Microsoft Interview Question for Software Engineers


Country: United States
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
2
of 2 vote

2. in ruby

def team_schedule(teams)
    count = 0
    day = 1
    teams << 'bye' if teams.length % 2 == 1
    while(count < teams.length - 1) do
        teams.first(teams.length/2).each_with_index do |team, index|
            puts "#{team} will play #{teams[teams.length - 1 - index]} day #{day}"
        end
        teams = [teams.first] + [teams.pop] + teams[1...teams.length]
        count += 1
        day += 1
    end
end

- ryangurney October 23, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

1.
O(n)

public static void main(String[] args){
  	SegT root = null;
    root = add(root, 2, 10, 5);
    add(root, 5, 7, 2);
    add(root, 12, 15, 4);
    add(root, 6, 10, 12);
    
    int n = getMax(root, Integer.MIN_VALUE);
    System.out.println(n);
  }
  
  
  
  public static int getMax(SegT node, int max){
  	if(node == null)
      return max;
    max = Math.max(max, node.bwidth);
    max = getMax(node.left, max);
    max = getMax(node.right, max);
    return max;
  }
  
  
  public static SegT add(SegT node, int start, int end, int bwidth){
    if(node == null){
       node = new SegT(start, end, bwidth);	
       return node;
    }
    merge(node, start, end, bwidth);
    return null;
  }
  
  public static void merge(SegT node, int start, int end, int bwidth){
  	if(node == null)
      return;
    
    if(start >= node.start && end <= node.end){
    	node.bwidth += bwidth;
      	return;
    }else if(start > node.start && start < node.end && end > node.end){
    	node.end = end;
      	node.bwidth += bwidth;
      	return;
    }else if(start < node.start && end < node.end && end > node.start){
    	node.start = start;
      	node.bwidth += bwidth;
      	return;
    }else if(start < node.start && node.left == null){
    	SegT n = new SegT(start, end, bwidth);
        node.left = n;
      	return;
    }else if(start > node.start && node.left == null){
    	SegT n = new SegT(start, end, bwidth);
        node.right = n;
      	return;
    }else if(start > node.end){
    	merge(node.right, start, end, bwidth);
    }else{
    	merge(node.left, start, end, bwidth);
    }
      
  }
  
  
  
  static class SegT{
  	int start;
    int end;
    int bwidth;
    SegT left;
    SegT right;
    
    public SegT(int start, int end, int bwidth){
    	this.start = start;
      	this.end = end;
      	this.bwidth = bwidth;
    }
    
  }

- sudip.innovates October 14, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

2.

public static void main(String[] args){
  	int N = 3;
    
    Map<Integer, List<List<Integer>>> sch = new TreeMap<Integer, List<List<Integer>>>();
    int[][] teams = new int[N][N];
    int[] tn = new int[N];
    schedule(sch, 0, 0, 0, teams, tn);
    for(Map.Entry entry: sch.entrySet()){
    	System.out.println("Day-"+entry.getKey() + " " + entry.getValue());
    }
  }
  
  public static int schedule(Map<Integer, List<List<Integer>>> sch, int day, int p, int q, int[][] teams, int[] tn){

    int n = teams.length;
    if(p != q){
      int dtext = day;
      List<List<Integer>> games = sch.get(dtext);
      List<Integer> game = new ArrayList<Integer>();
      game.add(p);
      game.add(q);

      if(games == null){
          games = new ArrayList<List<Integer>>();
      }
      games.add(game);
      sch.put(dtext, games);
    }
    
    for(int i  = 0; i < n; i++){
    	for(int j  = 0; j < n; j++){
          if(i == j || teams[i][j] == 1 || tn[i] == 1 || tn[j] == 1)
              continue;
          	teams[i][j] = 1;
          	teams[j][i] = 1;
          	tn[i] = 1;
          	tn[j] = 1;
          	day = schedule(sch, day, i, j, teams, tn);
          	tn[i] = 0;
          	tn[j] = 0;
        }
    }
    return day+1;
  }

- sudip.innovates October 14, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

/* when processes run in parallel - their 
bandwidth requirements add up */

List all start times and end times
Sort the list
Scan from left to right

maxBandwidth = 0
currentBandwidth =0

if start time
  currentBandwidth += bandWidth for this interval
  if ( maxBandwidth < currentBandwidth )
     maxBandwidth = currentBandwidth 

if end time
  currentBandwidth -= bandWidth for this interval
  if ( maxBandwidth < currentBandwidth )
     maxBandwidth = currentBandwidth 

print maxBandwidth

end

- Makarand October 15, 2017 | Flag Reply


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