CareerCup

  • Questions
  • Forum
  • Resume Tips
  • RSS
  • Sign In

Microsoft Interview Question for Software Engineers


Team: bing
Country: United States



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

This can be solved by DFS+Memo.

- qingqingmm October 10, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
#include <cmath>
#include <string>
#include <stdio.h>
#include <queue>
using namespace std;
 
const int INF = 100000;

void print_adj (int** adj) {
	//Print adj
	for (int i=0;i<sizeof(adj)-1;i++) {
		for (int k=0;k<sizeof(adj)-1;k++) {
			cout<<adj[i][k]<<" ";
		}
		cout<<endl;
	}
}

void _recurse (int** adj, int** augmented_adj, int ref, int cur, int dist, int* visited) {
	if (visited[cur]==1) return;
	visited[cur] = 1;
	augmented_adj[ref][cur] = dist;
	for (int k=0; k<sizeof(adj)-1;k++) { //loop on neighbors
		if (adj[cur][k] != -1) {
			_recurse (adj, augmented_adj, ref, k, dist+adj[cur][k], visited);
		}
	}
}

int** compute_all_distances (int** adj) {
	int** augmented_adj = new int* [sizeof(adj)];
	for (int k=0;k<sizeof(adj);k++) {
		augmented_adj[k] = new int [sizeof(adj)];
		for (int j=0; j<sizeof(adj);j++) augmented_adj[k][j] = adj[k][j];
	}

	for (int i=0;i<sizeof(adj)-1; i++) {	//Loop on all cities
		int* visited = new int [sizeof(adj)];
		for (int k=0;k<sizeof(adj);k++) visited[k] = 0;
		_recurse (adj, augmented_adj, i, i, 0, visited); 

		//Free memory
		delete [] visited;
	}
	return augmented_adj;
}

void update_distances (int** adj, int** augmented_adj, int city1, int city2, int dist) {
	int old_dist = augmented_adj[city1][city2];

	//Separate nodes into 2 clusters
	vector<int> cluster1, cluster2;

	queue<int> q;
	q.push (city1);
	vector<int> visited (sizeof(adj), 0);

	while (!q.empty()) {
		int front = q.front();
		q.pop();
		if (visited[front]) continue;
		visited[front] = 1;

		cluster1.push_back(front);
		for (int k=0;k<sizeof(adj)-1; k++) {
			if (adj[front][k]!=-1 && front!=k && k!=city2) {
				q.push(k);
			}
		}
	}
	visited.clear();
	visited.resize(sizeof(adj), 0);
	q.push(city2);

	while (!q.empty()) {
		int front = q.front();
		q.pop();
		if (visited[front]) continue;
		visited[front] = 1;

		cluster2.push_back(front);
		for (int k=0;k<sizeof(adj)-1; k++) {
			if (adj[front][k]!=-1 && front!=k && k!=city1) {
				q.push(k);
			}
		}
	}

	//Update on clusters
	for (int i=0;i<cluster1.size();i++) {
		for (int j=0;j<cluster2.size();j++) {
			int node1 = cluster1[i];
			int node2 = cluster2[j];
			augmented_adj[node1][node2] = augmented_adj [node2][node1] = augmented_adj[node1][node2] + dist - old_dist;
		}
	}
}

int calculate_metric (int** adj) {
	int tot = 0;
	for (int i=0;i<sizeof(adj)-1;i++) {
		for (int j=0;j<sizeof(adj)-1;j++) {
			tot += adj[i][j];
		}
	}
	return tot/2;
}



int main () {
	freopen ("in.txt", "r", stdin);
	int cities, queries;
	cin>>cities>>queries;

	int** adj = new int* [cities+1];
	for (int k=0;k<=cities;k++) {
		adj[k] = new int[cities+1];
		for (int j=0;j<=cities;j++) adj[k][j] = -1;
	}

	for (int k=0;k<cities-1;k++) {
		int city1, city2, dist;
		cin>>city1>>city2>>dist;
		city1--;	city2--;
		adj[city1][city2] = adj[city2][city1] = dist;
	//	printf ("dist between %d & %d = %d\n", city1, city2, dist);
	}

	int** augmented_adj = compute_all_distances (adj);
//	print_adj (augmented_adj);
	
	for (int k=0;k<queries;k++) {
		string s;
		cin>>s;
		if (s=="EDIT") {
			int city1, city2, dist;
			cin>>city1>>city2>>dist;
			city1--; city2--;
			update_distances (adj, augmented_adj, city1, city2, dist);
		}

		else {
	//		print_adj(augmented_adj);
			cout<<calculate_metric(augmented_adj)<<endl;
		}
	}
	return 0;
}

