# Print all permutations of a number N greater than itself

Given a **number N**, our task is to print those permutations of integer N which are greater than N.

**Examples:**

Input:N = 534

Output:543

Input:N = 324

Output:342, 423, 432

**Approach:** To solve this problem, we can obtain all the lexicographically larger permutations of N using **next_permutation()** method in C++. After getting all such numbers, print them.

For other languages, find the permutations of number N and print the numbers which are greater than N.

Below is the implementation of above approach:

`// C++ implementation to print all the ` `// permutation greater than the integer N ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to print all the permutation ` `// which are greater than N itself ` `void` `printPermutation(` `int` `N) ` `{ ` ` ` `int` `temp = N, count = 0; ` ` ` ` ` `// Iterate and count the ` ` ` `// number of digits in N ` ` ` `while` `(temp > 0) { ` ` ` `count++; ` ` ` `temp /= 10; ` ` ` `} ` ` ` ` ` `// vector to print the ` ` ` `// permutations of N ` ` ` `vector<` `int` `> num(count); ` ` ` ` ` `// Store digits of N ` ` ` `// in the vector num ` ` ` `while` `(N > 0) { ` ` ` `num[count-- - 1] = N % 10; ` ` ` `N = N / 10; ` ` ` `} ` ` ` ` ` `// Iterate over every permutation of N ` ` ` `// which is greater than N ` ` ` `while` `(next_permutation( ` ` ` `num.begin(), num.end())) { ` ` ` ` ` `// Print the current permutaion of N ` ` ` `for` `(` `int` `i = 0; i < num.size(); i++) ` ` ` `cout << num[i]; ` ` ` ` ` `cout << ` `"\n"` `; ` ` ` `} ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `N = 324; ` ` ` ` ` `printPermutation(N); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

342 423 432

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