Dear followers,
Tonight, I came across a nice problem for post-dinner exercise, about generating all possible valid parentheses sequences given the number of brackets to use in total.
PROBLEM LINK: https://leetcode.com/problems/generate-parentheses/
PROBLEM SOLVING APPROACH: Recursion (using Choice Diagram and Decision tree)
TIME-COMPLEXITY: O(2^n) in the worst case but the input n is given to be max 8, so very much doable :)
Here's a sample Java solution for your reference:
class Solution { public List<String> generateParenthesis(int n) { ArrayList<String> generatedParentheses = new ArrayList<>(); genParenthesesRecur(n, n, "", generatedParentheses); return generatedParentheses; } private void genParenthesesRecur (int remOpenBrackets, int remClosingBrackets, String outputFromCaller, List<String> generatedParentheses) { // Base Cases if (remOpenBrackets < 0 || remClosingBrackets < 0) return; if (remClosingBrackets < remOpenBrackets) { return; } if (remOpenBrackets == 0 && remClosingBrackets == 0) { generatedParentheses.add(outputFromCaller); } // Recursive Case String opUsingAnOpeningBracket = outputFromCaller + "("; genParenthesesRecur(remOpenBrackets-1, remClosingBrackets, opUsingAnOpeningBracket, generatedParentheses); if (remClosingBrackets > remOpenBrackets) { String opUsingAClosingBracket = outputFromCaller + ")"; genParenthesesRecur(remOpenBrackets, remClosingBrackets-1, opUsingAClosingBracket, generatedParentheses); } } }
Thanks for Reading!
Happy Programming!!
Love! ❤️
#Lets #Code
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