CareerCup

My answer in JavaScript

(function(){
    window.dedupe = function(str){
        var i, curr = 0,
            res = "",
            hash = {},
            len = str.length;

        for (i=0; i < len; i++){
            if (hash[str[i]]){
                continue;
            }
            res += str[i];
            hash[str[i]] = true;
        }
        return res;
    }
})();

My solution in JavaScript - not sure about the time complexity.

(function(){
   window.subsetToZero = function(arr){
       var i,
           j,
           k,
           subsets = [],
           sums = [];

       for (i=0; i < arr.length; i++){
           sums.push([0]);
           for (j=0; j < sums.length; j++){
               for (k=0; k <= i; k++){
                   sums[j][k] +=  arr[i];
                   if (sums[j][k] === 0){
                       subsets.push([j, i]);
                   }
               }
           }
       }
       return subsets;
   }
})();

Same solution in JavaScript:

(function(){
    var res,
        arr;

    var display = function(n) {
        var str = "";
        for(var i=0; i < n ; i++)
        {
            str += arr[i];
            if (i < n-1) {
                str += ",";
            }
        }
        console.log(str);
    }

    window.subsetToN = function(n, i, kk){
        i = i || 0;
        kk = kk || n - 1;

        if (i == 0){
            res = [];
            arr = [];
        }

        if (n == 0) {
            display(i);
            res.push(arr.slice(0, i));
        }
        else if(n > 0) {
            var k;
            for (k = 1; k <= kk; k++)
            {
                arr[i]= k;
                subsetToN(n-k, i+1,k);
            }
        }
        return res;
    }
})();