr[0]);
 
    for (int i=0; i<n; i++)
        isLucky(arr[i])? cout << arr[i] << " is Lucky \n":
                         cout << arr[i] << " is not Lucky \n";
    return 0;
}

Java




// Java program to check if
// a given number is lucky
 
class GFG
{
    // This function returns true if n is lucky
    static boolean isLucky(int n)
    {
        // Create an array of size 10 and initialize all
        // elements as false. This array is used to check
        // if a digit is already seen or not.
        boolean arr[]=new boolean[10];
        for (int i = 0; i < 10; i++)
            arr[i] = false;
     
        // Traverse through all digits
        // of given number
        while (n > 0)
        {
            // Find the last digit
            int digit = n % 10;
     
            // If digit is already seen,
            // return false
            if (arr[digit])
            return false;
     
            // Mark this digit as seen
            arr[digit] = true;
     
            // Remove the last digit from number
            n = n / 10;
        }
        return true;
    }
     
    // Driver code
    public static void main (String[] args)
    {
    int arr[] = {1291, 897, 4566, 1232, 80, 700};
        int n = arr.length;
     
        for (int i = 0; i < n; i++)
            if(isLucky(arr[i]))
                System.out.print(arr[i] + " is Lucky \n");
            else
            System.out.print(arr[i] + " is not Lucky \n");
    }
}
 
// This code is contributed by Anant Agarwal.

Python3




# python program to check if a
# given number is lucky
 
import math
 
# This function returns true
# if n is lucky
def isLucky(n):
     
    # Create an array of size 10
    # and initialize all elements
    # as false. This array is
    # used to check if a digit
    # is already seen or not.
    ar = [0] * 10
     
    # Traverse through all digits
    # of given number
    while (n > 0):
         
        #Find the last digit
        digit = math.floor(n % 10)
 
        # If digit is already seen,
        # return false
        if (ar[digit]):
            return 0
 
        # Mark this digit as seen
        ar[digit] = 1
 
        # REmove the last digit
        # from number
        n = n / 10
     
    return 1
 
# Driver program to test above function.
arr = [1291, 897, 4566, 1232, 80, 700]
n = len(arr)
 
for i in range(0, n):
    k = arr[i]
    if(isLucky(k)):
        print(k, " is Lucky ")
    else:
        print(k, " is not Lucky ")
     
# This code is contributed by Sam007.

C#




// C# program to check if
// a given number is lucky
using System;
 
class GFG {
     
    // This function returns true if
    // n is lucky
    static bool isLucky(int n)
    {
         
        // Create an array of size 10
        // and initialize all elements
        // as false. This array is used
        // to check if a digit is
        // already seen or not.
        bool []arr = new bool[10];
         
        for (int i = 0; i < 10; i++)
            arr[i] = false;
     
        // Traverse through all digits
        // of given number
        while (n > 0)
        {
            // Find the last digit
            int digit = n % 10;
     
            // If digit is already seen,
            // return false
            if (arr[digit])
                return false;
     
            // Mark this digit as seen
            arr[digit] = true;
     
            // Remove the last digit
            // from number
            n = n / 10;
        }
         
        return true;
    }
     
    // Driver code
    public static void Main ()
    {
    int []arr = {1291, 897, 4566, 1232,
                               80, 700};
        int n = arr.Length;
     
        for (int i = 0; i < n; i++)
            if(isLucky(arr[i]))
                Console.Write(arr[i] +
                        " is Lucky \n");
            else
            Console.Write(arr[i] +
                    " is not Lucky \n");
    }
}
 
// This code is contributed by sam007.

PHP




<?php
// PHP program to check if a given
// number is lucky
 
// This function returns true
// if n is lucky
function isLucky($n)
{
    // Create an array of size 10 and
    // initialize all elements as false.
    // This array is used to check if a
    // digit is already seen or not.
    $arr = array();
    for ($i = 0; $i < 10; $i++)
        $arr[$i] = false;
 
    // Traverse through all digits
    // of given number
    while ($n > 0)
    {
        // Find the last digit
        $digit = $n % 10;
 
        // If digit is already seen,
        // return false
        if ($arr[$digit])
        return false;
 
        // Mark this digit as seen
        $arr[$digit] = true;
 
        // Remove the last digit
        // from number
        $n = (int)($n / 10);
    }
    return true;
}
 
// Driver Code
$arr = array(1291, 897, 4566,
             1232, 80, 700);
$n = sizeof($arr);
 
for ($i = 0; $i < $n; $i++)
    if(isLucky($arr[$i]))
        echo $arr[$i] , " is Lucky \n";
    else
        echo $arr[$i] , " is not Lucky \n";
 
// This code is contributed by jit_t
?>

Javascript




<script>
 
// Javascript program to check if a given number is lucky
 
// This function returns true if n is lucky
function isLucky(n)
{
    // Create an array of size 10 and initialize all
    // elements as false. This array is used to check
    // if a digit is already seen or not.
    var arr=Array(10).fill(0);
    for (var i=0; i<10; i++)
        arr[i] = false;
 
    // Traverse through all digits of given number
    while (n > 0)
    {
        // Find the last digit
        var digit = n%10;
 
        // If digit is already seen, return false
        if (arr[digit])
        return false;
 
        // Mark this digit as seen
        arr[digit] = true;
 
        // REmove the last digit from number
        n = parseInt(n/10);
    }
    return true;
}
 
// Driver program to test above function.
var arr = [1291, 897, 4566, 1232, 80, 700]
var n = arr.length;
for (var i=0; i<n; i++)
    isLucky(arr[i])? document.write( arr[i] + " is Lucky<br>"):
                    document.write(arr[i] + " is not Lucky<br>");
 
</script>

Output: 
 

1291 is not Lucky
897 is Lucky
4566 is not Lucky
1232 is not Lucky
80 is Lucky
700 is not Lucky

Time Complexity: O(d) where d is a number of digits in the input number. 
Auxiliary Space: O(1)
This article is contributed by Himanshu. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

Method 2:(using STL and sorting)

 In this method we will firstly convert the number into a string. Then we will sort the whole string. Then we will compare every index element with the next index element. If both are equal then we will stop that position and will print that the given number is not lucky.

