Максимальное преобразование веса данной строки

Опубликовано: 21 Января, 2022

Дана строка, состоящая только из A и B. Мы можем преобразовать данную строку в другую, переключая любой символ. Таким образом, возможны многие преобразования данной строки. Задача - найти вес максимального преобразования веса.
Вес укуса рассчитывается по следующей формуле.

 Вес строки = Вес всего пар + 
                   вес одиночных знаков - 
                   Общее количество переключателей.

Два последовательных символа считаются парой, только если они 
разные. 
Вес одной пары (оба символа разные) = 4
Вес одного символа = 1

Примеры :

 Ввод: str = "AA"
Выход: 3
Преобразования данной строки: «AA», «AB», «BA» и «BB». 
Максимальное преобразование веса - «AB» или «BA». И вес
"Одна пара - один тумблер" = 4-1 = 3.

Ввод: str = "ABB"
Выход: 5
Преобразования: «ABB», «ABA», «AAB», «AAA», «BBB», 
«ББА», «БАБ» и «БАД»
Максимальный вес исходной струны 4 + 1 (одна пара + 1
персонаж)

Рекомендуется: сначала попробуйте свой подход в {IDE}, прежде чем переходить к решению.

Если (n == 1)
   maxWeight (str [0..n-1]) = 1

Иначе, если str [0]! = Str [1]
// Максимум два случая: первый символ рассматривается отдельно
// Первая пара рассматривается отдельно 
maxWeight (str [0..n-1]) = Max (1 + maxWeight (str [1..n-1]),
                              4 + getMaxRec (str [2..n-1])
Еще
// Максимум два случая: первый символ рассматривается отдельно
// Первая пара рассматривается отдельно 
// Поскольку первые два символа одинаковые, а переключатель - 
// требуется для формирования пары, вместо пары добавляется 3 
// из 4         
maxWeight (str [0..n-1]) = Max (1 + maxWeight (str [1..n-1]),
                              3 + getMaxRec (str [2..n-1])

If we draw the complete recursion tree, we can observer that many subproblems are solved again and again. Since same suproblems are called again, this problem has Overlapping Subprolems property. So min square sum problem has both properties (see thisand this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems.
Below is a memoization based solution. A lookup table is used to see if a problem is already computed.

C++

// C++ program to find maximum weight
// transformation of a given string
#include<bits/stdc++.h>
using namespace std;
 
// Returns weight of the maximum
// weight transformation
int getMaxRec(string &str, int i, int n,
                           int lookup[])
{
    // Base case
    if (i >= n) return 0;
 
    //If this subproblem is already solved
    if (lookup[i] != -1) return lookup[i];
 
    // Don"t make pair, so
    // weight gained is 1
    int ans = 1 + getMaxRec(str, i + 1, n,
                                  lookup);
 
    // If we can make pair
    if (i + 1 < n)
    {
    // If elements are dissmilar,
    // weight gained is 4
    if (str[i] != str[i+1])
        ans = max(4 + getMaxRec(str, i + 2,
                                n, lookup), ans);
 
    // if elements are similar so for
    // making a pair we toggle any of them.
    // Since toggle cost is 1 so
    // overall weight gain becomes 3
    else ans = max(3 + getMaxRec(str, i + 2,
                                 n, lookup), ans);
    }
 
    // save and return maximum
    // of above cases
    return lookup[i] = ans;
}
 
// Initializes lookup table
// and calls getMaxRec()
int getMaxWeight(string str)
{
    int n = str.length();
 
    // Create and initialize lookup table
    int lookup[n];
    memset(lookup, -1, sizeof lookup);
 
    // Call recursive function
    return getMaxRec(str, 0, str.length(),
                                 lookup);
}
 
// Driver Code
int main()
{
    string str = "AAAAABB";
    cout << "Maximum weight of a transformation of "
          << str << " is " << getMaxWeight(str);
    return 0;
}

Java

// Java program to find maximum
// weight transformation of a
// given string
class GFG {
     
    // Returns wieght of the maximum
    // weight transformation
    static int getMaxRec(String str, int i,
            int n, int[] lookup)
    {
        // Base case
        if (i >= n)
        {
            return 0;
        }
 
        // If this subproblem is already solved
        if (lookup[i] != -1)
        {
            return lookup[i];
        }
 
        // Don"t make pair, so
        // weight gained is 1
        int ans = 1 + getMaxRec(str, i + 1,
                            n, lookup);
 
        // If we can make pair
        if (i + 1 < n)
        {
             
            // If elements are dissmilar,
            // weight gained is 4
            if (str.charAt(i) != str.charAt(i + 1))
            {
                ans = Math.max(4 + getMaxRec(str, i + 2,
                                n, lookup), ans);
            }
             
            // if elements are similar so for
            // making a pair we toggle any of
            // them. Since toggle cost is
            // 1 so overall weight gain becomes 3
            else
            {
                ans = Math.max(3 + getMaxRec(str, i + 2,
                                n, lookup), ans);
            }
        }
 
        // save and return maximum
        // of above cases
        return lookup[i] = ans;
    }
 
    // Initializes lookup table
    // and calls getMaxRec()
    static int getMaxWeight(String str)
    {
        int n = str.length();
 
