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Remove Covered Intervals in C++
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
- if stack is empty or a[i] and stack top interval is intersecting,
- return size of the st.
Example(C++)
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h>
using namespace std;
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);
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();
}
};
main(){
vector<vector<int>> v = {{1,4},{3,6},{2,8}};
Solution ob;
cout << (ob.removeCoveredIntervals(v));
}Input
[[1,4],[3,6],[2,8]]
Output
2
Updated on: 30-Apr-2020
214 Views
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