How to remove extra bracket from expressions like ((((A+B))*C)) to give (A+B)*C in the most optimized way?

Remember the input expression will be valid expression and output expression generated must also be valid and should give the same result as the input expression.

This can be done in O(n) time complexity using stack data structure.

The algorithm

- Start traversing the expression from left examining every bracket.
- If you encounter "(" such that there is another "(" to the right of it, this might be one of the extra brackets to be removed. However we can't be sure right now. So put the minus times the index of this bracket in the integer stack.(we put minus times the index to distinguish brackets having similar brackets to their right from the ones not having )
- If you encounter "(" such that there is a character other than "(" to its right , put its index in the stack.
- If you encounter ")" such that there is another ")" to its left and we have a negative number on the top of the stack, pop the stack and replace current ")" and bracket at the index minus times top of the stack with "$"(so that these brackets can be removed later and are useless).
- If you encounter ")" such that there is a character other than ")" to its left and top of the stack has the positive number than you need that bracket so just pop the stack.

At the end we will have useless brackets replaced with "$" sign, which can be removed in a single traversal. Thus giving us an O(n) complexity. Here is the running code using above explained algorithm.

import java.util.Stack;

public class ExpressionValidate {

public static String validateExpression(String expr) {

char r[] = expr.toCharArray();

char s[] = expr.toCharArray();

Stack<Integer> st = new Stack<Integer>();

int i = 0;

while (i < s.length)

{

if (s[i] == '(') {

if (s[i + 1] == '(') {

st.push(-i);

} else {

st.push(i);

}

i++;

} else if (s[i] != ')' && s[i] != '(') {

i++;

} else if (s[i] == ')') {

int top = st.peek();

if (s[i - 1] == ')' && top < 0) {

r[-top] = '$';

r[i] = '$';

st.pop();

}

else if (s[i - 1] == ')' && top > 0) {

System.out.println("Something is wrong!!");

}

else if (s[i - 1] != ')' && top > 0)

st.pop();

i++;

}

}

StringBuffer result = new StringBuffer();

for (i = 0; i<r.length; i++) {

if (r[i] == '$') {

continue;

}

result.append(r[i]);

}

return result.toString();

}

public static void main(String[] args) {

String expr = "((((A+B))C))";

System.out.println("Validate expression"+validateExpression(expr));

}

}

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