Given an expression, find and mark matched and unmatched parenthesis in it. We need to replace all balanced opening parenthesis with 0, balanced closing parenthesis with 1 and all unbalanced with -1.
Input : ((a) Output : -10a1 Input : (a)) Output : 0a1-1 Input : (((abc))((d))))) Output : 000abc1100d111-1-1
The idea is based on stack. We run a loop from the start of the string upto end and for every ‘(‘, we push it into stack. If stack is empty and we encounter an closing bracket ‘)’ the we replace -1 at that index of the strig. Else we replace all opening brackets ‘(‘ with 0 and closing brackets with 1. Then pop from the stack.
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