Skip to main content

Minimum moves for balancing brackets

You’re given a non-empty string made in its entirety from opening and closing braces. Your task is to find the minimum number of “operations” needed to make the string stable. The definition of being stable is as follows:
  1. An empty string is stable.
  2. If S is stable, then {S} is also stable.
  3. If S and T are both stable, then ST (the concatenation of the two) is also stable.
All of these strings are stable: {}, {}{}, and {{}{}}; But none of these: }{, {{}{, nor {}{.
The only operation allowed on the string is to replace an opening brace with a closing brace, or visa-versa.

 Solution Heuristics: Traverse the string. When the number of opening braces( { ) matches the number of closing braces ( } ), the string is balanced. Else it's not balanced. So count both of them. Make a list. Insert only opening braces. If a closing brace is found delete one brace( if the size of the list is > 0 ) from the list. Because they would make a pair. If the list is empty then we'll convert the just found closing brace to opening brace and insert into the list. This will count as a move. When the iteration is finished, we will reverse half of the braces in the list and they will also count as moves. Now, find out the total summation of the moves for result.
Practice Problem: SPOJ ANARC09A - Seinfeld
Labels:

Comments

Post a Comment

Popular posts from this blog

SPOJ - GSS1

Considering input series: { 4 , -10 , 3 , 100 , -20 , 1 } Query(x,y) = Max { a[i]+a[i+1]+...+a[j] ; x ≤ i ≤ j ≤ y } A node contains-  [ START & END is a node's segment limit ] Prefix is the maximum sum starting at START, end can be anywhere. There are two possibilities of the maximum. One, node's leftChild's prefix or two, adding leftChild's sum + rightChild's prefix. (which will make the prefix contiguous) Suffix is the maximum sum ending at END, start can be anywhere. There's two possibility of the maximum. One, node's rightChild's already calculated suffix or two, add rightChild's sum + leftChild's suffix   (which will make the suffix contiguous). Sum : leftChild's sum + rightChild's sum. MAX Maximum of  -  prefix (result is in the node, starts from START but doesn't end in END ) suffix  (result is in the node, doesn't start from START but surely ends in END ) leftChild's max ( result is in left
14 comments
Read more

Trailing Zeroes of n!

Find the number of trailing zeroes in 122! ?  That's a big number right? The solution is pretty simple. Ask yourself when does any multiplication results in a trailing zero? The answer is (2x5) = 10, leaves one zero behind. The factorial of 122 is a multiplication which you already know. The expansion is, 98750442008336013624115798714482080125644041369 78359605958470050267671457205014364903379642774 50422940710230505796264047365129395968426948958 2137821062001338805474721479524352 0000000000000 000000000000000 So, we have to find the number of multiples of 5 from the expansion and they are 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120. So there are 24 multiples of 5. then again, 25= 5 x 5 , which has another extra factor of  5 to count. So how many multiples of 25 from 1 to 122? (122 / 25) = 4 (taking the only integer) So, add that 4 to the earlier count. 24 + 4 = 28 , so we have 28 trailing zero in 122! Let's take a numb
Post a Comment
Read more

SPOJ 4300 - Rectangles

Problem Statement Hint: If the input is 15 squares rectangles are-                 1x1 1x2 1x3 ...................................1x15                        2x2 2x3 ............................2x7                                   3x3   ....................3x5                          can't make 3x2 cause 3x2 and 2x3 are same . can't make 2x8 cause that requires 16 squares. can't make 4x4 cause that also requires 16 square.  Solution: #include<stdio.h> int main() {     int n,c=0,i,j;     scanf("%d",&n);     for(i=1;i*i<= n;i++)     {         c+= (n/i)-i+1;     }     printf("%d\n",c); } * If there's any query or mistake, let me know.
Post a Comment
Read more