CareerCup

/*
Mix all Ei and Si
Sort this Mix (ascending)
now scan this sorted E/S list
Let visitors=0
  for every E encountered: visitors +=1
  for every S encountered: visitors -=1
  If there is a pair (consecutive <E,S> with the same time stamp, you can drop this   tuple ).The visitor count is going to remain the same

  Now you have a timeline tagged with #visitors on site and
  for any time T in the query, you can read out the visitor value

*/

public static void main(String args[]) {
        IT it = null;
    
        it = add(it, new IT(15, 20));
        it = add(it, new IT(10, 30));
        it = add(it, new IT(17, 19));
        it = add(it, new IT(5, 20));
        it = add(it, new IT(12, 15));
        it = add(it, new IT(30, 40));
        
        System.out.println(getCount(it, 18, 0));
    }
    
    public static int getCount(IT it, int t, int count){
        if(it == null)
            return count;
       if(t < it.st || t <= it.max)
           count = getCount(it.left, t, count);
        if(t > it.st || t >= it.max)
           count = getCount(it.right, t, count);
        if(t >= it.st && t <= it.en)
            count += 1;
        return count;
    }
    
    public static IT add(IT it, IT node){
        if(it == null){
            it = node;
            it.max = node.en;
        }
        else
            addIn(it, node);
       update(it);
       return it;
       
    }    
        
    public static void addIn(IT it, IT node){
        if(it.st > node.st && it.left == null){
            it.left = node;
            node.max = node.en;
        }
        if(it.st < node.st && it.right == null){
            it.right = node;
            node.max = node.en;
        }
        int lm = -1;
        int rm = -1;
        if(it.st > node.st)
            addIn(it.left, node);
        if(it.st < node.st)
            addIn(it.right, node);
        node.max = Math.max(lm, rm);
        node.max = Math.max(node.max, node.en);
    }
    
    public static int update(IT it){
        if(it.left == null && it.right == null)
            return it.max;
        if(it.left == null)
            return it.right.max;
        if(it.right == null)
            return it.left.max;
        
        int lm = update(it.left);
        int rm = update(it.right);
        it.max = Math.max(lm, rm);
        it.max = Math.max(it.max, it.en);
        return it.max;
    }
    
    static class IT{
        int st;
        int en;
        int max;
        
        IT left;
        IT right;
        
        public IT(int st, int en){
            this.st = st;
            this.en = en;
        }
    }

This question can be converted into following question:

For given time during [TS, TE], how many user user active time(Si, Ei) has overlap with the given time during. In another word, count the number of (Si, Ei) that overlapping with [TS, TE].

If (Si, Ei) NOT overlapping with [TS, TE], there are TWO scenarios:
1. (Si, Ei) is ahead of [TS,TE] as a whole. Denote: Ei > TS
2. (Si, Ei) is behind of[TS, TE] as a whole. Denote: TE > Si

If none of above two scenarios exist, we know these two duration overlapping.

Python solution:

# input is list[][]
def activeUsers(self, input, TS, TE):
	count = 0
	for i in range(input):
		if not (input[i][0] < TE or input[i][1] > TS):
			count += 1
	return count

Sort all tuples based on end time. Then iterate the sorted list and use stack to push new user when seeing start time and pop user when seeing end time. Keep track of max user count in the stack which is the answer.

int usersActive(int s[], int e[], int n)
{
   sort(s, s+n);
   sort(e, e+n);
 
   int users = 1, result = 1;
   int i = 1, j = 0;
 
   while (i < n && j < n)
   {
      if (s[i] < e[j])
      {
          users++;
          i++;
 
          if (users > result) 
              result = users;
      }
      else
      {
          users--;
          j++;
      }
   }
 
   return result;
}

int usersActive(int s[], int e[], int n)
{
   sort(s, s+n);
   sort(e, e+n);
 
   int users = 1, result = 1;
   int i = 1, j = 0;
 
   while (i < n && j < n)
   {
      if (s[i] < e[j])
      {
          users++;
          i++;
 
          if (users > result) 
              result = users;
      }
      else
      {
          users--;
          j++;
      }
   }
 
   return result;

}