Wednesday, July 19th

 Finding non-overlapping intervals is the subject of today's problem

Given an array of intervals, find the minimum number of intervals needed to remove in order to have the remaining intervals non-overlapping

The idea is to sort the intervals based on ending time, then greedily choose each intervals with the interval with the earlier end time

First we sort the intervals.  To sort on end time we will need a custom comparator

    bool customCompare(vector<int>& v1, vector<int>& v2) {
    return v1[1] < v2[1];
    }

Then call sort

    sort( intervals.begin(), intervals.end(), customCompare );     

Initialize 2 variables, one to count the number of intervals to remove, and a second to track the end of the most recent interval

int count=0;
int end=INT_MIN;

Then iterate through each interval.  Get the start and end of the current interval.  If the end of the start of the current interval is after the end of the previously tracked interval, update end.  Otherwise, increment the number of intervals to remove (since the interval overlaps with a previous ending).

    int x,y;
for(int i=0; i<intervals.size(); i++) {
x=intervals[i][0];
y=intervals[i][1];

     x>=end ? end=y : count++;
}
return count;

Thanks to the Editorial for clearing the sorting of the intervals

Comments

Post a Comment

Popular posts from this blog

Wednesday, September 20th

Image
 Reduce X to Zero Given an integer array and value x, use elements at the right and left of the array to reduce the x value to zero A collection of windows, by Daderot .  Shared under the Creative Commons Attribution-Share Alike 3.0 Unported License .  No changes made A key observation is that if elements at the ends of the array are used to sum up to a target value x , then elements in a contiguous subarray will equal the total_sum - x .  So we will look for the largest window equal to  total_sum - x Since all elements in the array are positive, then we can solve the problem using 2 pointers that form a sliding window. The general idea is to begin 2 pointers right and left at index 0.  Continue incrementing the right index by 1, adding up the elements it passes into the current_window  sum.  If the sum equals the target value, update the length.  Otherwise, if the sum is greater than the target sum, increment the left index until the window...
Read more

Monday, September 11th

Image
 Groups There are a given number of people, each of whom is assigned to a group.  The number identifier of the group is also its size.  The size of a group may be a multiple of its size, e.g. a group of 3 may have 6 people.  Return the  collection of all groups . Team USA Softball (a group).  Shared by USASoftball , via Wikipedia , distributed under the Creative Commons Attribution-Share Alike 3.0 Unported license.  No changes made. This problem is pretty straightforward.  The idea is to create a map (or a vector) keyed by the group, storing an array of individuals belonging to the group. First we create the vector-of-vectors to store our solution, then we initialize a vector of vectors that will store individuals according to their group numbers.  The size of the solution vector is n+1, because contrary to what the beginning of the problem statement said, the indices/group sizes go from 1 to n, according to the stated constraints. vect...
Read more

Tuesday, September 19th

Image
 Find the duplicate number Given an array of integers of length n+1, which contains numbers within the range [1,n], and contains on element that is duplicated at least once, find the duplicate element , considering use of only constant space and no changes to the array "How They Met Themselves" by Dante Rossetti, depicting doppelgängers The problem of finding duplicates is made difficult by the constraints of no modifications to the original array, and use of only constant extra space.  Still, despite these constraints there are still several ways to solve the problem.   Since yesterday's solution involved binary search, we will hone our BS skills by applying it to today's problem The general idea is based on the observation that without duplicates, the number of elements less than or equal an element is equal to the element itself.  E.g., in the collection {1,2,3,4}, the number of elements less than or equal to 2 is 2, less than or equal to 3 is 3, etc.   Howe...
Read more