Given a string **S, **the task is to change the string id it doesn’t follow any of the rules given below and print the updated string. The rules for the proofreading are:

- If there are three consecutive characters, then its a wrong spell. Remove one of the character.
**For Example:**string**“ooops”**can be change to**“oops”**. - If two pairs of same character (AABB) are connected together, it’s a wrong spell. Delete one of the character of the second pair.
**For Example:**string**“helloo”**can be changed to**“hello”**. - The rules follow the priority from left to right.
- Initialize a stack to store the characters and to compare the last characters of the string.
- Traverse the string and add the character to the stack.
- Check the last 3 characters of the stack, if the same then pop the character at the top of the stack.
- Check the last 4 characters of the stack, if the same then pop the character at the top of the stack.
- Finally return the characters of the stack.
- Expand the string according to the given conditions
- Construct the Cypher string based on the given conditions
- Generate a string consisting of characters 'a' and 'b' that satisfy the given conditions
- Find the last remaining Character in the Binary String according to the given conditions
- Minimum number of substrings the given string can be splitted into that satisfy the given conditions
- Change gender of a given string
- Change string to a new character set
- Count of Binary Strings possible as per given conditions
- Find numbers a and b that satisfy the given conditions
- Generate elements of the array following given conditions
- Generate N integers satisfying the given conditions
- Count of N digit numbers possible which satisfy the given conditions
- Minimum cost to reduce the integer N to 1 as per given conditions
- Count of numbers in the range [L, R] which satisfy the given conditions
- Calculate weight of parenthesis based on the given conditions
- Find an integer in the given range that satisfies the given conditions
- Find an array of size N that satisfies the given conditions
- Count pairs of strings that satisfy the given conditions
- Generate an array of size K which satisfies the given conditions
- Maximum number of objects that can be created as per given conditions

**Examples:**

Input:S = “helloo”Output:helloExplanation:

As per the Rule #2

helloo => hello

Input:S = “woooow”Output:woowExplanation:

As per the Rule #2

woooow => wooow

As per the Rule #1

wooow => woow

**Approach:** The idea is to traverse the string and if there is a wrong spelling, remove the extra characters according to the given conditions. As the priority of errors is from left to right, and according to the rules given, it can be seen that the judgment of spelling errors will not conflict. Consider traversing from left to right, adding the already legal characters to the result. Below are the steps:

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to proofread the spells ` `string proofreadSpell(string& str) ` `{ ` ` ` `vector<` `char` `> result; ` ` ` ` ` `// Loop to iterate over the ` ` ` `// characters of the string ` ` ` `for` `(` `char` `c : str) { ` ` ` ` ` `// Push the current character c ` ` ` `// in the stack ` ` ` `result.push_back(c); ` ` ` ` ` `int` `n = result.size(); ` ` ` ` ` `// Check for Rule 1 ` ` ` `if` `(n >= 3) { ` ` ` `if` `(result[n - 1] ` ` ` `== result[n - 2] ` ` ` `&& result[n - 1] ` ` ` `== result[n - 3]) { ` ` ` `result.pop_back(); ` ` ` `} ` ` ` `} ` ` ` `n = result.size(); ` ` ` ` ` `// Check for Rule 2 ` ` ` `if` `(n >= 4) { ` ` ` `if` `(result[n - 1] ` ` ` `== result[n - 2] ` ` ` `&& result[n - 3] ` ` ` `== result[n - 4]) { ` ` ` `result.pop_back(); ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` ` ` `// To store the resultant string ` ` ` `string resultStr = ` `""` `; ` ` ` ` ` `// Loop to iterate over the ` ` ` `// characters of stack ` ` ` `for` `(` `char` `c : result) { ` ` ` `resultStr += c; ` ` ` `} ` ` ` ` ` `// Return the resultant string ` ` ` `return` `resultStr; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `// Given string str ` ` ` `string str = ` `"helloo"` `; ` ` ` ` ` `// Function Call ` ` ` `cout << proofreadSpell(str); ` `} ` |

*chevron_right*

*filter_none*

**Output:**

hello

**Time Complexity:** *O(N)*

**Auxiliary Space:** *O(N)*

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

## Recommended Posts:

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.