Nearest greater number by interchanging the digits

Given two integers A and B. The task is to find the nearest greater value to B by interchanging the digits of A. If no such permutataion possible then print -1.

Examples:

Input: A = 459, B = 500
Output: 549
549 is the nearest greater.



Input: A = 321, B = 567
Output: -1

Input: A = 231, B = 125
Output: 132

Prerequisites: All permutations of a string

Approach:

  • Set the minimum value of min1 by using the Integer.MAX_VALUE
  • Interchange the digit of A by using above mentioned permutation method.
  • Check if the permutataion of A is less than min1 or not. If less then update min1 as A.
  • Repeat this for all permutataions of A and find the minimum greater value

Below is the implementation of the above approach :

C++

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// C++ program to find nearest greater value
#include <bits/stdc++.h>
using namespace std;
  
int min1 = INT_MAX;
int _count = 0;
  
// Find all the possible permutation of Value A.
int permutation(string str1, int i, int n, int p)
{
    if (i == n)
    {
        // Convert into Integer
        int q = stoi(str1);
  
        // Find the minimum value of A by interchanging
        // the digit of A and store min1.
        if (q - p > 0 && q < min1)
        {
            min1 = q;
            _count = 1;
        }
    }
    else
    {
        for (int j = i; j <= n; j++)
        {
            // Swap two string character
            swap(str1[i], str1[j]);
            permutation(str1, i + 1, n, p);
            swap(str1[i], str1[j]);
        }
    }
    return min1;
}
  
// Driver code
int main()
{
    int A = 213;
    int B = 111;
  
    // Convert integer value into string to
    // find all the permutation of the number
    string str1 = to_string(A);
    int len = str1.length();
    int h = permutation(str1, 0, len - 1, B);
  
    // count=1 means number greater than B exists
    _count ? cout << h << endl : cout << -1 << endl;
  
    return 0;
}
  
// This code is contributed by
// sanjeev2552

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Java

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// JAVA program to find nearest greater value
import java.io.*;
import java.util.*;
  
class GFG {
    static int min1 = Integer.MAX_VALUE;
    static int count = 0;
  
    // Find all the possible permutation of Value A.
    public int permutation(String str1, int i, int n, int p)
    {
  
        if (i == n) {
  
            // Convert into Integer
            int q = Integer.parseInt(str1);
  
            // Find the minimum value of A by interchanging
            // the digit of A and store min1.
            if (q - p > 0 && q < min1) {
                min1 = q;
                count = 1;
            }
        }
  
        else {
            for (int j = i; j <= n; j++) {
                str1 = swap(str1, i, j);
                permutation(str1, i + 1, n, p);
                str1 = swap(str1, i, j);
            }
        }
        return min1;
    }
  
    // Swap two string character
    public String swap(String str, int i, int j)
    {
        char ch[] = str.toCharArray();
        char temp = ch[i];
        ch[i] = ch[j];
        ch[j] = temp;
        // Return the string after
        // swapping between two character.
        return String.valueOf(ch);
    }
  
    // Driver code
    public static void main(String[] args)
    {
  
        int A = 213;
        int B = 111;
  
        GFG gfg = new GFG();
  
        // Convert integer value into string to
        // find all the permutation of the number
        String str1 = Integer.toString(A);
        int len = str1.length();
        int h = gfg.permutation(str1, 0, len - 1, B);
  
        // count=1 means number greater than B exists
        if (count == 1)
            System.out.println(h);
        else
            System.out.println(-1);
    }
}

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Python3

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# Python3 program to find nearest greater value
min1 = 10**9
_count = 0
  
# Find all the possible permutation of Value A.
def permutation(str1, i, n, p):
    global min1, _count
    if (i == n):
          
        # Convert into Integer
        str1 = "".join(str1)
        q = int(str1)
  
        # Find the minimum value of A 
        # by interchanging the digits
        # of A and store min1.
        if (q - p > 0 and q < min1):
            min1 = q
            _count = 1
    else:
        for j in range(i, n + 1):
              
            # Swap two character)
            str1[i], str1[j] = str1[j], str1[i]
            permutation(str1, i + 1, n, p)
            str1[i], str1[j] = str1[j], str1[i]
  
    return min1
  
# Driver code
A = 213
B = 111
  
# Convert integer value into to
# find all the permutation of the number
str2 = str(A)
str1 = [i for i in str2]
le = len(str1)
  
h = permutation(str1, 0, le - 1, B)
  
# count=1 means number greater than B exists
if _count == 1:
    print(h)
else:
    print(-1)
  
# This code is contributed by
# mohit

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C#

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// C# program to find nearest greater value
using System;
      
class GFG
{
    static int min1 = int.MaxValue;
    static int count = 0;
  
    // Find all the possible permutation of Value A.
    public int permutation(String str1, int i, 
                                 int n, int p)
    {
        if (i == n) 
        {
  
            // Convert into Integer
            int q = int.Parse(str1);
  
            // Find the minimum value of A by interchanging
            // the digit of A and store min1.
            if (q - p > 0 && q < min1)
            {
                min1 = q;
                count = 1;
            }
        }
  
        else 
        {
            for (int j = i; j <= n; j++) 
            {
                str1 = swap(str1, i, j);
                permutation(str1, i + 1, n, p);
                str1 = swap(str1, i, j);
            }
        }
        return min1;
    }
  
    // Swap two string character
    public String swap(String str, int i, int j)
    {
        char []ch = str.ToCharArray();
        char temp = ch[i];
        ch[i] = ch[j];
        ch[j] = temp;
          
        // Return the string after
        // swapping between two character.
        return String.Join("", ch);
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        int A = 213;
        int B = 111;
  
        GFG gfg = new GFG();
  
        // Convert integer value into string to
        // find all the permutation of the number
        String str1 = A.ToString();
        int len = str1.Length;
        int h = gfg.permutation(str1, 0, len - 1, B);
  
        // count=1 means number greater than B exists
        if (count == 1)
            Console.WriteLine(h);
        else
            Console.WriteLine(-1);
    }
}
  
// This code is contributed by PrinciRaj1992 

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Output:

123


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