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++
// 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 itselfvoid 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 Codeint main(){ int N = 324; printPermutation(N); return 0;} |
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Java
// Java implementation to print all the// permutation greater than the integer Nimport java.util.*;class GFG{static 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 int[] num = new int[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)) { // Print the current permutaion of N for (int i = 0; i < num.length; i++) System.out.print(num[i]); System.out.print("\n"); } } // Function to print all the permutation // which are greater than N itself static boolean next_permutation(int[] p) { for (int a = p.length - 2; a >= 0; --a) if (p[a] < p[a + 1]) for (int b = p.length - 1;; --b) if (p[b] > p[a]) { int t = p[a]; p[a] = p[b]; p[b] = t; for (++a, b = p.length - 1; a < b; ++a, --b) { t = p[a]; p[a] = p[b]; p[b] = t; } return true; } return false; }// Driver Codepublic static void main(String[] args) { int N = 324; printPermutation(N);}}// This code contributed by sapnasingh4991 |
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C#
// C# implementation to print all the// permutation greater than the integer Nusing System;class GFG{static 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 int[] num = new int[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)) { // Print the current permutaion of N for (int i = 0; i < num.Length; i++) Console.Write(num[i]); Console.Write("\n"); } } // Function to print all the permutation // which are greater than N itself static bool next_permutation(int[] p) { for (int a = p.Length - 2; a >= 0; --a) if (p[a] < p[a + 1]) for (int b = p.Length - 1;; --b) if (p[b] > p[a]) { int t = p[a]; p[a] = p[b]; p[b] = t; for (++a, b = p.Length - 1; a < b; ++a, --b) { t = p[a]; p[a] = p[b]; p[b] = t; } return true; } return false; }// Driver Codepublic static void Main(String[] args) { int N = 324; printPermutation(N);}}// This code is contributed by Rohit_ranjan |
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Output:
342 423 432
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