And if we do not get any index as discussed above then we will print that given number is lucky.

This whole task will be completed in O(K) time where K is the total number of digits of N.

Below is the implementation of the above approach:

C++




// C++ code to check is the number
// lucky or not.
#include <bits/stdc++.h>
using namespace std;
void checklucky(string s,int n)
{
    bool x = false;
 
    // traversing the whole string
    for (int i = 0; i < s.length() - 1; i++)
    {
        // checking next element whether
        // it is equal or not
        if (s[i] == s[i + 1])
        {
            cout << n << " is not lucky number"<<endl;
            x = true;
            break;
        }
    }
    if (!x)
    {
        cout << n << " is lucky number"<<endl;
    }
}
int main()
{
    int n1 = 1234,n2=5868;
    // converting the number from
    // integer to string using
    // C++ STL function.
    string s1 = to_string(n1);
    string s2 = to_string(n2);
     
    // sorting the string
    sort(s1.begin(), s1.end());
    sort(s2.begin(), s2.end());
    //function calling
    checklucky(s1,n1);
    checklucky(s2,n2);
    return 0;
}
 
//this code is contribut by Machhaliya Muhammad

Java




// Java code to check is the number
// lucky or not.
import java.util.*;
 
class GFG {
    static void checklucky(char[] s, int n)
    {
        boolean x = false;
 
        // traversing the whole string
        for (int i = 0; i < s.length - 1; i++)
        {
           
            // checking next element whether
            // it is equal or not
            if (s[i] == s[i + 1]) {
                System.out.println(
                    n + " is not lucky number");
                x = true;
                break;
            }
        }
        if (!x) {
            System.out.println(n + " is lucky number");
        }
    }
 
    public static void main(String[] args)
    {
        int n1 = 1234, n2 = 5868;
       
        // converting the number from
        // integer to string using
        // library function.
        char[] s1 = String.valueOf(n1).toCharArray();
        char[] s2 = String.valueOf(n2).toCharArray();
 
        // sorting the string
        Arrays.sort(s1);
        Arrays.sort(s2);
 
        // function calling
        checklucky(s1, n1);
        checklucky(s2, n2);
    }
}
 
// this code is contribut by phasing17

Python3




# Python3 code to check is the number
# lucky or not.
def checklucky(s, n):
    x = False
 
    # traversing the whole string
    for i in range(len(s) - 1):
 
        # checking next element whether
        # it is equal or not
        if (s[i] == s[i + 1]):
            print(n, "is not a lucky number")
            return
 
    if not x:
        print(n, "is a lucky number")
 
 
# Driver Code
n1 = 1234
n2 = 5868
 
# converting the number from
# integer to string
s1 = str(n1)
s2 = str(n2)
 
# sorting the string
s1 = "".join(sorted(s1))
s2 = "".join(sorted(s2))
 
# function calling
checklucky(s1, n1)
checklucky(s2, n2)
 
 
# this code is contribut by phasing17

C#




// C# code to check is the number
// lucky or not.
 
using System;
using System.Collections.Generic;
 
class GFG {
    static void checklucky(char[] s, int n)
    {
        bool x = false;
 
        // traversing the whole string
        for (int i = 0; i < s.Length - 1; i++) {
            // checking next element whether
            // it is equal or not
            if (s[i] == s[i + 1]) {
                Console.WriteLine(n
                                  + " is not lucky number");
                x = true;
                break;
            }
        }
        if (!x) {
            Console.WriteLine(n + " is lucky number");
        }
    }
 
    public static void Main(string[] args)
    {
        int n1 = 1234, n2 = 5868;
        // converting the number from
        // integer to string using
        // library function.
        char[] s1 = Convert.ToString(n1).ToCharArray();
        char[] s2 = Convert.ToString(n2).ToCharArray();
 
        // sorting the string
        Array.Sort(s1);
        Array.Sort(s2);
 
        // function calling
        checklucky(s1, n1);
        checklucky(s2, n2);
    }
}
 
// this code is contribut by phasing17

Javascript




// JavaScript code to check is the number
// lucky or not.
 
function checklucky(s, n)
{
    let x = false;
 
    // traversing the whole string
    for (var i = 0; i < s.length - 1; i++)
    {
        // checking next element whether
        // it is equal or not
        if (s[i] == s[i + 1])
        {
            console.log(n + " is not lucky number");
            x = true;
            break;
        }
    }
    if (!x)
    {
        console.log(n + " is lucky number");
    }
}
 
let n1 = 1234,n2=5868;
 
// converting the number from
// integer to string   
let s1 = n1.toString();
let s2 = n2.toString();
 
// sorting the string
s1 = s1.split("");
s2 = s2.split("");
s1.sort();
s2.sort();
s1 = s1.join("");
s2 = s2.join("");
 
// function calling
checklucky(s1,n1);
checklucky(s2,n2);
 
// this code is contribut by phasing17

Output

1234 is lucky number
5868 is not lucky number

Time Complexity: O(n1*logn1 + n2*logn2), where n1 and n2 represents the lengths of the given strings.
Auxiliary Space: O(1), no extra space is required, so it is a constant.


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