- Youssef October 06, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
#include <cmath>
#include <string>
#include <stdio.h>
#include <queue>
using namespace std;
 
const int INF = 100000;

void print_adj (int** adj) {
	//Print adj
	for (int i=0;i<sizeof(adj)-1;i++) {
		for (int k=0;k<sizeof(adj)-1;k++) {
			cout<<adj[i][k]<<" ";
		}
		cout<<endl;
	}
}

void _recurse (int** adj, int** augmented_adj, int ref, int cur, int dist, int* visited) {
	if (visited[cur]==1) return;
	visited[cur] = 1;
	augmented_adj[ref][cur] = dist;
	for (int k=0; k<sizeof(adj)-1;k++) { //loop on neighbors
		if (adj[cur][k] != -1) {
			_recurse (adj, augmented_adj, ref, k, dist+adj[cur][k], visited);
		}
	}
}

int** compute_all_distances (int** adj) {
	int** augmented_adj = new int* [sizeof(adj)];
	for (int k=0;k<sizeof(adj);k++) {
		augmented_adj[k] = new int [sizeof(adj)];
		for (int j=0; j<sizeof(adj);j++) augmented_adj[k][j] = adj[k][j];
	}

	for (int i=0;i<sizeof(adj)-1; i++) {	//Loop on all cities
		int* visited = new int [sizeof(adj)];
		for (int k=0;k<sizeof(adj);k++) visited[k] = 0;
		_recurse (adj, augmented_adj, i, i, 0, visited); 

		//Free memory
		delete [] visited;
	}
	return augmented_adj;
}

void update_distances (int** adj, int** augmented_adj, int city1, int city2, int dist) {
	int old_dist = augmented_adj[city1][city2];

	//Separate nodes into 2 clusters
	vector<int> cluster1, cluster2;

	queue<int> q;
	q.push (city1);
	vector<int> visited (sizeof(adj), 0);

	while (!q.empty()) {
		int front = q.front();
		q.pop();
		if (visited[front]) continue;
		visited[front] = 1;

		cluster1.push_back(front);
		for (int k=0;k<sizeof(adj)-1; k++) {
			if (adj[front][k]!=-1 && front!=k && k!=city2) {
				q.push(k);
			}
		}
	}
	visited.clear();
	visited.resize(sizeof(adj), 0);
	q.push(city2);

	while (!q.empty()) {
		int front = q.front();
		q.pop();
		if (visited[front]) continue;
		visited[front] = 1;

		cluster2.push_back(front);
		for (int k=0;k<sizeof(adj)-1; k++) {
			if (adj[front][k]!=-1 && front!=k && k!=city1) {
				q.push(k);
			}
		}
	}

	//Update on clusters
	for (int i=0;i<cluster1.size();i++) {
		for (int j=0;j<cluster2.size();j++) {
			int node1 = cluster1[i];
			int node2 = cluster2[j];
			augmented_adj[node1][node2] = augmented_adj [node2][node1] = augmented_adj[node1][node2] + dist - old_dist;
		}
	}
}

int calculate_metric (int** adj) {
	int tot = 0;
	for (int i=0;i<sizeof(adj)-1;i++) {
		for (int j=0;j<sizeof(adj)-1;j++) {
			tot += adj[i][j];
		}
	}
	return tot/2;
}



int main () {
	freopen ("in.txt", "r", stdin);
	int cities, queries;
	cin>>cities>>queries;

	int** adj = new int* [cities+1];
	for (int k=0;k<=cities;k++) {
		adj[k] = new int[cities+1];
		for (int j=0;j<=cities;j++) adj[k][j] = -1;
	}

	for (int k=0;k<cities-1;k++) {
		int city1, city2, dist;
		cin>>city1>>city2>>dist;
		city1--;	city2--;
		adj[city1][city2] = adj[city2][city1] = dist;
	//	printf ("dist between %d & %d = %d\n", city1, city2, dist);
	}

	int** augmented_adj = compute_all_distances (adj);
//	print_adj (augmented_adj);
	
	for (int k=0;k<queries;k++) {
		string s;
		cin>>s;
		if (s=="EDIT") {
			int city1, city2, dist;
			cin>>city1>>city2>>dist;
			city1--; city2--;
			update_distances (adj, augmented_adj, city1, city2, dist);
		}

		else {
	//		print_adj(augmented_adj);
			cout<<calculate_metric(augmented_adj)<<endl;
		}
	}
	return 0;
}

- Youssef October 06, 2015 | Flag Reply


Add a Comment
Name:

Writing Code? Surround your code with {{{ and }}} to preserve whitespace.

Books

is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.

Learn More

Videos

CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.

Learn More

Resume Review

Most engineers make critical mistakes on their resumes -- we can fix your resume with our custom resume review service. And, we use fellow engineers as our resume reviewers, so you can be sure that we "get" what you're saying.

Learn More

Mock Interviews

Our Mock Interviews will be conducted "in character" just like a real interview, and can focus on whatever topics you want. All our interviewers have worked for Microsoft, Google or Amazon, you know you'll get a true-to-life experience.

Learn More