Google Interview Question for Software Engineers


Country: United States




Comment hidden because of low score. Click to expand.
1
of 3 vote

Solution:

int moveCoins(TreeNode root) {
        return dfs(root, new HashMap<>());
    }

    int dfs(TreeNode n, Map count) {
        if(!count.containsKey(n)) {
            count.put(n, n.val);
        }
        int coinsNum = count.get(n);
        int res = 0;
        for(TreeNode kid : n.children) {
            res += dfs(kid, count);
            coinsNum += count.get(kid);
            res += Math.abs(count.get(kid));
        }
        count.put(n, coinsNum - 1);
        return res;
    }

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- aonecoding4 November 18, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Question is ambiguous. Not given whether it is a binary tree, BST or any such thing.

- pkp November 18, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

This can be solved for a general tree. Perhaps a follow-up question would be to refine the algorithm for a specific kind of tree.

- abedillon November 19, 2018 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

Each sub-tree will require a certain number of "moves" to ensure one coin in each node leaving and some "remainder" of extra or deficit coins at the root of the sub-tree. You count up all the moves of each sub-tree and add the absolute value of the "remainder":

class Tree:
  def __init__(self, value, *children):
    super().__init__()
    self.value = value
    self.children = children

  def moves(self):
    return abs(self.remainder()) + sum(child.moves() for child in self.children)

  def remainder(self):
    return 1 - self.value + sum(child.remainder() for child in self.children)

if __name__ == "__main__":
  if __name__ == "__main__":
  bbt = \
  Tree(1, 
    Tree(1, 
      Tree(1),
      Tree(1)),
    Tree(1, 
      Tree(1), 
      Tree(1)))
  assert bbt.moves() == 0, "balanced binary tree"
  print("bbt good")

  ll = \
  Tree(0, 
    Tree(0, 
      Tree(0, 
        Tree(0, 
          Tree(5)))))
  assert ll.moves() == 10, "linked list"
  print("ll good")

  mix = \
  Tree(5,
    Tree(0,
      Tree(0),
      Tree(0)
    ),
    Tree(0,
      Tree(0),
      Tree(0)
    ),
    Tree(0,
      Tree(0),
      Tree(5)
    )
  )
  assert mix.moves() == 17, "mix"
  print("mix good")

  mix2 = \
  Tree(5,
    Tree(0,
      Tree(0),
      Tree(0)
    ),
    Tree(0,
      Tree(0),
      Tree(0)
    ),
    Tree(5,
      Tree(0),
      Tree(0)
    )
  )
  assert mix2.moves() == 14, "mix2"
  print("mix2 good")

- abedillon November 19, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

This is how I'd go about it
a node having excess coins or deficit of coins requires that many moves of coins from the connected node
so the deficit or excess is to be added to the no of moves the excess or deficit is moved to or from the parent
Note: assumptions made:
only one coin can be moved per move
only up or down one step of the tree

private static Node tree;
private static int moves;

public static void main(String[] args) {
    // init tree here
    moves = 0;
    traverse(tree);

    System.out.println("The number of moves required : " + moves);
}

private static int traverse(Node tmp) {
    int ret = 0;
    for (Node n : tmp.kids) {
        int t = traverse(n);
        ret += t;
        moves += Math.abs(t);
    }
    ret += tmp.val - 1;
    return ret;
}

- PeyarTheriyaa November 22, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

I don't understand the Math.abs(t)
When does a value ever go negative here?

- aabb November 27, 2018 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

struct Node {
    
    let coins: Int
    
    let children: [Node]
    
    init(coins: Int = 0, children: [Node] = []) {
        self.coins = coins
        self.children = children
    }

}

/* Tree:
          0
        2   1
      0  3 0 0
        1 0 0
 */

let tree = Node(coins: 0, children:
    [Node(coins: 2, children: [Node(), Node()]),
     Node(coins: 1, children: [
        Node(),
        Node(coins: 3, children: [Node(), Node(), Node()])
        ])
    ])


func traverse(node: Node) -> Int {
    var value = node.coins == 0 ? 0 : (node.coins - 1)
    for child in node.children {
        value += traverse(node: child)
    }
    return value
}

print(traverse(node: tree)) // 3

- eugene.kazaev November 23, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include <bits/stdc++.h>
#define ll long long 
#define F first
#define S second
#define pb push_back 
#define pii pair <int,int >
#define pll pair <ll,ll >
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
using namespace std;
const int N=1111;
const int MOD=1000000000+7;


std::vector<int > g[N];
int val[N];

pii dfs(int u,int par){
	int sum=0,no_of_alive=0;
	for(auto it: g[u]){
		if(it==par)
			continue;
		pii temp=dfs(it,u);
		sum+=temp.F+abs(temp.S);
		no_of_alive+=temp.S;
	}
	return {sum,no_of_alive+1-val[u]};
}

int main(){
	int n;
	si(n);
	for(int i=1;i<=n;i++){
		si(val[i]);
	}
	for(int i=1;i<n;i++){
		int a,b;
		si(a),si(b);
		g[a].pb(b);
		g[b].pb(a);
	}

	pii f=dfs(1,1);	
	cout<<f.F;
	return 0;
}

- Siddharth Yadav December 12, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
-2
of 2 vote

Question asks ofr minimum number of moves. That's 0 if the coins are already distributed.

- Anonymous November 18, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

No. It's asking for the minimum moves *given a tree* so it's a function that takes a tree and outputs the number of moves it takes to evenly distribute the coins on that tree.

- abedillon November 19, 2018 | Flag


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