Encuentre el número más grande más pequeño que N con el mismo conjunto de dígitos

Dado un número N en forma de string. La tarea es encontrar el número más grande que tenga el mismo conjunto de dígitos que N y sea más pequeño que N. Si no es posible encontrar ese número, imprima «no es posible».

Ejemplos : 

Input:  N = "218765"
Output: 218756

Input:  N = "1234"
Output: Not Possible

Input: N = "262345"
Output: 256432

Las siguientes son algunas observaciones sobre el mayor número menor que N:  

1. Si todos los dígitos están ordenados en orden ascendente, la salida siempre es «No posible». Por ejemplo, 1234.
2. Si todos los dígitos están ordenados en orden descendente, entonces necesitamos intercambiar los dos últimos dígitos. Por ejemplo, 4321.
3. Para otros casos, necesitamos procesar el número del lado derecho (¿por qué? Porque necesitamos encontrar el mayor de todos los números más pequeños).

Algoritmo :  

• Recorra el número dado desde el dígito más a la derecha, siga recorriendo hasta que encuentre un dígito que sea mayor que el dígito recorrido anteriormente. Por ejemplo, si el número de entrada es «262345», nos detenemos en 6 porque 6 es mayor que el dígito anterior 5. Si no encontramos ese dígito, la salida es «No posible».
• Ahora busque en el lado derecho del dígito anterior ‘d’ el dígito más grande menor que ‘d’. Para “262345?, el lado derecho de 6 contiene “2345”. El mayor dígito menor que 6 es 5.
• Intercambie los dos dígitos encontrados arriba, obtenemos «252346» en el ejemplo anterior.
• Ahora ordene todos los dígitos desde la posición junto a ‘d’ hasta el final del número en orden descendente. El número que obtenemos después de ordenar es la salida. Para el ejemplo anterior, ordenamos los dígitos en negrita 25 2346 . Obtenemos 256432 , que es el número más pequeño anterior para la entrada 262345 .

A continuación se muestra la implementación del enfoque anterior:  

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find previous number
void findPrevious(string number, int n)
{
    int i, j;
 
    // I) Start from the right most digit
    // and find the first digit that is
    // smaller than the digit next to it.
    for (i = n - 1; i > 0; i--)
        if (number[i] < number[i - 1])
            break;
 
    // If no such digit is found
    // then all digits are in ascending order
    // means there cannot be a smallest number
    // with same set of digits
    if (i == 0) {
        cout << "Previous number is not possible";
        return;
    }
 
    // II) Find the greatest digit on
    // right side of (i-1)'th digit that is
    // smaller than number[i-1]
    int x = number[i - 1], greatest = i;
    for (j = i; j < n; j++)
        if (number[j] < x && number[j] > number[greatest])
            greatest = j;
 
    // III) Swap the above found smallest digit with number[i-1]
    swap(number[greatest], number[i - 1]);
 
    // IV) Sort the digits after (i-1) in descending order
    sort(number.begin() + i, number.begin() + n, greater<char>());
 
    cout << "Greatest smaller number with same set of digits is " << number;
 
    return;
}
 
// Driver code
int main()
{
    string digits = "262345";
    int n = digits.length();
 
    findPrevious(digits, n);
 
    return 0;
}

// Java implementation of the above approach
import java.util.*;
 
class GFG
{
 
    // Function to find previous number
    static void findPrevious(char[] number, int n)
    {
        int i, j;
 
        // I) Start from the right most digit
        // and find the first digit that is
        // smaller than the digit next to it.
        for (i = n - 1; i > 0; i--)
        {
            if (number[i] < number[i - 1])
            {
                break;
            }
        }
 
        // If no such digit is found
        // then all digits are in ascending order
        // means there cannot be a smallest number
        // with same set of digits
        if (i == 0)
        {
            System.out.print("Previous number is not possible");
            return;
        }
 
        // II) Find the greatest digit on
        // right side of (i-1)'th digit that is
        // smaller than number[i-1]
        int x = number[i - 1], greatest = i;
        for (j = i; j < n; j++)
        {
            if (number[j] < x && number[j] > number[greatest])
            {
                greatest = j;
            }
        }
 
        // III) Swap the above found smallest digit with number[i-1]
        swap(number, greatest, i - 1);
 
        // IV) Sort the digits after (i-1) in descending order
        Arrays.sort(number, i, n);
        reverse(number, i, n - 1);
        System.out.print("Greatest smaller number with" +
              "same set of digits is " + String.valueOf(number));
 
        return;
    }
 
    static String swap(char[] ch, int i, int j)
    {
        char temp = ch[i];
        ch[i] = ch[j];
        ch[j] = temp;
        return String.valueOf(ch);
    }
 
    static void reverse(char str[], int start, int end)
    {
 
        // Temporary variable to store character
        char temp;
        while (start <= end)
        {
            // Swapping the first and last character
            temp = str[start];
            str[start] = str[end];
            str[end] = temp;
            start++;
            end--;
        }
    }
     
    // Driver code
    public static void main(String[] args)
    {
        String digits = "262345";
        int n = digits.length();
 
        findPrevious(digits.toCharArray(), n);
    }
}
 
// This code has been contributed by 29AjayKumar

# Python3 implementation of the above approach
 
# Function to find previous number
def findPrevious(number, n):
 
    # This is necessary as strings
    # do not support item assignment
    number = list(number)
 
    i, j = -1, -1
 
    # I) Start from the right most digit
    # and find the first digit that is
    # smaller than the digit next to it.
    for i in range(n - 1, 0, -1):
        if number[i] < number[i - 1]:
            break
     
