add: bracket combinations👌

Author: supercodestart

README.md

@@ -1,5 +1,6 @@
 # 📓JavaScript Assessments
 
+- [Bracket Combinations](bracket-combinations)
 - [Calculate the sum of the numbers in a nested array](calculate-the-sum-of-the-numbers-in-a-nested-array)
 - [Convert string to group](convert-string-to-group)
 - [Delete Overlapping Intervals](delete-overlapping-intervals)

bracket-combinations/README.md

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+# Bracket Combinations
+
+Have the function `BracketCombinations(num)` read `num` which will be an integer greater than or equal to zero, and return the number of valid combinations that can be formed with num pairs of parentheses. For example, if the input is 3, then the possible combinations of 3 pairs of parenthesis, namely: `()()()`, are `()()()`, `()(())`, `(())()`, `((()))`, and `(()())`. There are 5 total combinations when the input is 3, so your program should return 5.
+
+## Examples
+
+```javascript
+BracketCombinations(2); // should == 2
+BracketCombinations(3); // should == 5
+```
+
+## Execute
+
+```bash
+node solution.js
+```

bracket-combinations/solution.js

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+function BracketCombinations(num) {
+  if (num === 0) {
+    return 1;
+  }
+  // By Doing sum search i found a formula that can achieve what this problem want
+  // called Catalan number (Catalan Formula)
+  // where catalan formula is ==> (2n!) / (n+1)! n!
+
+  // first i will calculate the factorial of the num
+  let factorial = (n) => {
+    let k = 1;
+    for (var i = n; i >= 2; i--) {
+      k *= i;
+    }
+    return k;
+  };
+
+  // formula going down
+  const result = factorial(2 * num) / (factorial(num + 1) * factorial(num));
+  return result;
+}
+
+(() => {
+  [2, 3, 5, 8].forEach((num) => {
+    console.log(JSON.stringify(num), '==', BracketCombinations(num));
+  });
+})();