Google Interview Question for SDE1s


Country: United States




Comment hidden because of low score. Click to expand.
0
of 0 vote

For arrays we can use a trie (the code below). O(n^2 + m * n) time and O(min(n, m)^2) space.

In case of a 2d array, I assume we are looking for the largest common rectangle. We can use sum of the rectangle values as a hash. For each cell, we can find the sum of values of rectangle with upper left corner 0,0, and lower right corner at the cell, for O(n*m) time. Then, we iterate through each rectangle in Matrix 1 in O(n1*m1*min(n1, m1)) time. For each rectangle we find sum of its values in O(1) time (using the precomputed sums of rectangles starting from 0,0), and append a hashmap at key = sum with coordinates fo the rectangle.
In Matrix 2 we iterate through all rectangles and get from the hashmap possible matches.
The worst case time is O(m1 * n1 * min(m1, n1) * n2 * m2), but the expected time feels like O(m1 * n1 * min(m1, n1)).

#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

class Node {
	public:
		~Node()
		{
			for (auto &c : children_) {
				delete c.second;
			}
		}
		Node *Child(int k) const
		{
			auto it = children_.find(k);
			return it != children_.end() ? it->second : NULL;
		}
		Node *AddChild(int k)
		{
			Node *n = Child(k);
			if (!n) {
				children_[k] = n = new Node();
			}
			return n;
		}

	private:
		unordered_map<int, Node *> children_;
};

class Trie {
	public:
		Node *Root()
		{
			return &root_;
		}

	private:
		Node root_;
};

pair<int, int> LargestCommonSubarray(vector<int> const &a1, vector<int> const &a2)
{
	Trie trie;

	for (int i = 0; i < a2.size(); ++i) {
		Node *n = trie.Root();
		for (int j = i; j < a2.size(); ++j) {
			n = n->AddChild(a2[j]);
		}
	}

	pair<int, int> out;
	int max_size = 0;
	for (int i = 0; i < a1.size(); ++i) {
		Node *n = trie.Root();
		int size = 0;
		for (int j = i; j < a1.size(); ++j) {
			n = n->Child(a1[j]);
			if (!n) {
				break;
			}
			if (++size > max_size) {
				max_size = size;
				out = pair<int, int>(i, i + size - 1);
			}
		}
	}
	return out;
}

int main()
{
	auto out = LargestCommonSubarray({1, 2, 3, 4, 5}, {2, 3, 4, 5, 6});
	cout << out.first << " -> " << out.second << "\n";
}

- Alex December 07, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

If we have only 2 arrays, this is another example of KMP pattern searching. The answer can be found in O(n):

function findLongestSuffixPrefix(nums1, nums2) {
  return Math.max(
      findLongestSuffix(nums1, nums2),
      findLongestSuffix(nums2, nums1));
}

function findLongestSuffix(nums1, nums2) {
  var tf = buildTf(nums2);
  var state = 0;
  for (var i = 0; i < nums1.length; i++) {
    state = tf[state].get(nums1[i]) || 0;
  }
  return state;
}

function buildTf(nums) {
  var result = [];
  var lps = -1;
  for (var i = 0; i <= nums.length; i++) {
    result[i] = lps >= 0 ? new Map(result[lps]) : new Map();
    if (i < nums.length) {
      result[i].set(nums[i], i + 1);
      lps = lps >= 0 && result[lps].get(nums[i]) || 0;
    }
  }
  return result;
}

2d matrix is harder as you will need to split the second matrix into multiple rows, finding 1 row in the first matrix is O(n^2), finding k rows in the first matrix is O(kn^2), or max O(n^3).

- inthecottonfield January 21, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

O(n) using longest prefix which is also a suffix KMP algo

#include <iostream>
using namespace std;

int GetValue(int* a, int* b, int m, int n, int index)
{
	if(index < n)
	{
		return b[index];
	}
	else
	{
		return a[index-n];
	}
	
}
void FindSubfixPrefix(int* a, int* b, int m, int n)
{
	int lps[m+n];
	lps[0] = 0;
	for(int i=1; i<m+n; i++)
	{
		int length = lps[i-1];
		while(length > 0)
		{
			if(GetValue(a, b, m, n, length) == GetValue(a, b, m, n, i))
			{
				lps[i] =  length + 1;
				break;
			}
			else
			{
				length = lps[length-1];
			}
		}
		if(length == 0)
		{
			if(GetValue(a, b, m, n, 0) == GetValue(a, b, m, n, i))
			{
				lps[i] = 1;
			}
			else
			{
				lps[i] = 0;
			}
		}
	}
	int result = lps[m+n-1];
	if(result > m || result > n)
	{
		result = m < n ? m : n;
	}
	for(int i=0; i<result; i++)
	{
		cout<<b[i]<<" ";
	}
}
int main() {
	int n;
	int m;
	cin>>m;
	cin>>n;
	int a[m];
	int b[n];
	for(int i=0; i<m; i++)
	{
		cin>>a[i];
	}
	for(int i=0; i<n; i++)
	{
		cin>>b[i];
	}
	FindSubfixPrefix(a, b, m, n);
}

- Nishagrawa March 31, 2019 | Flag Reply


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