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 permutation 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 permutation of A is less than min1 or not. If less then update min1 as A.
- Repeat this for all permutations of A and find the minimum greater value
Below is the implementation of the above approach :
C++
// 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 |
Java
// 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 );
}
} |
Python3
# 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 |
C#
// 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 |
Javascript
<script> // JAVAscript program to find nearest greater value let min1 = Number.MAX_VALUE; let count = 0; // Find all the possible permutation of Value A. function permutation(str1,i,n,p)
{ if (i == n) {
// Convert into Integer
let q = 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 (let 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
function swap(str,i,j)
{ let ch = str.split( "" );
let temp = ch[i];
ch[i] = ch[j];
ch[j] = temp;
// Return the string after
// swapping between two character.
return (ch).join( "" );
} // Driver code let A = 213; let B = 111;
// Convert integer value into string to
// find all the permutation of the number
let str1 = A.toString();
let len = str1.length;
let h = permutation(str1, 0, len - 1, B);
// count=1 means number greater than B exists
if (count == 1)
document.write(h);
else
document.write(-1);
// This code is contributed by unknown2108 </script> |
Output:
123