# Write a program to maximize the cost of removing all occurrences of substrings “ab” and “ba”.

Suppose a string S and integers P and Q, which denotes the cost of removal of substrings “ab” and “ba” respectively from S.

So, the task is to find the maximum cost of removing all occurrences of substrings “ab” and “ba”.

Take an example:

``````Input: S = “cbbaabbaab”, P = 6, Q = 4
Output: 22
Explanation:
Removing substring “ab” from “cbbaabbaab”, the string obtained is “cbbabaab”. Cost = 6.
Removing substring “ab” from “cbbabaab”, the string obtained is “cbbaab”. Cost = 6.
Removing substring “ba” from “cbbaab”, the string obtained is “cbab”. Cost = 4.
Removing substring “ab” from “cbab“, the string obtained is “cb”. Cost = 6.
Total cost = 6 + 6 + 4 + 6 = 22

Input: S = “bbaanaybbabd”, P = 3, Q = 5
Output: 15
Explanation:
Removing substring “ba” from “bbaanaybbabd”, the string obtained is “banaybbabd”. Cost = 5.
Removing substring “ba”, the string obtained is “banaybbabd”, the string obtained is “naybbabd”. Cost = 5.
Removing substring “ba” from “naybbabd”, the string obtained is “naybbd”. Cost = 5.
Total cost = 5 + 5 + 5 = 15``````

## How to solve this problem?

This problem can be solved by using Greedy technique. So, follow the steps to solve this problem.

Step-1: In the first step traverse the string and remove one type of substring. This can be done using the greedy approach as:

• If P >= Q, remove all occurrences of the “ab” substring and then remove all occurrences of the “ba” substring.
• Otherwise, remove all occurrences of the “ba” substring and then remove all occurrences of the “ab” substring.

Step-2: Here we used stack data structure can be used.

Step-3: Initialize our higher cost and lower cost string as “ab” or “ba” according to the value of P and Q as character arrays maxstr[] and minstr[] of size 2 and initialize maximum cost and minimum cost as maxp and minp respectively.

Step-4: Initialize variable, say cost, to store maximum cost.

Step-5: Then Traverse the string and perform the following steps:

• When the stack is not empty and the top of the stack and the current character forms maxstr[], then pop the stack top and add maxp to cost.
• Otherwise, add the current character to the stack.

Step-6: Traverse the remaining string.

• If stack is not empty and top of stack and the current character forms the minstr[], then pop the stack top and add minp to cost.
• Otherwise, add the current character to the stack.

Print cost as the maximum cost.

Now below is the implementation of above algorithm:

```// Java program for the above approach:

import java.util.*;

class GFG {

// Function to find the maximum cost of
// removing substrings "ab" and "ba" from S
public static int MaxCollection(
String S, int P, int Q)
{
// MaxStr is the substring char
// array with larger cost
char maxstr[]
= (P >= Q ? "ab" : "ba").toCharArray();

// MinStr is the substring char
// array with smaller cost;
char minstr[]
= (P >= Q ? "ba" : "ab").toCharArray();

// Denotes larger point
int maxp = Math.max(P, Q);

// Denotes smaller point
int minp = Math.min(P, Q);

// Stores cost scored
int cost = 0;

// Removing all occurrences of
// maxstr from the S

// Stack to keep track of characters
Stack<Character> stack1 = new Stack<>();
char[] s = S.toCharArray();

// Traverse the string
for (char ch : s) {

// If the substring is maxstr

if (!stack1.isEmpty()
&& (stack1.peek() == maxstr[0]
&& ch == maxstr[1])) {

// Pop from the stack
stack1.pop();

cost += maxp;
}

// Push the character to the stack
else {

stack1.push(ch);
}
}

// Remaining string after removing maxstr
StringBuilder sb = new StringBuilder();

// Find remaining string
while (!stack1.isEmpty())
sb.append(stack1.pop());

// Reversing the string
sb = sb.reverse();
String remstr = sb.toString();

// Removing all occurences of minstr
for (char ch : remstr.toCharArray()) {

// If the substring is minstr
if (!stack1.isEmpty()
&& (stack1.peek() == minstr[0]
&& ch == minstr[1])) {

// Pop from the stack
stack1.pop();

// Add minp to the cost
cost += minp;
}

// Otherwise
else {
stack1.push(ch);
}
}

// Return the maximum cost
return cost;
}

// Driver Code
public static void main(String[] args)
{

// Input String
String S = "cbbaabbaab";

// Costs
int P = 6;
int Q = 4;

System.out.println(MaxCollection(S, P, Q));
}
}```

Output:

``22``

Time Complexity : O(N)
Auxiliary Space : O(N)