Convert a string to a form that has abcd..z as a subsequence

String conversion (also known as string conversion) is an operation in C that stores the result in an output array after the entire process is executed. In C, there is a function called "transform()", which exists in the directory of the C environment, through which we can convert a string into a new string.

There are two forms of conversion functions −

• Unary operation

• The operation is applied to each element of the input array.

• After the operation is completed, the results will be stored in an output array.

• Binary operations

• The operation applies to each element of a specific array.

• The first input element and the second corresponding input element participate in the operation.

• The output data will be stored in an output array.

A subsequence string is a brand new string generated by performing various operations on the input string (for example: deletion). For subsequence strings, the operation occurs without affecting the remaining characters.

For string conversion, the input contains an operation string of length n 1. The original characters belong to the series a to z. The length of the print string is treated as n here, which is an output string.

In this article, we will learn how to convert a string in C environment to have abcd….z as a subsequence.

Recursive algorithm generates subsequent strings

By using a recursive approach, the following is a possible algorithm for a subsequent string. This is a specific string and T is the time it takes to complete the operation.

• Step 1 - Count the number of occurrences.

• Step 2 - If i = length(s) and j = length(T).

• The third step − then returns 1.

• Step 4 - End.

• Step 5 - If i = length(S).

• Step 6 - Then return 0.

• Step 7 - End.

• Step 8 − Count

• Step 9 - If, j

• Step 10 − Count

• Step 11 - End.

• Step 12 - Count

• Step 13 - Return the count.

• Step 14 - End.

Syntax for subsequent arrays

Here, we have two given sequences. X and Y.
Initialize a table with a dimension of X.length * Y.length
X.label1 = X
Y.label2 = Y
CS1[0][] = 0
CS2[][0] = 0
Start from CS[1][1]
Compare X[i] and Y[j]
   If
      X[i] = Y[j]
      CS[i][j] = 1 + CS[i-1, j-1]
      Point an arrow to CS[i][j]
   Else
      CS[i][j] = max(CS[i-1][j], CS[i][j-1])
      Point an arrow to max(CS[i-1][j], CS[i][j-1])

Here we create a basic working syntax for subsequent arrays. When there are two sequences, we have to follow the following steps to get the output.

Following method

• Method 1−Use C to convert string

• Method 2 for unary operations on strings using C

• Method 3 of using C to perform binary operations on strings

• Use C to print all possible subsequent strings

• Method to convert string to having abcd….z as subsequence using C 5

Convert string using C

In this C code, we create a new string and remove all vowels from the input string. # is added in place of these vowels.

Example 1

#include <bits/stdc++.h>
using namespace std;
string change_case(string r) {
   int l = r.length();
   for(int i = 0 ; i < l ; i++) {
      if(r[i] >= 'a' && r[i] <= 'z')
      r[i] = r[i] - 32;
      else if(r[i] >= 'A' && r[i] <= 'Z')
      r[i] = r[i] + 32;
   }
   return r;
}
string delete_vowels(string a) {
   string temp = "";
   int l = a.length();
   for(int i = 0 ; i < l ; i++) {
      if(a[i] != 'a' && a[i] != 'e' &&
      a[i] != 'i' && a[i] != 'o' &&
      a[i] != 'u' && a[i] != 'A' &&
      a[i] != 'E' && a[i] != 'O' &&
      a[i] != 'U'&& a[i] != 'I')
      temp += a[i];
   }
   return temp;
}
string insert_hash(string a) {
   string temp = "";
   int l = a.length();
   for(int i = 0 ; i < l ; i++) {
      if((a[i] >= 'a' && a[i] <= 'z') ||
      (a[i] >= 'A' && a[i] <= 'Z'))
      temp = temp + '#' + a[i];
      else
      temp = temp + a[i];
   }
   return temp;
}
void transformSting(string a) {
   string b = delete_vowels(a);
   string c = change_case(b);
   string d = insert_hash(c);
   if(d=="")
   cout<<"-1"<<endl;
   else
   cout << d<<endl;
}
int main() {
   string a = "RudraDevDas!!";
   string b = "aeiou";
   transformSting(a);
   transformSting(b);
   return 0;
}

Output

#r#D#R#d#V#d#S!!
-1

Use C to perform unary operations on strings

In this particular code, we show how to perform unary operations on the input array. This function accepts a pointer to the start and end position of a single input. And operate at the beginning of the output array.

