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Maximize cost of removing all occurrences of substrings “ab” and “ba”

  • Difficulty Level : Hard
  • Last Updated : 25 Aug, 2021

Given a string S and integers P and Q, which denotes the cost of removal of substrings “ab” and “ba” respectively from S, the task is to find the maximum cost of removing all occurrences of substrings “ab” and “ba”.

Examples:

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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

Approach: The problem can be solved using Greedy technique. Follow the steps below to solve the problem:

  • Traverse the string and remove one type of substring. This can be done using greedy approach as:
    • If P >= Q, remove all occurrences of “ab” substring and then remove all occurrences of “ba” substring.
    • Otherwise, remove all occurrences of “ba” substring and then remove all occurrences of “ab” substring.
  • Stack Data Structure can be used
  • 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.
  • Initialize variable, say cost, to store maximum cost
  • Traverse the string and perform the following steps:
    • If stack is not empty and top of 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.
  • 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.

Below is the implementation of the above approach:

C++




// C++ program for the above approach:
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum cost of
// removing substrings "ab" and "ba" from S
int MaxCollection(string S, int P, int Q)
{
    // MaxStr is the substring char
    // array with larger cost
    char maxstr[2];
    string x = (P >= Q ? "ab" : "ba");
    strcpy(maxstr, x.c_str());
 
    // MinStr is the substring char
    // array with smaller cost;
    char minstr[2];
    x = (P >= Q ? "ba" : "ab");
    strcpy(minstr, x.c_str());
 
    // Denotes larger point
    int maxp = max(P, Q);
 
    // Denotes smaller point
    int minp = min(P, Q);
 
    // Stores cost scored
    int cost = 0;
 
    // Removing all occurrences of
    // maxstr from the S
 
    // Stack to keep track of characters
    stack<char> stack1;
    char s[S.length()];
    strcpy(s, S.c_str());
 
    // Traverse the string
    for (auto &ch : s) {
 
        // If the substring is maxstr
 
        if (!stack1.empty()
            && (stack1.top() == maxstr[0]
                && ch == maxstr[1])) {
 
            // Pop from the stack
            stack1.pop();
 
            // Add maxp to cost
            cost += maxp;
        }
 
        // Push the character to the stack
        else {
 
            stack1.push(ch);
        }
    }
     
    // Remaining string after removing maxstr
    string sb = "";
         
    // Find remaining string
    while (stack1.size() > 0)
    {
        sb = sb + stack1.top();
        stack1.pop();
    }
 
    // Reversing the string
    // retrieved from the stack
    reverse(sb.begin(), sb.end());
 
    // Removing all occurences of minstr
    for (auto &ch : sb) {
 
        // If the substring is minstr
        if (!stack1.empty()
            && (stack1.top() == 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;
}
     
int main()
{
    // Input String
    string S = "cbbaabbaab";
  
    // Costs
    int P = 6;
    int Q = 4;
  
    cout << MaxCollection(S, P, Q);
 
    return 0;
}
 
// This code is contributed by decode2207.

Java




// 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();
 
                // Add maxp to cost
                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
        // retrieved from the stack
        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));
    }
}

Python3




# Python3 program for the above approach:
 
# Function to find the maximum cost of
# removing substrings "ab" and "ba" from S
def MaxCollection(S, P, Q):
   
    # MaxStr is the substring char
    # array with larger cost
    maxstr = [i for i in ("ab" if P >= Q else "ba")]
     
    # MinStr is the substring char
    # array with smaller cost;
    minstr = [i for i in ("ba" if P >= Q else "ab")]
 
    # Denotes larger point
    maxp = max(P, Q)
 
    # Denotes smaller point
    minp = min(P, Q)
 
    # Stores cost scored
    cost = 0
 
    # Removing all occurrences of
    # maxstr from the S
 
    # Stack to keep track of characters
    stack1 = []
    s = [i for i in S]
    # Traverse the string
    for ch in s:
 
        # If the substring is maxstr
 
        if (len(stack1)>0 and (stack1[-1] == maxstr[0] and ch == maxstr[1])):
 