        // Create and initialize lookup table
        int[] lookup = new int[n];
        for (int i = 0; i < n; i++)
        {
            lookup[i] = -1;
        }
 
        // Call recursive function
        return getMaxRec(str, 0, str.length(),
                            lookup);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        String str = "AAAAABB";
        System.out.println("Maximum weight of a"
                        + " transformation of "
                        + str + " is "
                        + getMaxWeight(str));
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to find maximum weight
# transformation of a given string
 
# Returns weight of the maximum
# weight transformation
def getMaxRec(string, i, n, lookup):
     
    # Base Case
    if i >= n:
        return 0
 
    # If this subproblem is already solved
    if lookup[i] != -1:
        return lookup[i]
 
    # Don"t make pair, so
    # weight gained is 1
    ans = 1 + getMaxRec(string, i + 1, n,
                        lookup)
 
    # If we can make pair
    if i + 1 < n:
         
        # If elements are dissimilar
        if string[i] != string[i + 1]:
            ans = max(4 + getMaxRec(string, i + 2,
                                    n, lookup), ans)
        # if elements are similar so for
        # making a pair we toggle any of them.
        # Since toggle cost is 1 so
        # overall weight gain becomes 3
        else:
            ans = max(3 + getMaxRec(string, i + 2,
                                    n, lookup), ans)
    # save and return maximum
    # of above cases
    lookup[i] = ans
    return ans
 
# Initializes lookup table
# and calls getMaxRec()
def getMaxWeight(string):
 
    n = len(string)
 
    # Create and initialize lookup table
    lookup = [-1] * (n)
 
    # Call recursive function
    return getMaxRec(string, 0,
                     len(string), lookup)
 
# Driver Code
if __name__ == "__main__":
    string = "AAAAABB"
    print("Maximum weight of a transformation of",
           string, "is", getMaxWeight(string))
 
# This code is contributed by vibhu4agarwal

C#

// C# program to find maximum
// weight transformation of a
// given string
using System;
 
class GFG
{
// Returns wieght of the maximum
// weight transformation
static int getMaxRec(string str, int i,
                     int n, int []lookup)
{
    // Base case
    if (i >= n) return 0;
 
    //If this subproblem is already solved
    if (lookup[i] != -1) return lookup[i];
 
    // Don"t make pair, so
    // weight gained is 1
    int ans = 1 + getMaxRec(str, i + 1,
                            n, lookup);
 
    // If we can make pair
    if (i + 1 < n)
    {
    // If elements are dissmilar,
    // weight gained is 4
    if (str[i] != str[i + 1])
        ans = Math.Max(4 + getMaxRec(str, i + 2,
                                     n, lookup), ans);
 
    // if elements are similar so for
    // making a pair we toggle any of
    // them. Since toggle cost is
    // 1 so overall weight gain becomes 3
    else ans = Math.Max(3 + getMaxRec(str, i + 2,
                                      n, lookup), ans);
    }
 
    // save and return maximum
    // of above cases
    return lookup[i] = ans;
}
 
// Initializes lookup table
// and calls getMaxRec()
static int getMaxWeight(string str)
{
    int n = str.Length;
 
    // Create and initialize lookup table
    int[] lookup = new int[n];
    for(int i = 0 ; i < n ; i++)
    lookup[i] = -1;
 
    // Call recursive function
    return getMaxRec(str, 0, str.Length,
                                 lookup);
}
 
// Driver Code
public static void Main()
{
    string str = "AAAAABB";
    Console.Write("Maximum weight of a" +
                  " transformation of " +
                           str + " is " +
                      getMaxWeight(str));
}
}
 
// This code is contributed by Sumit Sudhakar

Javascript

<script>
    // Javascript program to find maximum
    // weight transformation of a
    // given string
     
    // Returns wieght of the maximum
    // weight transformation
    function getMaxRec(str, i, n, lookup)
    {
        // Base case
        if (i >= n)
        {
            return 0;
        }
   
        // If this subproblem is already solved
        if (lookup[i] != -1)
        {
            return lookup[i];
        }
   
        // Don"t make pair, so
        // weight gained is 1
        let ans = 1 + getMaxRec(str, i + 1, n, lookup);
   
        // If we can make pair
        if (i + 1 < n)
        {
               
            // If elements are dissmilar,
            // weight gained is 4
            if (str[i] != str[i + 1])
            {
                ans = Math.max(4 + getMaxRec(str, i + 2,
                                n, lookup), ans);
            }
               
            // if elements are similar so for
            // making a pair we toggle any of
            // them. Since toggle cost is
            // 1 so overall weight gain becomes 3
            else
            {
                ans = Math.max(3 + getMaxRec(str, i + 2,
                                n, lookup), ans);
            }
        }
   
        // save and return maximum
        // of above cases
        return lookup[i] = ans;
    }
   
    // Initializes lookup table
    // and calls getMaxRec()
    function getMaxWeight(str)
    {
        let n = str.length;
   
        // Create and initialize lookup table
        let lookup = new Array(n);
        lookup.fill(0);
        for (let i = 0; i < n; i++)
        {
            lookup[i] = -1;
        }
   
        // Call recursive function
        return getMaxRec(str, 0, str.length, lookup);
    }
     
    let str = "AAAAABB";
    document.write("Maximum weight of a"
                  &nb
            

                    
                                    
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