70 lines
1.8 KiB
Java
70 lines
1.8 KiB
Java
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// Problem statement
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/*
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Given a string of balanced expression, find if it contains a redundant parenthesis or not.
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A set of parenthesis are redundant if the same sub-expression is surrounded by unnecessary or multiple brackets.
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Print True if redundant, else False.
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Note: Expression may contain ‘+‘, ‘*‘, ‘–‘ and ‘/‘ operators. Given expression is valid and there are no white spaces present.
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For example:
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((a+b)) can reduced to (a+b), this is Redundant
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(a+b*(c-d)) doesn't have any redundant or multiple
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brackets
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*/
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// Solution
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import java.util.Stack;
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class Main {
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public static void main(String[] args){
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Redundant_Parenthesis obj = new Redundant_Parenthesis();
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boolean ans1 = obj.problem("((a+b))");
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if(ans1) {
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System.out.println("True");
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} else {
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System.out.println("False");
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}
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boolean ans2 = obj.problem("(a+b*(c-d))");
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if(ans2) {
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System.out.println("True");
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} else {
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System.out.println("False");
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}
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}
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}
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class Redundant_Parenthesis {
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public boolean problem(String input_string){
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Stack<Character> st = new Stack<>();
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for(int i = 0; i < input_string.length(); i++) {
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char current_char = input_string.charAt(i);
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if(current_char == ')') {
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char top_char = st.pop();
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boolean flag = true;
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while(top_char != '('){
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if(top_char == '+' || top_char == '-' || top_char == '/' || top_char == '*'){
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flag = false;
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}
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top_char = st.pop();
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}
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if(flag){
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return true;
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}
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} else {
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st.push(current_char);
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}
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}
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return false;
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}
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}
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