Maximum weight transformation of a given string

Given a string consisting of only A’s and B’s. We can transform the given string to another string by toggling any character. Thus many transformations of the given string are possible. The task is to find Weight of the maximum weight transformation.

Weight of a sting is calculated using below formula.


Weight of string = Weight of total pairs + 
                   weight of single characters - 
                   Total number of toggles.

Two consecutive characters are considered as pair only if they 
are different. 
Weight of a single pair (both character are different) = 4
Weight of a single character = 1 

Examples :

Input: str = "AA"
Output: 3
Transformations of given string are "AA", "AB", "BA" and "BB". 
Maximum weight transformation is "AB" or "BA".  And weight
is "One Pair - One Toggle" = 4-1 = 3.

Input: str = "ABB"
Output: 5
Transformations are "ABB", "ABA", "AAB", "AAA", "BBB", 
"BBA", "BAB" and "BAA"
Maximum weight is of original string 4+1 (One Pair + 1
character)



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

Else If str[0] != str[1]
// Max of two cases: First character considered separately
//                   First pair considered separately 
maxWeight(str[0..n-1]) = Max (1 + maxWeight(str[1..n-1]),
                              4 + getMaxRec(str[2..n-1])
Else
// Max of two cases: First character considered separately
//                   First pair considered separately 
// Since first two characters are same and a toggle is 
// required to form a pair, 3 is added for pair instead 
// of 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 this and 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;
}

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


Output:

Maximum weight of a transformation of AAAAABB is 11 

Thanks to Gaurav Ahirwar for providing above solution.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



My Personal Notes arrow_drop_up


Improved By : Sam007

Article Tags :
Practice Tags :



Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.