Given two numbers A and B, the task is to find the arrangement of digits of A such that it is just greater than the given number B, i.e., to find the minimum value permutation of A greater than B. If no such permutation is possible then print -1
Examples:
Input: A = 9236, B = 3125
Output: 3269
Explanation:
The minimum number greater than 3125 formed from the digits of A is 3269.
Input: A = 1234, B = 9879
Output: -1
Approach: The idea is to use next_permutation() and stol(). The following steps can be followed to compute the answer:
- Take both the numbers as String input to make use of next_permutation().
- Use stol() to find the long value of B.
- Then find the lowest permutation of the number A.
- For each permutation of A, check whether the number is greater than B.
- If any permutation is greater than the number B then it is one of the possible answers. Select the minimum of all the possible answers.
- If no such number is present print -1.
Below is the implementation of the above approach:
CPP
// C++ program to find the greater permutation#include <bits/stdc++.h>using namespace std;#define ll long long#define inf 999999999999999999// Function to find the greater permutationll solve(string a, string b){ ll n, val, ans = inf; // Convert the string B to long val = stol(b); n = a.length(); // To find the lowest permutation // of the number sort(a.begin(), a.end()); // Find if the lowest permutation of A is // greater than the given number B if (stol(a) > val) { ans = min((ll)stol(a), ans); } // Find all the permutations of A while (next_permutation(a.begin(), a.end())) { if (stol(a) > val) { ans = min((ll)stol(a), ans); } } // If ans is not the initial value // then return ans if (ans != inf) { return ans; } // Else return -1 else { return -1; }}// Driver codeint main(){ string a, b; ll ans; a = "9236"; b = "3145"; ans = solve(a, b); cout << ans;} |
Java
// Java program to find the greater permutationimport java.util.*;class GFG { private static ArrayList<String> _permutations = new ArrayList<String>(); static long inf = 99999999; private static void GetPer(char[] list) { int x = list.length - 1; GetPer(list, 0, x); } private static void GetPer(char[] list, int k, int m) { if (k == m) { _permutations.add(new String(list)); } else for (int i = k; i <= m; i++) { char temp = list[k]; list[k] = list[i]; list[i] = temp; GetPer(list, k + 1, m); temp = list[k]; list[k] = list[i]; list[i] = temp; } } // Function to find the greater permutation static long solve(String a, String b) { long n, val, ans = inf; // Convert the string B to long val = Integer.valueOf(b); n = a.length(); // To find the lowest permutation // of the number char[] a1 = a.toCharArray(); Arrays.sort(a1); a = new String(a1); // Find if the lowest permutation of A is // greater than the given number B if (Integer.valueOf(a) > val) { ans = Math.min(Integer.valueOf(a), ans); } GetPer(a.toCharArray()); // Find all the permutations of A for (String permut : _permutations) { if (Integer.valueOf(permut) > val) { ans = Math.min(Integer.valueOf(permut), ans); } } // If ans is not the initial value // then return ans if (ans != inf) { return ans; } // Else return -1 else { return -1; } } // Driver code public static void main(String[] args) { String a, b; long ans; a = "9236"; b = "3145"; ans = solve(a, b); System.out.println(Long.valueOf(ans)); }}// This code is contributed by phasing17 |
Python3
# Python program to find the greater permutationfrom itertools import *inf = 9999999999999# Function to find the greater permutationdef solve(a, b): a = list(a) ans = inf; # Convert the string B to long val = int(b); n = len(a); # To find the lowest permutation # of the number a.sort() # Find if the lowest permutation of A is # greater than the given number B if (int("".join(a)) > val) : ans = min(int(a), ans); # Find all the permutations of A flag = False; permuts = list(permutations(a)) for permut in permuts: curr = int("".join(permut)) if (curr > val) : ans = min(curr, ans); # If ans is not the initial value # then return ans if (ans != inf): return ans; # Else return -1 else: return -1; # Driver codea = "9236";b = "3145";ans = solve(a, b);print(ans)# This code is contributed by phasing17 |
C#
// C# program to find the greater permutationusing System;using System.Collections.Generic;class GFG { private static List<string> _permutations = new List<string>(); static long inf = 99999999999999; private static void Swap(ref char a, ref char b) { if (a == b) return; a ^= b; b ^= a; a ^= b; } private static void GetPer(char[] list) { int x = list.Length - 1; GetPer(list, 0, x); } private static void GetPer(char[] list, int k, int m) { if (k == m) { _permutations.Add(new string(list)); } else for (int i = k; i <= m; i++) { Swap(ref list[k], ref list[i]); GetPer(list, k + 1, m); Swap(ref list[k], ref list[i]); } } // Function to find the greater permutation static long solve(string a, string b) { long n, val, ans = inf; // Convert the string B to long val = Convert.ToInt64(b); n = a.Length; // To find the lowest permutation // of the number char[] a1 = a.ToCharArray(); Array.Sort(a1); a = new string(a1); // Find if the lowest permutation of A is // greater than the given number B if (Convert.ToInt64(a) > val) { ans = Math.Min(Convert.ToInt64(a), ans); } GetPer(a.ToCharArray()); // Find all the permutations of A foreach(string permut in _permutations) { if (Convert.ToInt64(permut) > val) { ans = Math.Min(Convert.ToInt64(permut), ans); } } // If ans is not the initial value // then return ans if (ans != inf) { return ans; } // Else return -1 else { return -1; } } // Driver code public static void Main(string[] args) { string a, b; long ans; a = "9236"; b = "3145"; ans = solve(a, b); Console.WriteLine(ans); }}// This code is contributed by phasing17 |
Javascript
// JS program to find the greater permutationlet inf = 9999999999999var nextPermutation = function(N) { const swap = (i, j) => [N[i],N[j]] = [N[j],N[i]] let len = N.length - 1, i for (i = len - 1; N[i] >= N[i+1];) i-- let j = i + 1, k = len while (j < k) swap(j++,k--) if (i >= 0) { for (j = i + 1; N[i] >= N[j];) j++ swap(i,j) }};// Function to find the greater permutationfunction solve(a, b){ a = a.split("") let n, val, ans = inf; // Convert the string B to long val = parseInt(b); n = a.length; // To find the lowest permutation // of the number a.sort() // Find if the lowest permutation of A is // greater than the given number B if (parseInt(a) > val) { ans = Math.min(parseInt(a), ans); } // Find all the permutations of A a = a.map(function(item) { return parseInt(item);}); x = [...a]; x.reverse(); let flag = false; while (true) { nextPermutation(a); let curr = parseInt(a.join("")) if (curr > val) { ans = Math.min(curr, ans); } if (flag) break; if (a.join("") == x.join("")) flag = true; } // If ans is not the initial value // then return ans if (ans != inf) { return ans; } // Else return -1 else { return -1; }}// Driver codelet a, b;let ans;a = "9236";b = "3145";ans = solve(a, b);console.log(ans)// This code is contributed by phasing17 |
Output:
3269
Time Complexity:O(n * log n)
Space Complexity: O(1) as no extra space has been taken.
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