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All Valid Parentheses Combinations

Certification: Advanced Level Accuracy: 100% Submissions: 1 Points: 15

Write a Python program to generate all valid combinations of parentheses for a given number N pairs. A valid combination means that every opening parenthesis has a corresponding closing parenthesis in the correct order. For example, for N=2, valid combinations are ["(())", "()()"]. The program should return all possible valid combinations as a list of strings.

Example 1
    
      
  • Input: n = 3
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  • Output: ["((()))", "(()())", "(())()", "()(())", "()()()"]
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  • Explanation:
    Step 1: For n=3, we need to generate all valid combinations using 3 opening and 3 closing parentheses.
    Step 2: Each valid combination must have exactly 3 '(' and 3 ')' characters.
    Step 3: At any point while reading from left to right, the count of ')' should not exceed the count of '('.
    Step 4: The final result contains 5 valid combinations for n=3.
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Example 2
    
      
  • Input: n = 2
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  • Output: ["(())", "()()"]
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  • Explanation:
    Step 1: For n=2, we need to generate combinations using 2 opening and 2 closing parentheses.
    Step 2: "(())" is valid because the inner parentheses are properly nested within the outer ones.
    Step 3: "()()" is valid because it represents two separate pairs of parentheses in sequence.
    Step 4: ")(" would be invalid because it starts with a closing parenthesis before any opening one.
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Constraints
    
      
  • 1 ≤ n ≤ 8
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  • The output should contain all unique valid combinations
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  • Each combination should be a string of length 2*n
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  • Time Complexity: O(4^n / √n) - Catalan number complexity
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  • Space Complexity: O(4^n / √n) for storing all valid combinations
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StringsBacktracking KPMGSamsung
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Hint
Solution

Solution Hints

    
      
  • Use backtracking approach to generate all possible combinations
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  • Keep track of the count of opening and closing parentheses used so far
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  • Add an opening parenthesis if the count of opening parentheses is less than n
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  • Add a closing parenthesis if the count of closing parentheses is less than opening parentheses count
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  • When the total length reaches 2*n, add the current combination to the result
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  • Use recursion to explore all possible paths and backtrack when necessary
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  • Ensure that at any point, closing parentheses count never exceeds opening parentheses count
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Steps to solve by this approach:

 Step 1: Initialize an empty result list to store all valid combinations and define a recursive backtracking function.
 Step 2: In the backtracking function, maintain current string being built and counts of opening and closing parentheses used.
 Step 3: Check base case - if current string length equals 2*n, add it to result and return.
 Step 4: If opening parentheses count is less than n, recursively add an opening parenthesis and increment open count.
 Step 5: If closing parentheses count is less than opening count, recursively add a closing parenthesis and increment close count.
 Step 6: The recursive calls automatically handle backtracking by exploring all valid paths through the decision tree.
 Step 7: Return the final result list containing all valid parentheses combinations.

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