Order of combinations shouldn't matter

• ``````/**
* Since the order of each element in the output array shouldn't really matter, I won't bother outputting exactly the same result as the expected answer. Anyway. Correct me if I'm wrong.
* This solution applies the Decrease and Conquer algorithm to the problem.
* To develop this solution, consider each pair of "()" as a container. We can either place another container on one's left, right, or inside of it.
* Then we simplify this problem to finding out the combination of different ways you place N containers.
* At the end of the method, don't forget to get rid of all duplicate ways.
**/
/**
* @param {number} n
* @return {string[]}
*/
var generateParenthesis = function(n) {
if(n <= 1){
return ["()"];
}
var a = [], r = [];
//recursively decrease this problem into a smaller one
r = generateParenthesis(n - 1);
//expand the current result by adding 3 combinations on the ['left', 'right', 'inside'] pattern to each current combination.
for(i = 0; i < r.length; i ++){
a = a.concat(["()"+r[i], r[i]+"()", "("+r[i]+")"]);
}
//filter out duplicates and return
return a.filter(function(value, index, b){
return b.indexOf(value) === index;
});
};
``````

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