CareerCup

This can be solved using a stack, by pushing all elements until a closing bracket is encountered, then you start to pop the elements until you hit a starting bracket then multiply the result by the order of the nested brackets (3,2,1), then push that result into the stack until the whole array of characters is processed.

class Element {
    int val;
    List<Element> list;
    boolean isList;
}

 private long getValue(Element e, int factor) {
        long sum = 0;
        if(e == null)
            return sum;
        if(!e.isList)
            return factor * e.val;
            
        for(Element item : e.list) {
            sum += getValue(item, factor + 1);
        }
        return sum;
    }

struct NestedList{
    bool isList;
    int val;
    vector<NestedList> internal;

    NestedList(int v):isList(false),val(v){
    }

    NestedList(const vector<NestedList> &v):isList(true),internal(v){}
};

int nestedListSum(const vector<NestedList> &v,int multFactor){
    if(v.empty()){
        return 0;
    }

    int sum = 0;
    for(auto &l : v){
        if(l.isList){
            int res = nestedListSum(l.internal,multFactor + 1);
            sum += res*multFactor;
        }else{
            sum += multFactor*l.val;
        }
    }

    return sum;
}

int nestedListSum(const vector<NestedList> &v){
    return nestedListSum(v,1);
}

public class Sum {
	public static int calculate(List<Object> a, int multiplier) {
		int sum = 0;
		for (int i = 0; i < a.size(); i++) {
			if (a.get(i) instanceof ArrayList) {
				List l = (ArrayList) a.get(i);
				if (l.size() >= 1) {
					multiplier++;
					sum += multiplier * ((int) calculate(l, multiplier));
				}
			} else {
				sum += ((int) a.get(i));
			}
		}
		return sum;
	}

	public static void main(String[] args) {
		List<Object> list  = new ArrayList<>();
        
       		list.add(8);
        	list.add(3);
        	list.add(2);
        
        	List<Object> inner1 = new ArrayList<>();
        	inner1.add(5);
        	inner1.add(6);
        
        	List<Object> inner2 = new ArrayList<>();
        	inner2.add(9);
        	inner1.add(inner2);  
        
        	list.add(inner1);
        
        	list.add(6);
        
        	System.out.println(calculate(list, 1));
	}
}

{{
private static void addSum(List in) {
int res = addSumHelper(in, 0 , 1);
System.out.println(res);
}

private static int addSumHelper(List in, int sum, int level) {
for (Object list : in) {
if(list instanceof List)
sum += addSumHelper((List)list, 0, level+1);
else
sum += (Integer)list;
}
return sum*level;
}
}}

l=[8,3,2,[5,6,[9]],6]
count=1
ans=0
def fun(l,count,ans):
    for e in l:
        if(type(e) is list):
            ans=fun(e,count*(count+1),ans)
        else:
            ans+=count*e
    return ans
ans=fun(l,count,ans)
print(ans)

Javascript Solution
{{
var input = [8,3,2,[5,6,[9]],6];

var mySum = function(inputX, level){
var result = 0;
var temp;
level = level + 1;
for(var i = 0; i < inputX.length; i++){
if(Array.isArray(inputX[i])){
temp = mySum(inputX[i], level);
result = result + (level * (temp));
}else{
result = result + inputX[i];
}
}
return result;
}

var result = mySum(input, 1);
console.log(result);
}}

type NType = (number | NType)[];
function add(what : NType) {
   let value = 0;
   let last = Number.NaN;
   for (let e of what) {
      if (typeof e == "number") {
         value += last;
         last = e;
      } else {
         if (Number.isNaN(last))
            last = 1;
         last = last * add(e);
      }
   }
   if (Number.isNaN(last)) {
      assert(value == 0);
      return 0;
   }
   return value + last;
}

arr = [8,3,2,[5,6,[9]],6]

def arr_calc(arr):
    
    def helper(arr, i):
        s = 0
        for el in arr:
            if type(el) == list:
                s += i*helper(el,i+1)
            else:
                print el
                s += el
        
        print s
        return s
    
    return helper(arr, 2)
    
    

arr_calc(arr)

```
def sum_depth_based2(l):
return helper2(l, 1)

def helper2(l, depth):
ret = 0
for i in range(len(l)):
if type(l[i]) is list:
ret += depth * (helper2(l[i], depth+1))
else:
ret += depth * l[i]
return ret
```