Skip to main content

Balanced brackets in C++

It was a question on coursera course in Data structure and I spent some time on the program. Though I had taught the program to students earlier.

But somehow coursera is not accepting my submission:( So why not blog it?

The question is Write a program to balance brackets using stack. This is fundamental requirement in any programming language.

Let us look at some valid and invalid  strings.

{[]}()   valid
[]{}()[] valid
{([([])])} valid

{}) invalid
([)] invalid
({[]} invalid

You are getting the gist, right? When ever there is an opening bracket, there should be a corresponding closing bracket of matching type. And order is important, you can not have round bracket , square bracket followed by closing round bracket first and square bracket next.

So the algorithm is very simple
  1. create a stack of characters
  2. repeat for each i from 0 to length of string
    1. ch = ith character
    2. if ch is either { or [ or (, push it to stack
    3. if ch is closing bracket i.e. } or ] or } 
      1. get stack top charcter ch2
      2. if ch and ch2 are of same type pop the  char
      3. if not return a false
  3. if stack is empty all opening brackets are popped. Return true
  4. if stack is not empty, there are some extra opening brackets. Return false

Now for the code. I used the stack from stl in c++. Here is the code.


#include<stack>
#include<iostream>
using namespace std;
bool isBalanced(string str,char &wrongChar);
int main()
{

string str;
cout<<"Enter your string";
getline(cin,str,'\n');
char ch;
bool bal = isBalanced(str,ch);
if(bal)
cout<<"success";
else
cout<<ch;
}

bool isBalanced(string str,char &wrongChar)
{
stack<char> brackStack;
int len = str.length();
for(int i=0;i<len;i++)
{
char ch= str[i];
if(ch=='(' || ch=='[' || ch=='{')
brackStack.push(ch);
else if ( ch==')' || ch==']' || ch=='}')
{
char topChar = brackStack.top();
if( (ch==')' && topChar=='(') ||
(ch==']' && topChar=='[') ||
(ch=='}' && topChar=='{') )
{
brackStack.pop();
continue;
}
else{
/* not matching*/
wrongChar = ch;
return false;
}
}
}
if(brackStack.empty())
return true;
else
{
wrongChar = brackStack.top();
return false;
}
}

Not so tough? Isn't  it?

For more such questions and notes on all advanced topics, download my app c++ notes

Comments

Popular posts from this blog

In order traversal of nodes in the range x to y

Question : Write a function for in-order traversal of nodes in the range x to y from a binary search tree. This is quite a simple function. As a first solution we can just traverse our binary search tree in inorder and display only the nodes which are in the range x to y. But if the current node has a value less than x, do we have to traverse its left subtree? No. Because all the nodes in left subtree will be smaller than x. Similarly if the current node has a key value more than y, we need not visit its right subtree. Now we are ready to write our algorithm.     if nd is NOT NULL  if nd->val >=x then visit all the nodes of left subtree of nd recursively display nd->val if nd->val <y then visit all the nodes of right subtree of nd recursively  That's all. We have our function ready. void in_order_middle (NODEPTR nd, int x, int y) { if (nd) { if (nd -> val >= x) in_order_middle(nd...

Josephus problem

Question: Write a function to delete every k th node from circular linked list until only one node is left. This has a story associated with it. Flavius Josephus was Jewish Historian from 1st century. He and 40 other soldiers were trapped in a cave by Romans. They decided to kill themselves rather than surrendering to Romans. Their method was like this. All the soldiers will stand in a circle and every k th soldier will be shot dead. Josephus said to have calculated the starting point so that he would remain alive. So we have similar problem at hand. We delete every kth node in a circular list. Eventually only one node will be left. e.g. Let us say this is our list And we are deleting every third node.  We will delete 30. Then we delete 60. Next we delete 10. Next it will be 50. Next to be deleted is 20. Next 80. This continues. Implementation   We can count k-1 nodes and delete next node. This can be repeated in  a loop. What must be the termina...

Delete a node from doubly linked list

Deletion operation in DLL is simpler when compared to SLL. Because we don't have to go in search of previous node of to-be-deleted node.  Here is how you delete a node Link previous node of node of to-be-deleted to next node. Link next node of node of to-be-deleted to previous node. Free the memory of node of to-be-deleted Simple, isn't it. The code can go like this. prevnode = delnode->prev; nextnode = delnode->next; prevnode->next = nextnode; nextnode->prev = prevnode; free(delnode); And that is it. The node delnode is deleted. But we should always consider boundary conditions. What happens if we are trying to delete the first node or last node? If first node is to be deleted, its previous node is NULL. Hence step 3 should not be used.  And also, once head is deleted, nextnode becomes head . Similarly if last node is to be deleted, nextnode is NULL. Hence step 4 is as strict NO NO. And we should set prevnode to tail. After we put these things together, we have...