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# 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++

 `// CPP program to find a number less than``// n such that all its digits are distinct``#include ``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;``}`

## Java

 `// 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`

## Python3

 `# 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`

## C#

 `// 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`

## PHP

 `= 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``?>`

## Javascript

 ``

Output:

`8479`

Time Complexity: O(N*log10N) where n is number of elements in given array. As, we are using a loop to traverse N times so it will cost us O(N) time and we are using a while loop which will cost O (logN) as we are decrement by floor division of 10 each time.
Auxiliary Space: O(1), as we are not using any extra space.

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