# Data Structure Questions and Answers-Decimal to Binary using Stacks

## Data Structure Questions and Answers-Decimal to Binary using Stacks

 Question 6 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
Write a piece of code which returns true if the string contains balanced parenthesis, false otherwise.
 A `public boolean isBalanced(String exp) { int len = exp.length(); Stack stk = new Stack(); for(int i` B `public boolean isBalanced(String exp) { int len = exp.length(); Stack stk = new Stack(); for(int i <` C `public boolean isBalanced(String exp) { int len = exp.length(); Stack stk = new Stack(); for(int i <` D `public boolean isBalanced(String exp) { int len = exp.length(); Stack stk = new Stack(); for(int i <`
Question 6 Explanation:
Whenever a '(' is encountered, push it into the stack, and when a ')' is encountered check the top of the stack to see if there is a matching '(', if not return false, continue this till the entire string is processed and then return true.

 Question 7 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
What is the time complexity of the above code?
 A O(logn) B O(n) C O(1) D O(nlogn)
Question 7 Explanation:
All the characters in the string have to be processed, hence the complexity is O(n).

 Question 8 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
For every matching parenthesis, print their indices.
 A `public void dispIndex(String exp) { Stack stk = new Stack(); for (int i = 0; i < len; i++` B `public void dispIndex(String exp) { Stack stk = new Stack(); for (int i = 0; i < len; i++` C `public void dispIndex(String exp) { Stack stk = new Stack(); for (int i = 0; i < len; i++` D None of the mentioned
Question 8 Explanation:
Whenever a '(' is encountered, push the index of that character into the stack, so that whenever a corresponding ')' is encountered, you can pop and print it.

There are 8 questions to complete.