Merge pull request #800 from ishikasinha-d/redundant-braces

Create redundant_braces.cpp

Author: noviicee

Data Structures/Stacks/redundant_braces.cpp

@@ -0,0 +1,69 @@
+// Given a string A of length n denoting an expression. It contains the following operators ’+’, ‘-‘, ‘*’, ‘/’.
+// Check whether A has redundant braces or not.
+
+#include <bits/stdc++.h> 
+using namespace std; 
+
+int redundantBraces(string A) {
+ 
+    int n= A.length();
+    stack <char> s;
+    
+    for(int i=0; i< n; i++)
+    {
+        //we'll push the operators and opening braces in the stack
+        if(A[i]=='+' || A[i]=='-'|| A[i]=='/'|| A[i]=='*' || A[i]=='(' )
+            s.push(A[i]);
+        
+     //when we encounter closing braces, we need to start popping
+        if(A[i]==')')
+        { 
+         //if there are no operators, it means we have redundant braces for eg. ()
+            if(s.top()=='(')
+                return 1;
+         
+         //else we start popping till we encounter opening braces
+         //count variable will keep count of number of operators enclosed within the braces
+            int count=0;
+            while(!s.empty() && s.top()!='(')
+            {
+                s.pop();
+                count++;
+            }
+         //count=0 means we have something like (a) which has redundant braces
+            if(count==0) 
+                return 1;
+            s.pop();
+        }
+    }
+ 
+        return 0;
+} 
+
+int main() 
+{
+    string str;
+    cin>>str;
+    if(redundantBraces(str))
+        cout<<str+" has redundant braces";
+    else 
+        cout<<str+" doesn't have any redundant braces";
+return 0;
+
+}  
+
+// Test Cases
+// Input 1:
+//     A = "((a + b))"
+// Output 1:
+//     1
+//     Explanation 1:
+//         ((a + b)) has redundant braces so answer will be 1.
+// Input 2:
+//     A = "(a + (a + b))"
+// Output 2:
+//     0
+//     Explanation 2:
+//         (a + (a + b)) doesn't have have any redundant braces so answer will be 0.
+
+// Time complexity: O(n)