Print a number strictly less than a given number such that all its digits are distinct.

Given a positive number n, print a number less than n such that all its digits are distinct.

Examples:

Input : 1134
Output : 1098
1098  is the largest number smaller than
1134 such that all digits are distinct.

Input : 4559
Output : 4539


The problem can easily be solved by using counting. Firstly, loop through numbers less than n and for each number count the frequency of the digits using count array. If all the digits occur only once than we print that number. The answer always exists so there is no problem of infinite loop.

C++

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// CPP program to find a number less than
// n such that all its digits are distinct
#include <bits/stdc++.h>
using namespace std;
  
// Function to find a number less than
// n such that all its digits are distinct
int findNumber(int n)
{
    // looping through numbers less than n
    for (int i = n - 1; >=0 ; i--) {
  
        // initializing a hash array
        int count[10] = { 0 }; 
  
        int x = i; // creating a copy of i
  
        // initializing variables to compare lengths of digits
        int count1 = 0, count2 = 0;
  
        // counting frequency of the digits
        while (x) {
            count[x % 10]++; 
            x /= 10;
            count1++;
        }
  
        // checking if each digit is present once
        for (int j = 0; j < 10; j++) {
            if (count[j] == 1) 
                count2++;
        }         
        if (count1 == count2)
            return i;
    }
}
  
// Driver code
int main()
{
    int n = 8490;
    cout << findNumber(n);
    return 0;
}

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Java

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// Java program to find a number less than
// n such that all its digits are distinct
  
class GFG{
// Function to find a number less than
// n such that all its digits are distinct
static int findNumber(int n)
{
    // looping through numbers less than n
    for (int i = n - 1;i >=0 ; i--) {
  
        // initializing a hash array
        int[] count=new int[10]; 
  
        int x = i; // creating a copy of i
  
        // initializing variables to compare lengths of digits
        int count1 = 0, count2 = 0;
  
        // counting frequency of the digits
        while (x>0) {
            count[x % 10]++; 
            x /= 10;
            count1++;
        }
  
        // checking if each digit is present once
        for (int j = 0; j < 10; j++) {
            if (count[j] == 1
                count2++;
        }         
        if (count1 == count2)
            return i;
    }
    return -1;
}
  
// Driver code
public static void main(String[] args)
{
    int n = 8490;
    System.out.println(findNumber(n));
}
}
// This code is contributed by mits

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Python3

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# python 3 program to find a number less than
# n such that all its digits are distinct
  
# Function to find a number less than
# n such that all its digits are distinct
def findNumber(n):
    # looping through numbers less than n
    i = n-1
    while(i>=0):
        # initializing a hash array
        count = [0 for i in range(10)] 
  
        x = i
        # creating a copy of i
  
        # initializing variables to compare lengths of digits
        count1 = 0
        count2 = 0
  
        # counting frequency of the digits
        while (x):
          
            count[x%10] += 1
            x = int(x / 10)
            count1 += 1
          
        # checking if each digit is present once
        for j in range(0,10,1):
            if (count[j] == 1):
                count2 += 1
                  
        if (count1 == count2):
            return i
        i -= 1
      
# Driver code
if __name__ == '__main__':
  
    n = 8490
    print(findNumber(n))
  
# This code is implemented by
# Surendra_Gangwar 

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C#

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// C# program to find a number less than
// n such that all its digits are distinct
using System;
  
class GFG
{
      
// Function to find a number less than
// n such that all its digits are distinct
static int findNumber(int n)
{
    // looping through numbers less than n
    for (int i = n - 1; i >= 0; i--)
    {
  
        // initializing a hash array
        int[] count = new int[10]; 
  
        int x = i; // creating a copy of i
  
        // initializing variables to compare
        // lengths of digits
        int count1 = 0, count2 = 0;
  
        // counting frequency of the digits
        while (x > 0)
        {
            count[x % 10]++; 
            x /= 10;
            count1++;
        }
  
        // checking if each digit is
        // present once
        for (int j = 0; j < 10; j++) 
        {
            if (count[j] == 1) 
                count2++;
        }     
        if (count1 == count2)
            return i;
    }
    return -1;
}
  
// Driver code
static public void Main ()
{
    int n = 8490;
    Console.WriteLine(findNumber(n));
}
}
  
// This code is contributed by akt_mit

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PHP

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<?php
// PHP program to find a number less than
// n such that all its digits are distinct
  
// Function to find a number less than
// n such that all its digits are distinct
function findNumber($n)
{
    // looping through numbers less than n
    for ($i = $n - 1;$i >= 0 ; $i--) 
    {
  
        // initializing a hash array
        $count = array_fill(0, 10, 0); 
  
        $x = $i; // creating a copy of i
  
        // initializing variables to 
        // compare lengths of digits
        $count1 = 0; $count2 = 0;
  
        // counting frequency of the digits
        while ($x
        {
            $count[$x % 10]++; 
            $x = (int)($x / 10);
            $count1++;
        }
  
        // checking if each digit 
        // is present once
        for ($j = 0; $j < 10; $j++) 
        {
            if ($count[$j] == 1) 
                $count2++;
        }     
        if ($count1 == $count2)
            return $i;
    }
}
  
// Driver code
$n = 8490;
echo findNumber($n);
  
// This code is contributed 
// by Akanksha Rai
?>

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Output:

8479


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