For a given number, n, generate all combinations of balanced parentheses.
Constraints:
n ≤10Our solution using the subset pattern has the following main parts:
Adding parentheses:
Determine whether to add a left parenthesis '(' or a right parenthesis ')' based on certain conditions:
If the count of left parentheses ('left_count') is less than 'n', a left parenthesis can be added to make a valid combination.
If the count of right parentheses ('right_count') is less than that of left parentheses ('left_count'), a right parenthesis can be added to make a valid combination.
Constructing the result: As the algorithm explores different combinations of parentheses, when 'left_count' and 'right_count' both reach 'n', indicating that 'n' pairs of parentheses have been added and the combination is complete, it appends the current combination in the 'output' list to the 'result' list.
Backtracking:
After exploring a possible path by adding a parenthesis and making recursive calls, the algorithm backtracks to explore other possible combinations.
Backtracking involves removing the last added parenthesis from the 'output' list using the 'pop' operation.
This reverts the 'output' list to its state before the current recursive call, allowing the algorithm to explore other possible combinations without missing any.
The basic workflow of the algorithm goes as follows:
Create a list to store all possible combinations of parentheses.
Call the back_track function for n pairs of parentheses, count of opening parentheses, count of closing parentheses, an empty list for output, and an empty list for the result.
If the count of opening and closing parentheses equals n, then we have a valid combination of parentheses, so we will append the string to our list.
Otherwise, we will check if the number of opening parentheses is less than n. If yes, we will add opening parentheses to our string and increment the count.
Lastly, we will check the count of closing parentheses. If the count is less than the count of opening parentheses, then we will add closing parentheses to our string and increment its count.
Let’s see how this algorithm works for an example, where
Let’s look at the code for this solution below:
As visualized in the slides above, it’s easier to understand the problem by creating a tree. Given a tree with branching of b and a maximum depth of m, the maximum number of nodes in this tree would be the following:
...
For a given number, n, generate all combinations of balanced parentheses.
Constraints:
n ≤10Our solution using the subset pattern has the following main parts:
Adding parentheses:
Determine whether to add a left parenthesis '(' or a right parenthesis ')' based on certain conditions:
If the count of left parentheses ('left_count') is less than 'n', a left parenthesis can be added to make a valid combination.
If the count of right parentheses ('right_count') is less than that of left parentheses ('left_count'), a right parenthesis can be added to make a valid combination.
Constructing the result: As the algorithm explores different combinations of parentheses, when 'left_count' and 'right_count' both reach 'n', indicating that 'n' pairs of parentheses have been added and the combination is complete, it appends the current combination in the 'output' list to the 'result' list.
Backtracking:
After exploring a possible path by adding a parenthesis and making recursive calls, the algorithm backtracks to explore other possible combinations.
Backtracking involves removing the last added parenthesis from the 'output' list using the 'pop' operation.
This reverts the 'output' list to its state before the current recursive call, allowing the algorithm to explore other possible combinations without missing any.
The basic workflow of the algorithm goes as follows:
Create a list to store all possible combinations of parentheses.
Call the back_track function for n pairs of parentheses, count of opening parentheses, count of closing parentheses, an empty list for output, and an empty list for the result.
If the count of opening and closing parentheses equals n, then we have a valid combination of parentheses, so we will append the string to our list.
Otherwise, we will check if the number of opening parentheses is less than n. If yes, we will add opening parentheses to our string and increment the count.
Lastly, we will check the count of closing parentheses. If the count is less than the count of opening parentheses, then we will add closing parentheses to our string and increment its count.
Let’s see how this algorithm works for an example, where
Let’s look at the code for this solution below:
As visualized in the slides above, it’s easier to understand the problem by creating a tree. Given a tree with branching of b and a maximum depth of m, the maximum number of nodes in this tree would be the following:
...