    # If no such digit is found
    # then all digits are in ascending order
    # means there cannot be a smallest number
    # with same set of digits
    if i == 0:
        print("Previous number is not possible")
        return
     
    x, greatest = number[i - 1], i
 
    # II) Find the greatest digit on
    # right side of(i-1)'th digit that is
    # smaller than number[i-1]
    for j in range(i, n):
        if (number[j] < x and
            number[j] > number[greatest]):
            greatest = j
     
    # III) Swap the above found smallest digit
    # with number[i-1]
    (number[greatest],
     number[i - 1]) = (number[i - 1],
                       number[greatest])
 
    l = number[i:]
    del number[i:]
 
    # IV) Sort the digits after(i-1)
    # in descending order
    l.sort(reverse = True)
 
    number += l
 
    # Again join the list to make it string
    number = '' . join(number)
    print("Greatest smaller number with",
          "same set of digits is", number)
 
    return
 
# Driver Code
if __name__ == "__main__":
    digits = "262345"
    n = len(digits)
 
    findPrevious(digits, n)
 
# This code is contributed by sanjeev2552

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to find previous number
    static void findPrevious(char[] number, int n)
    {
        int i, j;
 
        // I) Start from the right most digit
        // and find the first digit that is
        // smaller than the digit next to it.
        for (i = n - 1; i > 0; i--)
        {
            if (number[i] < number[i - 1])
            {
                break;
            }
        }
 
        // If no such digit is found
        // then all digits are in ascending order
        // means there cannot be a smallest number
        // with same set of digits
        if (i == 0)
        {
            Console.Write("Previous number is not possible");
            return;
        }
 
        // II) Find the greatest digit on
        // right side of (i-1)'th digit that is
        // smaller than number[i-1]
        int x = number[i - 1], greatest = i;
        for (j = i; j < n; j++)
        {
            if (number[j] < x && number[j] > number[greatest])
            {
                greatest = j;
            }
        }
 
        // III) Swap the above found
        // smallest digit with number[i-1]
        swap(number, greatest, i - 1);
 
        // IV) Sort the digits after (i-1) in descending order
        Array.Sort(number, i, n-i);
        reverse(number, i, n - 1);
        Console.Write("Greatest smaller number with" +
            "same set of digits is " + String.Join("",number));
 
        return;
    }
 
    static String swap(char[] ch, int i, int j)
    {
        char temp = ch[i];
        ch[i] = ch[j];
        ch[j] = temp;
        return String.Join("",ch);
    }
 
    static void reverse(char []str, int start, int end)
    {
 
        // Temporary variable to store character
        char temp;
        while (start <= end)
        {
            // Swapping the first and last character
            temp = str[start];
            str[start] = str[end];
            str[end] = temp;
            start++;
            end--;
        }
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        String digits = "262345";
        int n = digits.Length;
 
        findPrevious(digits.ToCharArray(), n);
    }
}
 
/* This code contributed by PrinciRaj1992 */

<script>
 
// Javascript implementation of the above approach
 
// Function to find previous number
function findPrevious(number, n)
{
    let i, j;
     
    // I) Start from the right most digit
    // and find the first digit that is
    // smaller than the digit next to it.
    for(i = n - 1; i > 0; i--)
    {
        if (number[i] < number[i - 1])
        {
            break;
        }
    }
 
    // If no such digit is found then all
    // digits are in ascending order means
    // there cannot be a smallest number
    // with same set of digits
    if (i == 0)
    {
        document.write("Previous number " +
                       "is not possible");
        return;
    }
 
    // II) Find the greatest digit on right
    // side of (i-1)'th digit that is smaller
    // than number[i-1]
    let x = number[i - 1], greatest = i;
    for(j = i; j < n; j++)
    {
        if (number[j] < x && number[j] >
            number[greatest])
        {
            greatest = j;
        }
    }
 
    // III) Swap the above found smallest
    // digit with number[i-1]
    swap(number, greatest, i - 1);
 
    // IV) Sort the digits after (i-1)
    // in descending order
    number = number.slice(0, i).concat(
             number.slice(i, n).sort(
                 function(a, b){return b - a;}));
     
    document.write("Greatest smaller number with " +
                   "same set of digits is " +
                   (number).join(""));
 
    return;
}
 
function swap(ch, i, j)
{
    let temp = ch[i];
    ch[i] = ch[j];
    ch[j] = temp;
    return String.valueOf(ch);
}
 
function reverse(str, start, end)
{
     
    // Temporary variable to store character
    let temp;
     
    while (start <= end)
    {
         
        // Swapping the first and last character
        temp = str[start];
        str[start] = str[end];
        str[end] = temp;
        start++;
        end--;
    }
}
 
// Driver code
let digits = "262345";
let n = digits.length;
 
findPrevious(digits.split(""), n);
 
// This code is contributed by rag2127
 
</script>

Greatest smaller number with same set of digits is 256432

Complejidad de tiempo: O (nlogn), donde n representa la longitud de la string dada.
Espacio auxiliar: O(1), no se requiere espacio adicional, por lo que es una constante.