Translate the following into Chinese according to the given condition: Remove characters to get the index sum of the empty string according to the given condition

Concepts related to string manipulation, such as the sum of character indices removed to obtain an empty string, are often used in programming challenges and competitions. The result is then calculated using the sum of the eliminated character indices.

Removing the index sum of characters to obtain an empty string is a practical idea in string manipulation that can be used to solve various programming difficulties and challenges.

Problem handling

We must first understand the problem statement and the given criteria to find the total number of character indices that are eliminated to produce an empty string.

Given a string S, the goal is to determine the total number of characters that can be removed from S while still leaving the string empty. For example, if S = "code", the characters at positions 0, 4, 5, and 6 can be eliminated to obtain an empty string. These exponents add up to 0 4 5 6 = 15.

However, using a stack is a typical strategy for solving this problem. We can loop through the string S and determine whether each character can be removed on each iteration. If its index can be removed, we can add it to the stack. If it can't be removed, we can see if we can remove the character at the top of the stack along with the existing characters. If it can be eliminated, we do so and add its index to the index of the current character. This process can be repeated until all characters in the string have been processed.

The following pseudocode illustrates this strategy -

stack = []
sum = 0
for k in range(len(S)):
   if stack and S[k] == S[stack[-1]]:
      stack.pop()
      sum += k + stack[-1] if stack else k
   else:
      stack.append(k)
return sum

In this pseudocode, the sum variable and the empty stack are both initialized to 0. Then use a for loop to iterate over the string S repeatedly. Check each character to see if it can be removed along with the character on top. stack, if the stack is not empty. If possible, the character is removed from the stack and the sum of its index and the characters in the game is added to the sum variable. In this case we add its index to the stack and try to remove it. Then we return the sum variable.

The time complexity and space complexity of this method are both O(n), where n is the length of the string S and n is the maximum number of characters that can be deleted from S.

grammar

The C syntax for determining the total number of character indices eliminated by creating an empty string based on specified conditions is as follows -

illustrate

• We first get the string entered by the user.

• We set the starting value of n to the length of the string str.

• Next, we initialize cnt to 0, which will count the number of occurrences of the character "U".

• We set the initial value of sum to 0, which will store the total number of character indices eliminated.

• After that, we loop through str, checking each character as shown below -

• If the character is "U", we raise cnt and increase the sum by (n - i - 1) 2 * cnt.

• If the character is not "U", we increase the sum by adding i 2 * cnt.

• Finally, we output the value of sum.

NOTE - Since the specifics of this question are not explicitly stated, these conditions are assumed.

{
   string str;
   cin >> str;

   int n = str.size();
   int cnt = 0, sum = 0;
   for (int k = 0; i < n; k++) {
      if (str[k] == 'U') {
         sum += (n - k - 1) + 2 * cnt;
         cnt++;
      } else {
         sum += k + 2 * cnt;
      }
   }
   cout << sum << endl;
}

algorithm

A C algorithm for counting the total number of character indices eliminated by creating an empty string under defined conditions -

• Step 1 - First, define a string variable and enter the user-supplied string.

• Step 2 - Create a stack to hold the characters of the string.

• Step 3 - Input string loop character by character.

• Step 4 - If the current character is empty, push it onto the stack.

• Step 5 - If the current character and the top character of the stack are the same, pop the top character from the stack.

• Step 6 - If the current character is different from the character at the top of the stack, push it onto the stack.

• Step 7 - Only characters that cannot be erased will remain in the stack after the loop.

• Step 8 - Add the index of the character still on the stack.

• Step 9 - Display the total number of indexes.

Method to follow

method 1

Calculate the sum of character removal indices using the following conditions to produce an empty string -

In this example, the string "abacbdc" is used as input. The code uses two indices i and j to traverse the string from beginning to end. The conditions for removing characters from a string are as follows:

If s[i] and s[j] are equal, move both indices to the center of the string.

• If s[i] is less than s[j], delete the character at index j and increase the index sum by index i 1.

• If s[i] is greater than s[j], delete the character at index i and increase the index sum by index j 1.

After all characters have been eliminated, report the index total to the console.

Please keep in mind that this is just an illustration and the requirements for character removal may change depending on the nature of the problem.

示例 1

#include <iostream>
#include <string>

using namespace std;

int main() {
   string s = "abacbdc";
   int sum = 0;
   int i = 0;
   int j = s.length() - 1;
   while (i < j) {
      if (s[i] == s[j]) {
         i++;
         j--;
      } else if (s[i] < s[j]) {
         sum += i + 1;
         i++;
         s.erase(j, 1);
         j--;
      } else {
         sum += j + 1;
         j--;
         s.erase(i, 1);
         i++;
      }
   }
   cout << "Sum of indices of characters removed: " << sum << endl;
   return 0;
}

输出

Sum of indices of characters removed: 6

方法2

str 字符串和字符是 sum_of_indices 函数的输入。然后，迭代字符串，确定每个字符是否等于 c。如果是这样，函数会递减循环索引以考虑删除的字符，并将字符的索引添加到运行总数中，然后使用擦除技术从字符串中删除字符。然后该函数返回已消除的字符索引的总数。

示例字符串 str 和字符 c 在 main 函数中定义，这两个输入用于调用 sum_of_indices 函数。总数作为结果打印到控制台。

示例 2

#include <iostream>
#include <string>
using namespace std;
int sum_of_indices(string str, char c) {
   int sum = 0;
   for (int i = 0; i < str.length(); i++) {
      if (str[i] == c) {
         sum += i;
         str.erase(i, 1);
         i--;
      }
   }
   return sum;
}
int main() {
   string str = "abcbcdc";
   char c = 'c';
   int sum = sum_of_indices(str, c);
   cout << "Sum of indices of characters removed to obtain empty string: " << sum << endl;
   return 0;
}

输出

Sum of indices of characters removed to obtain empty string: 9

结论

需要操作字符串及其索引来解决根据提供的条件计算消除的字符索引之和以获得空字符串的问题。为了解决这个问题，循环遍历字符串，如果两个连续字符相同，则在更新索引之前删除它们。一旦我们有了一个空字符串，我们就可以添加被删除的字符的索引以产生一个空字符串。

有许多解决方案需要解决，例如利用堆栈或队列来跟踪要删除的字符，或者使用递归来迭代地从字符串中删除字符。

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