Modify a string by adding each character to the distance from the end of the word

When dealing with strings, sometimes we need to modify them in a specific way to meet certain requirements. One of the requirements is to modify the string by increasing the distance of each character from the end of the word. In this article, we will discuss ways to solve this problem using C.

Problem Statement

Given a string S, modify the string by increasing the distance of each character from the end of the word.

method

In order to solve this problem, we can follow the following steps:

• Cut the given string S into individual words.

• Iterate over each word, and for each character, add the position from the end to its ASCII value.

• Add the modified words to the final string, called res.

• Repeat steps 2 and 3 to operate on all words in the string.

• Return the final modified string.

Example

This is the code implementation in C:

#include 
#include 
#include 

using namespace std;

string modifyString(string S) {
   string res = "";
   vector words;
   
   // Tokenize the string into individual words
   istringstream ss(S);
   string word;
   while (ss >> word) {
      words.push_back(word);
   }
    
   // Iterate over each word
   for (int i = 0; i < words.size(); i++) {
      string word = words[i];
      string modified_word = "";
      
      // Iterate over each character in the word
      for (int j = 0; j < word.length(); j++) {
         int ascii_value = word[j] + (word.length() - 1 - j);
         modified_word += char(ascii_value);
      }
      
      // Add the modified word to the final string
      res += modified_word;
      
      // Add a space to the final string if there are more words to be added
      if (i != words.size() - 1) {
         res += " ";
      }
   }
    
   return res;
}

int main() {
   string S = "hello world";
   string modified_S = modifyString(S);
   cout << modified_S << endl; // Outputs "oekmo kmlqx"
   return 0;
}

Output

lhnmo {rtmd

time complexity

The time complexity of the solution is O(N*M), where N is the number of words in the string and M is the average length of the words.

Space complexity

The space complexity of the solution is O(N*M), where N is the number of words in the string and M is the average length of the words.

In the above example, we take the string "hello world" as input. The modified string is "oekmo kmlqx". In the modified string, the first character 'h' is modified to 'o' because its distance from the end of the word is 4. Likewise, other characters have been modified.

The code implementation first segments the given string S into words and stores them in a vector. It then goes through each word and for each character in the word, adds it from the end position to its ASCII value. The modified words are then added to the final string res. Finally, the code returns the modified string.

in conclusion

In summary, we successfully modify the given string by increasing the distance of each character from the end of the word. The above methods and implementations can be used to solve similar problems related to string operations.

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