The Chinese translation of

Example 2

is:

Example 2

#include <iostream>
#include <algorithm>
using namespace std;
int op_increment (int x) {
   x = x + 1;
   return x;
}
int main () {
   int n = 5;
   int input_array[] = {7, 16, 10, 97, 2001};
   int output_array[n];
   std::cout << "Input array present here:";
   for(int i=0; i<5; i++){
      cout << ' ' << input_array[i];
   }
   cout << '\n';
   transform (input_array, input_array+5, output_array, op_increment);
   std::cout << "The output array now contains with:";
   for(int i=0; i<5; i++){
      cout << ' ' << output_array[i];
   }
   cout << '\n';
   return 0;
}

Output

Input array present here: 7 16 10 97 2001
The output array now contains with: 8 17 11 98 2002

Use C to perform binary operations on strings

In this particular code, we show how to perform binary operations on the input array. The function transform() adds a pointer between the starting point and the first input array. Remember that binary operations always operate on two input data sets.

The Chinese translation of

Example 3

is:

Example 3

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int op_add (int i, int j) {
   return i+j;
}
int main () {
   int n = 5;
   int arr1[] = {7, 16, 10, 2001, 1997};
   int arr2[] = {1, 2, 3, 4, 5};
   int output[n];
   std::cout << "Input data in array1:";
   for(int i=0; i<n; i++){
      cout << ' ' << arr1[i];
   }
   cout << '\n';
   std::cout << "Input data in array2:";
   for(int i=0; i<n; i++){
      cout << ' ' << arr2[i];
   }
   cout << '\n';
   std::transform (arr1, arr1+n, arr2, output, op_add);
   std::cout << "Output array is here now:";
   for(int i=0; i<5; i++){
      cout << ' ' << output[i];
   }
   cout << '\n';
   return 0;
}

Output

Input data in array1: 7 16 10 2001 1997
Input data in array2: 1 2 3 4 5
Output array is here now: 8 18 13 2005 2002

Use C to print all subsequent strings

Apply the concepts of selection and non-selection to find all subsequences of a specific array. During this process, some characters may be removed without changing the order of the elements. Here, the time complexity of this process is O(2^n) and the space complexity is O(n).

Example 4

#include <bits/stdc++.h>
using namespace std;
void printSubsequence(string input, string output) {
   if (input.empty()) {
      cout << output << endl;
      return;
   }
   printSubsequence(input.substr(1), output + input[0]);
   printSubsequence(input.substr(1), output);
}
int main() {
   string output = "";
   string input = "rudraabonikoaa";
   printSubsequence(input, output);
   return 0;
}

Output

rudraabonikoaa
rudraabonikoa
rudraabonikoa
rudraaboniko
rudraabonikaa
rudraabonika
rudraabonika
rudraabonik
rudraabonioaa
rudraabonioa
rudraabonioa
rudraabonio
rudraaboniaa
rudraabonia
rudraabonia

Convert a string to have abcd...z as a subsequence

This is a specific procedure for converting a string into a form that has abcd...z as a subsequence.

• Initialization characters.

• If the length is less than 26, return false.

• Iterate the loop from 0 to s.size() - 1.

• If the character reaches z, break out of the loop.

• If the current character is less than s or equal to character.

• Replace the increment of the current character with 1.

• If the character is less than or equal to z, return false.

• Otherwise, return true.

在这个过程中，时间复杂度为O(n)，辅助空间为O(1)。这里，n是特定字符串的长度。

Example 5

的中文翻译为：

示例5

#include <bits/stdc++.h>
using namespace std;
bool transformString(string& s) {
   char ch = 'a';
   if (s.size() < 26)
   return false;
   for (int i = 0; i < s.size(); i++) {
      if (int(ch) > int('z'))
      break;
      if (s[i] <= ch) {
         s[i] = ch;
         ch = char(int(ch) + 1);
      }
   }
   if (ch <= 'z')
   return false;
   return true;
}
int main() {
   string str = "aaaaaaaaaaaaaaaaaaaaaaaaaaa";
   if (transformString(str))
   cout << str << endl;
   else
   cout << "Not Possible" << endl;
   return 0;
}

输出

abcdefghijklmnopqrstuvwxyza

结论

在本文中，我们学习了使用C++环境进行字符串转换及其不同形式。通过遵循特定的算法和语法，我们检查和构建了一些不同的C++代码，并了解了如何转换字符串，使其具有abcd...z作为子序列。

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