Permutación lexicográficamente más pequeña de una string con subsecuencias dadas

Posted on by Rudeus Greyrat

Dada una string que consta solo de dos caracteres en minúsculas  X y  y dos números  pags y  q . La tarea es imprimir la permutación lexicográficamente más pequeña de la string dada de modo que el recuento de subsecuencias de  xy es  pags y de  yx es  q . Si no existe tal string, imprima «Imposible» (sin comillas).
Ejemplos: 
 

Input: str = "yxxyx", p = 3, q = 3 
Output: xyxyx

Input: str = "yxxy", p = 3, q = 2
Output: Impossible

Enfoque: en primer lugar, por inducción se puede demostrar que el producto de la cuenta de ‘x’ y la cuenta de ‘y’ debe ser igual a la suma de la cuenta de una subsecuencia de ‘xy’ y ‘yx’ para cualquier string dada . Si esto no se cumple, la respuesta es ‘Imposible’, de lo contrario, la respuesta siempre existirá.
Ahora, ordene la string dada para que el recuento de una subsecuencia de ‘yx’ sea cero. Sea nx la cuenta de ‘x’ y ny la cuenta de ‘y’. sean a y b la cuenta de la subsecuencia ‘xy’ y ‘yx’ respectivamente, entonces a = nx*ny y b = 0 . Luego, desde el comienzo de la string, encuentre la ‘x’ que tiene al lado a ‘y’ e intercambie ambos hasta llegar al final de la string. En cada intercambio unse decrementa en 1 yb se incrementa en 1 . Repita esto hasta que se logre el conteo de una subsecuencia de ‘yx’, es decir , a se convierte en p y b se convierte en q .
A continuación se muestra la implementación del enfoque anterior: 
 

C++

// CPP program to find lexicographically smallest
// string such that count of subsequence 'xy' and
// 'yx' is p and q respectively.
#include <bits/stdc++.h>
using namespace std;
 
// function to check if answer exits
int nx = 0, ny = 0;
 
bool check(string s, int p, int q)
{
    // count total 'x' and 'y' in string
    for (int i = 0; i < s.length(); ++i) {
        if (s[i] == 'x')
            nx++;
        else
            ny++;
    }
 
    // condition to check existence of answer
    if (nx * ny != p + q)
        return 1;
    else
        return 0;
}
 
// function to find lexicographically smallest string
string smallestPermutation(string s, int p, int q)
{
    // check if answer exist or not
    if (check(s, p, q) == 1) {
        return "Impossible";
    }
 
    sort(s.begin(), s.end());
    int a = nx * ny, b = 0, i, j;
 
    // check if count of 'xy' and 'yx' becomes
    // equal to p and q respectively.
    if (a == p && b == q) {
        return s;
    }
 
    // Repeat until answer is found.
    while (1) {
        // Find index of 'x' to swap with 'y'.
        for (i = 0; i < s.length() - 1; ++i) {
            if (s[i] == 'x' && s[i + 1] == 'y')
                break;
        }
 
        for (j = i; j < s.length() - 1; j++) {
            if (s[j] == 'x' && s[j + 1] == 'y') {
                swap(s[j], s[j + 1]);
                a--; // 'xy' decrement by 1
                b++; // 'yx' increment by 1
 
                // check if count of 'xy' and 'yx' becomes
                // equal to p and q respectively.
                if (a == p && b == q) {
                    return s;
                }
            }
        }
    }
}
 
// Driver code
int main()
{
    string s = "yxxyx";
    int p = 3, q = 3;
 
    cout<< smallestPermutation(s, p, q);
     
    return 0;
}

Java

// Java program to find lexicographically
// smallest string such that count of
// subsequence 'xy' and 'yx' is p and
// q respectively.
import java.util.*;
 
class GFG
{
static int nx = 0, ny = 0;
     
static boolean check(String s,
                     int p, int q)
{
    // count total 'x' and 'y' in string
    for (int i = 0; i < s.length(); ++i)
    {
        if (s.charAt(i) == 'x')
            nx++;
        else
            ny++;
    }
 
