最大化从字符串 S 中移除字符串 P 或其反转的成本

2024年8月28日 | 阅读 4 分钟

我们给定几个变量，即a和b，以及两个字符串，即X和Y，它们持有成本。我们被要求执行一项任务，即在以a和b作为给定成本成功地从字符串Y中移除字符串X及其反转字符串后，获得的最小总价。

C++ 代码

// Here we are writing down the C++ programming language code to 
// demonstrate the concept of Maximize cost of repeated removal of string P
// or it's reverse from the string S
#include <bits/stdc++.h>
using namespace std;

// This is the function that helps us with finding out the maximum cost of
// removing the substrings "hg" and "gh" from the string input S for us
int MaxCollection(string S, int P, int Q)
{
	// The MaxStr function helps us with the function. It is the substring char
	// array with higher cost in our function code written
	char maxstr[2];
	string x = (P >= Q ? "ab" : "ba");
	strcpy(maxstr, x.c_str());

     // The MaxStr function helps us with the function. It is the substring char
	// array with a smaller cost in our function code written
	char minstr[2];
	x = (P >= Q ? "ba" : "ab");
	strcpy(minstr, x.c_str());

	// the below code snippet helps us with denoting the larger point code
	int maxp = max(P, Q);

	// the below code helps us with denoting the minor point in the code
	int minp = min(P, Q);

	// the below code snippet helps us with storing the cost-scored code
	int cost = 0;

// The below small code snippet helps us with Removing all occurrences of
	// the master from the string S given for us as in the form of input

	// the below code snippet helps us to create the Stack data structure // to, which keeps track of characters in the code functionality
	stack<char> stack1;
	char s[S.length()];
	strcpy(s, S.c_str());

	// here, we are trying to create a loop that can traverse the string
	// using the in-built auto code in the Standard Template Library(STL)
	for (auto &ch : s) {

		// this is a checking condition if the substring is the same as maxstr

		if (!stack1.empty()
			&& (stack1.top() == maxstr[0]
				&& ch == maxstr[1])) {

			// deleting using the Pop operation from the stack in our code
			stack1.pop();

			// Here, we are trying to add the map to the cost, ensuring code
			cost += maxp;
		}

		// the below conditional Pushes the character to the stack code
		else {

			stack1.push(ch);
		
		// Pushing operation has been completed
		    
		}
	}
	
	// Now, the Remaining part of the string after we have removed the 
	// max function, we are storing the result in sb
	string sb = "";
		
	// Here we are trying to find out the remaining string leftover code
	while (stack1.size() > 0)
	{
		sb = sb + stack1.top();
		stack1.pop();
	}

	// Here, we are trying to reverse the string and perform certainly 
	// operations which have been retrieved from the stack data structure
	reverse(sb.begin(), sb.end());

	// Here we are trying to Remove all occurrences of the string minster
	// the for loop which we have created below helps us with looping code
	for (auto &ch : sb) {

		// the if conditional checks if the substring is ministry or not
		if (!stack1.empty()
			&& (stack1.top() == minstr[0]
				&& ch == minstr[1])) {

			// performing the Pop operation from the stack data structure
			stack1.pop();

			// Here, we are trying to add the map to the cost in our code
			cost += maxp;
		}

		// the else conditional implements the below code if the above 
		// code fails using the small code snippet 
		else {
			stack1.push(ch);
		}
	}

	// Here, we are trying to return the maximum cost to the main driver
	// code functionality after the function completes its task
	return cost;
}

// The main driver code functionality starts from here, explaining the 
// concept of Maximize cost of repeated removal of string P or its reverse 
// from the string S 	
int main()
{
	// This is the input string where we will be performing the task on
	string S = "cbbaabbaab";

	// the steps to perform the string operation are written below
	int P = 6;
	int Q = 4;

     // Here we are trying to display the final result after we have 
    // performed the task of Maximize cost of repeated removal of string P
   // or it's reverse from the string S
	cout << MaxCollection(S, P, Q);

	return 0;
}

// End of the main driver code functionality explaining the concept of
// Maximize the cost of repeated removal of string P or its reverse from the 
// string S

输出

祝您编码愉快！

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