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 string 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)
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 subproblems are called again, this problem has Overlapping Subproblems 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 transformationint 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 dissimilar, // 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 Codeint 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 stringclass GFG { // Returns weight 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 dissimilar, // 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 Codeif __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 stringusing System;class GFG{// Returns weight of the maximum // weight transformationstatic 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 dissimilar, // 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 Codepublic 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 weight 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 dissimilar, // 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" + " transformation of " + str + " is " + getMaxWeight(str)); // This code is contributed by decode2207.</script> |
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
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