Added rendundant-parenthesis solution, Issue #933

algorithms/Java/stacks/redundant-parenthesis.java

@@ -0,0 +1,69 @@
+// Problem statement
+
+/*
+Given a string of balanced expression, find if it contains a redundant parenthesis or not. 
+A set of parenthesis are redundant if the same sub-expression is surrounded by unnecessary or multiple brackets. 
+Print True if redundant, else False.
+Note: Expression may contain ‘+‘, ‘*‘, ‘–‘ and ‘/‘ operators. Given expression is valid and there are no white spaces present.
+
+For example:
+
+((a+b)) can reduced to (a+b), this is Redundant
+(a+b*(c-d)) doesn't have any redundant or multiple
+brackets
+*/
+
+
+// Solution
+
+import java.util.Stack;
+
+class Main {
+
+    public static void main(String[] args){
+        Redundant_Parenthesis obj = new Redundant_Parenthesis();
+        boolean ans1 = obj.problem("((a+b))");
+        if(ans1) {
+            System.out.println("True");
+        } else {
+            System.out.println("False");
+        }
+
+        boolean ans2 = obj.problem("(a+b*(c-d))");
+        if(ans2) {
+            System.out.println("True");
+        } else {
+            System.out.println("False");
+        }
+    }
+}
+
+class Redundant_Parenthesis {
+
+    public boolean problem(String input_string){
+
+        Stack<Character> st = new Stack<>();
+        for(int i = 0; i < input_string.length(); i++) {
+            char current_char = input_string.charAt(i);
+            if(current_char == ')') {
+                char top_char = st.pop();
+                boolean flag = true;
+                while(top_char != '('){
+                    if(top_char == '+' || top_char == '-' || top_char == '/' || top_char == '*'){
+                        flag = false;
+                    }
+                    top_char = st.pop();
+                }
+                if(flag){
+                    return true;
+                }
+            } else {
+                st.push(current_char);
+            }
+        }
+        
+        return false;
+
+    }
+}
+