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 <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; ` `} ` |

## 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

`<?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 ` `?> ` |

**Output:**

8479

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