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Reduce a string of lowercase characters in range ascii[‘a’..’z’]by doing a series of operations. In each operation, select a pair of adjacent letters that match, and delete them.Delete as many characters as possible using this method and return the resulting string. If the final string is empty, return Empty String

Question

Reduce a string of lowercase characters in range ascii[‘a’..’z’]by doing a series of operations. In each operation, select a pair of adjacent letters that match, and delete them.Delete as many characters as possible using this method and return the resulting string. If the final string is empty, return Empty String

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Solution

To reduce a string of lowercase characters in the range ascii['a'..'z'], follow these steps:

  1. Initialize an empty stack.
  2. Iterate through each character in the string.
  3. If the stack is empty or the current character does not match the top element of the stack, push the current character onto the stack.
  4. If the current character matches the top element of the stack, pop the top element from the stack.
  5. After iterating through all the characters, the remaining elements in the stack represent the reduced string.
  6. If the stack is empty, return "Empty String". Otherwise, concatenate the remaining elements in the stack and return the resulting string.

Here is an example implementation in Python:

def reduce_string(s):
    stack = []
    for char in s:
        if not stack or char != stack[-1]:
            stack.append(char)
        else:
            stack.pop()
    if not stack:
        return "Empty String"
    return ''.join(stack)

# Example usage
string = "abbaca"
reduced_string = reduce_string(string)
print(reduced_string)  # Output: "ca"

This implementation uses a stack to keep track of the characters that have not been deleted yet. It compares each character with the top element of the stack and either pushes it onto the stack or pops the top element if they match. Finally, it checks if the stack is empty and returns the appropriate result.

This problem has been solved

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