Maximize cost of removing all occurrences of substrings "ab" and "ba" - GeeksforGeeks

09 Feb, 2022
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 = 4Output: 22Explanation: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 = 5Output: 15Explanation: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.
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.
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.
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++
Java
Python3
C#
Javascript
// 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 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 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# 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.
<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>
22
 
Time Complexity : O(N)Auxiliary Space : O(N)
mohit kumar 29
patel2127
mukesh07
decode2207
simmytarika5
Google
interview-preparation
substring
Greedy
Stack
Strings
Google
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Stack
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