            # Pop from the stack
            del stack1[-1]
 
            # Add maxp to cost
            cost += maxp
 
        # Push the character to the stack
        else:
            stack1.append(ch)
 
    # Remaining string after removing maxstr
    sb = ""
 
    # Find remaining string
    while (len(stack1) > 0):
        sb += stack1[-1]
        del stack1[-1]
 
    # Reversing the string
    # retrieved from the stack
    sb = sb[::-1]
    remstr = sb
 
    # Removing all occurences of minstr
    for ch in remstr:
 
        # If the substring is minstr
        if (len(stack1) > 0 and (stack1[-1] == minstr[0] and ch == minstr[1])):
 
            #  Pop from the stack
            del stack1[-1]
 
            # Add minp to the cost
            cost += minp
 
        # Otherwise
        else:
            stack1.append(ch)
 
    # Return the maximum cost
    return cost
 
# Driver Code
if __name__ == '__main__':
 
    # Input String
    S = "cbbaabbaab"
 
    # Costs
    P = 6;
    Q = 4;
 
    print(MaxCollection(S, P, Q));
 
# This code is contributed by mohit kumar 29.

C#




// C# program for the above approach:
using System;
using System.Collections;
class GFG {
     
    // Function to find the maximum cost of
    // removing substrings "ab" and "ba" from S
    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 stack1 = new Stack();
        char[] s = S.ToCharArray();
  
        // Traverse the string
        foreach(char ch in s) {
  
            // If the substring is maxstr
  
            if (stack1.Count > 0 && ((char)stack1.Peek() == maxstr[0] && ch == maxstr[1])) {
  
                // Pop from the stack
                stack1.Pop();
  
                // Add maxp to cost
                cost += maxp;
            }
  
            // Push the character to the stack
            else {
  
                stack1.Push(ch);
            }
        }
  
        // Remaining string after removing maxstr
        string sb = "";
  
        // Find remaining string
        while (stack1.Count > 0)
        {
            sb = sb + stack1.Peek();
            stack1.Pop();
        }
  
        // Reversing the string
        // retrieved from the stack
        char[] chars = sb.ToCharArray();
        Array.Reverse(chars);
        string remstr = new string(chars);
  
        // Removing all occurences of minstr
        foreach(char ch in remstr.ToCharArray()) {
  
            // If the substring is minstr
            if (stack1.Count > 0 && ((char)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;
    }
     
  static void Main()
  {
    // Input String
    string S = "cbbaabbaab";
 
    // Costs
    int P = 6;
    int Q = 4;
 
    Console.Write(MaxCollection(S, P, Q));
  }
}
 
// This code is contributed by mukesh07.

Javascript




<script>
 
// JavaScript program for the above approach:
 
// Function to find the maximum cost of
    // removing substrings "ab" and "ba" from S
function MaxCollection(S,P,Q)
{
    // MaxStr is the substring char
        // array with larger cost
        let maxstr
            = (P >= Q ? "ab" : "ba").split("");
  
        // MinStr is the substring char
        // array with smaller cost;
        let minstr
            = (P >= Q ? "ba" : "ab").split("");
  
        // Denotes larger point
        let maxp = Math.max(P, Q);
  
        // Denotes smaller point
        let minp = Math.min(P, Q);
  
        // Stores cost scored
        let cost = 0;
  
        // Removing all occurrences of
        // maxstr from the S
  
        // Stack to keep track of characters
        let stack1 = [];
        let s = S.split("");
  
        // Traverse the string
        for (let ch=0;ch< s.length;ch++) {
  
            // If the substring is maxstr
  
            if (stack1.length!=0
                && (stack1[stack1.length-1] == maxstr[0]
                    && s[ch] == maxstr[1])) {
  
                // Pop from the stack
                stack1.pop();
  
                // Add maxp to cost
                cost += maxp;
            }
  
            // Push the character to the stack
            else {
  
                stack1.push(s[ch]);
            }
        }
  
        // Remaining string after removing maxstr
        let sb = [];
  
        // Find remaining string
        while (stack1.length!=0)
            sb.push(stack1.pop());
  
        // Reversing the string
        // retrieved from the stack
        sb = sb.reverse();
        let remstr = sb.join("");
  
        // Removing all occurences of minstr
        for (let ch =0;ch<remstr.length;ch++) {
  
            // If the substring is minstr
            if (stack1.length!=0
                && (stack1[stack1.length-1] == minstr[0]
                    && remstr[ch] == minstr[1])) {
  
                // Pop from the stack
                stack1.pop();
  
                // Add minp to the cost
                cost += minp;
            }
  
            // Otherwise
            else {
                stack1.push(remstr[ch]);
            }
        }
  
        // Return the maximum cost
        return cost;
}
 
// Driver Code
// Input String
        let S = "cbbaabbaab";
  
        // Costs
        let P = 6;
        let Q = 4;
  
        document.write(MaxCollection(S, P, Q));
 
 
// This code is contributed by patel2127
 
</script>

 
 

Output: 
22

 

 

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

 




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