Maximize cost of removing all occurrences of substrings “ab” and “ba”
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:
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 = 22Input: 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 Sint 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 occurrences 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 occurrences 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 Sdef 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 occurrences 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 Codeif __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 occurrences 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 Sfunction 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 occurrences 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), since N extra space has been taken.
Please Login to comment...