    // condition to check
    // existence of answer
    if ((nx * ny) != (p + q))
        return true;
    else
        return false;
}
 
public static String smallestPermutation(String s,
                                         int p, int q)
{
    if (check(s, p, q) == true)
    {
        return "Impossible";
    }
         
    char tempArray[] = s.toCharArray();
    Arrays.sort(tempArray);
     
    String str = new String(tempArray);
    int a = nx * ny, b = 0, i = 0, j = 0;
         
    if (a == p && b == q)
    {
        return str;
    }
     
    while (1 > 0)
    {
         
    // Find index of 'x' to swap with 'y'.
    for (i = 0; i < str.length() - 1; ++i)
    {
        if (str.charAt(i) == 'x' &&
            str.charAt(i + 1) == 'y')
            break;
    }
 
    for (j = i; j < str.length() - 1; j++)
    {
        if (str.charAt(j) == 'x' &&
            str.charAt(j + 1) == 'y')
        {
        StringBuilder sb = new StringBuilder(str);
        sb.setCharAt(j, str.charAt(j + 1));
        sb.setCharAt(j + 1, str.charAt(j));
        str = sb.toString();
        /* char ch[] = str.toCharArray();
            char temp = ch[j+1];
            ch[j+1] = ch[j];
            ch[j] = temp;*/
             
            a--; // 'xy' decrement by 1
            b++; // 'yx' increment by 1
 
            // check if count of 'xy' and
            // 'yx' becomes equal to p
            // and q respectively.
            if (a == p && b == q)
            {
                return str;
            }
        }
    }
}
}
 
// Driver Code
public static void main (String[] args)
{
    String s = "yxxyx";
    int p = 3, q = 3;
     
    System.out.print(smallestPermutation(s, p, q));
}
}
 
// This code is contributed by Kirti_Mangal

Python3

# Python3 program to find lexicographically
# smallest string such that count of subsequence
# 'xy' and 'yx' is p and q respectively.
 
# Function to check if answer exits
def check(s, p, q):
     
    global nx
    global ny
     
    # count total 'x' and 'y' in string
    for i in range(0, len(s)):
        if s[i] == 'x':
            nx += 1
        else:
            ny += 1
 
    # condition to check existence of answer
    if nx * ny != p + q:
        return 1
    else:
        return 0
 
# Function to find lexicographically
# smallest string
def smallestPermutation(s, p, q):
 
    # check if answer exist or not
    if check(s, p, q) == 1:
        return "Impossible"
 
    s = sorted(s)
    a, b, i = nx * ny, 0, 0
 
    # check if count of 'xy' and 'yx' becomes
    # equal to p and q respectively.
    if a == p and b == q:
        return '' . join(s)
 
    # Repeat until answer is found.
    while True:
         
        # Find index of 'x' to swap with 'y'.
        for i in range(0, len(s) - 1):
            if s[i] == 'x' and s[i + 1] == 'y':
                break
 
        for j in range(i, len(s) - 1):
            if s[j] == 'x' and s[j + 1] == 'y':
                 
                s[j], s[j + 1] = s[j + 1], s[j]
                a -= 1 # 'xy' decrement by 1
                b += 1 # 'yx' increment by 1
 
                # check if count of 'xy' and 'yx' becomes
                # equal to p and q respectively.
                if a == p and b == q:
                    return '' . join(s)
 
# Driver code
if __name__ == "__main__":
     
    nx, ny = 0, 0
    s = "yxxyx"
    p, q = 3, 3
 
    print(smallestPermutation(s, p, q))
     
# This code is contributed by Rituraj Jain

C#

// C# program to find lexicographically
// smallest string such that count of
// subsequence 'xy' and 'yx' is p and
// q respectively.
using System;
using System.Text;
 
class GFG
{
     
static int nx = 0, ny = 0;
     
static Boolean check(String s,
                    int p, int q)
{
    // count total 'x' and 'y' in string
    for (int i = 0; i < s.Length; ++i)
    {
        if (s[i] == 'x')
            nx++;
        else
            ny++;
    }
 
