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; 
} 

Output:

342
423
432

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