Remove Covered Intervals in C++

C++Server Side ProgrammingProgramming

Suppose we have a list of intervals, we have to remove all intervals that are covered by another interval in the list. Here Interval [a,b) is covered by interval [c,d) if and only if c <= a and b <= d. So after doing so, we have to return the number of remaining intervals. If the input is like [[1,4],[3,6],[2,8]], then the output will be 2, The interval [3,6] is covered by [1,4] and [2,8], so the output will be 2.

To solve this, we will follow these steps −

  • Sort the interval list based on the ending time
  • define a stack st
  • for i in range 0 to size of a – 1
    • if stack is empty or a[i] and stack top interval is intersecting,
      • insert a[i] into st
    • otherwise
      • temp := a[i]
      • while st is not not empty and temp and stack top interval is intersecting
        • pop from stack
      • insert temp into st
  • return size of the st.

Example

Let us see the following implementation to get better understanding −

class Solution {
   public:
   bool intersect(vector <int>& a, vector <int>& b){
      return (b[0] <= a[0] && b[1] >= a[1]) || (a[0] <= b[0] && a[1] >= b[1]);
   }
   static bool cmp(vector <int> a, vector <int> b){
      return a[1] < b[1];
   }
   void printVector(vector < vector <int> > a){
      for(int i = 0; i < a.size(); i++){
         cout << a[i][0] << " " << a[i][1] << endl;
      }
      cout << endl;
   }
   int removeCoveredIntervals(vector<vector<int>>& a) {
      sort(a.begin(), a.end(), cmp);
      //printVector(a);
      stack < vector <int> > st;
      for(int i = 0; i < a.size(); i++){
         if(st.empty() || !intersect(a[i], st.top())){
            st.push(a[i]);
         } else {
            vector <int> temp = a[i];
            while(!st.empty() && intersect(temp, st.top())){
               st.pop();
            }
            st.push(temp);
         }
      }
      return st.size();
   }
};

Input

[[1,4],[3,6],[2,8]]

Output

2
raja
Published on 17-Mar-2020 12:12:30
Advertisements