    // condition to check
    // existence of answer
    if ((nx * ny) != (p + q))
        return true;
    else
        return false;
}
 
public static String smallestPermutation(String s,
                                        int p, int q)
{
    if (check(s, p, q) == true)
    {
        return "Impossible";
    }
         
    char []tempArray = s.ToCharArray();
    Array.Sort(tempArray);
     
    String str = new String(tempArray);
    int a = nx * ny, b = 0, i = 0, j = 0;
         
    if (a == p && b == q)
    {
        return str;
    }
     
    while (1 > 0)
    {
         
    // Find index of 'x' to swap with 'y'.
    for (i = 0; i < str.Length - 1; ++i)
    {
        if (str[i] == 'x' &&
            str[i + 1] == 'y')
            break;
    }
 
    for (j = i; j < str.Length - 1; j++)
    {
        if (str[j] == 'x' &&
            str[j + 1] == 'y')
        {
            StringBuilder sb = new StringBuilder(str);
            sb.Remove(j,1);
            sb.Insert(j, str[j + 1]);
            sb.Remove(j+1,1);
            sb.Insert(j + 1, str[j]);
            str = sb.ToString();
            /* char ch[] = str.toCharArray();
                char temp = ch[j+1];
                ch[j+1] = ch[j];
                ch[j] = temp;*/
             
            a--; // 'xy' decrement by 1
            b++; // 'yx' increment by 1
 
            // check if count of 'xy' and
            // 'yx' becomes equal to p
            // and q respectively.
            if (a == p && b == q)
            {
                return str;
            }
        }
    }
}
}
 
// Driver Code
public static void Main (String[] args)
{
    String s = "yxxyx";
    int p = 3, q = 3;
     
    Console.WriteLine(smallestPermutation(s, p, q));
}
}
 
// This code has been contributed by 29AjayKumar

Javascript

<script>
// Javascript program to find lexicographically
// smallest string such that count of
// subsequence 'xy' and 'yx' is p and
// q respectively.
 
let nx = 0, ny = 0;
function check(s, p, q)
{
    // count total 'x' and 'y' in string
    for (let i = 0; i < s.length; ++i)
    {
        if (s[i] == 'x')
            nx++;
        else
            ny++;
    }
   
    // condition to check
    // existence of answer
    if ((nx * ny) != (p + q))
        return true;
    else
        return false;
}
 
function smallestPermutation(s,p,q)
{
    if (check(s, p, q) == true)
    {
        return "Impossible";
    }
           
    let tempArray = s.split("");
    (tempArray).sort();
       
    let str = (tempArray).join("");
    let a = nx * ny, b = 0, i = 0, j = 0;
           
    if (a == p && b == q)
    {
        return str;
    }
       
    while (1 > 0)
    {
           
    // Find index of 'x' to swap with 'y'.
    for (i = 0; i < str.length - 1; ++i)
    {
        if (str[i] == 'x' &&
            str[i+1] == 'y')
            break;
    }
   
    for (j = i; j < str.length - 1; j++)
    {
        if (str[j] == 'x' &&
            str[j+1] == 'y')
        {
        let sb = (str).split("");
        sb[j] = str[j+1];
        sb[j + 1] = str[j];
        str = sb.join("");
        /* char ch[] = str.toCharArray();
            char temp = ch[j+1];
            ch[j+1] = ch[j];
            ch[j] = temp;*/
               
            a--; // 'xy' decrement by 1
            b++; // 'yx' increment by 1
   
            // check if count of 'xy' and
            // 'yx' becomes equal to p
            // and q respectively.
            if (a == p && b == q)
            {
                return str;
            }
        }
    }
}
}
 
// Driver Code
let s = "yxxyx";
let p = 3;
let q = 3;
 
document.write(smallestPermutation(s, p, q));
 
// This code is contributed by patel2127
</script>
Producción: 

xyxyx

 

Complejidad temporal: O(N 2 )
 

Publicación traducida automáticamente

Artículo escrito por Sanjit_Prasad y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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