Given a list of ranges with starting and end value, the task is to find the minimum number of ranges that are required to be removed to make remaining ranges non-overlapping. Examples:
Input : input = {{1, 2}, {4, 7}, {3, 8}} Output : 1 Explanation: Removal of {3, 8} makes {{1, 2} and {4, 7}} non-overlapping. Input : input = {{ 10, 20 }, { 10, 20 } , { 10, 20 }} Output : 2 Explanation: Removal of [10, 20] makes the remaining ranges non-overlapping. Input : input = {{1, 2}, {5, 10}, {18, 35}, {40, 45}} Output : 0 Explanation:All ranges are already non-overlapping.
Approach:Â
- Sort the ranges by their starting values.
- Traverse through the ranges and check if any range has a starting point less than the ending point of the previous (ie. there is an overlap).
- Remove the range with greater ending point.
Below code is the implementation of the above approach:
C++
#include <bits/stdc++.h>using namespace std;Â
int minRemovals(vector<vector<int> >& ranges){Â
    int size = ranges.size(), rem = 0;Â
    if (size <= 1)        return 0;Â
    // Sort by minimum starting point    sort(ranges.begin(), ranges.end(),        [](const vector<int>& a, const vector<int>& b)                    { return a[0] < b[0]; });Â
    int end = ranges[0][1];    for (int i = 1; i < ranges.size(); i++) {Â
        // If the current starting point is less than        // the previous interval's ending point        // (ie. there is an overlap)        if (ranges[i][0] < end) {            // increase rem            rem++;            // Remove the interval            // with the higher ending point            end = min(ranges[i][1], end);        }        else            end = ranges[i][1];    }Â
    return rem;}Â
// Driver codeint main(){    vector<vector<int> > input = { { 19, 25 },                        { 10, 20 }, { 16, 20 } };    cout << minRemovals(input) << endl;} |
Java
//Java equivalent of above codeimport java.util.ArrayList;import java.util.Collections;import java.util.Comparator;import java.util.List;Â
public class Main {Â Â public static int minRemovals(List<int[]> ranges) {Â Â Â Â int size = ranges.size(), rem = 0;Â
    if (size <= 1)      return 0;Â
    // Sort by minimum starting point    Collections.sort(ranges, new Comparator<int[]>() {      @Override      public int compare(int[] o1, int[] o2) {        return o1[0] - o2[0];      }    });Â
    int end = ranges.get(0)[1];    for (int i = 1; i < ranges.size(); i++) {      // If the current starting point is less than      // the previous interval's ending point      // (ie. there is an overlap)      if (ranges.get(i)[0] < end) {        // increase rem        rem++;        // Remove the interval        // with the higher ending point        end = Math.min(ranges.get(i)[1], end);      } else        end = ranges.get(i)[1];    }Â
    return rem;  }Â
  // Driver code  public static void main(String[] args) {    List<int[]> input = new ArrayList<int[]>();    input.add(new int[] { 19, 25 });    input.add(new int[] { 10, 20 });    input.add(new int[] { 16, 20 });    System.out.println(minRemovals(input));  }} |
Python3
def minRemovels (ranges):Â
    size = len(ranges)    rem = 0Â
    # Sort by minimum starting point    ranges.sort()Â
    end = ranges[0][1]    for i in range(1, size):Â
        # If the current starting point is less        # than the previous interval's ending        # point (ie. there is an overlap)        if (ranges[i][0] < end):Â
            # Increase rem            rem += 1Â
            # Remove the interval            # with the higher ending point            end = min(ranges[i][1], end)                     else:            end = ranges[i][1]Â
    return remÂ
# Driver Codeif __name__ == '__main__':         Input = [ [ 19, 25 ],            [ 10, 20 ],            [ 16, 20 ] ]                     print(minRemovels(Input))Â
# This code is contributed by himanshu77 |
C#
using System;using System.Linq;Â
class GFG {Â Â static int MinRemovals(int[][] ranges)Â Â {Â Â Â Â int size = ranges.Length;Â Â Â Â int rem = 0;Â
    // Sort by minimum starting point    Array.Sort(ranges, (a, b) => a[0] - b[0]);Â
    int end = ranges[0][1];    for (int i = 1; i < size; i++)    {             // If the current starting point is less than      // the previous interval's ending point (ie.      // there is an overlap)      if (ranges[i][0] < end)       {                 // Increase rem        rem += 1;Â
        // Remove the interval with the higher        // ending point        end = Math.Min(ranges[i][1], end);      }      else {        end = ranges[i][1];      }    }Â
    return rem;  }Â
  // Driver code  public static void Main(string[] args)  {    int[][] ranges      = new int[][] { new int[] { 19, 25 },                     new int[] { 10, 20 },                     new int[] { 16, 20 } };Â
Â
    // Function call    Console.WriteLine(MinRemovals(ranges));  }}Â
// This code is contributed by phasing17 |
Javascript
// JavaScript implementation to find the minimum// removals required to make intervals non-overlappingfunction minRemovels(ranges) {Â Â Â Â var size = ranges.length;Â Â Â Â var rem = 0;Â
    // Sort by minimum starting point    ranges.sort(function(a, b) {        return a[0] - b[0];    });         var end = ranges[0][1];    for (var i = 1; i < size; i++) {        // If the current starting point is less than the previous interval's ending        // point (ie. there is an overlap)        if (ranges[i][0] < end) {            // Increase rem            rem += 1;                 // Remove the interval with the higher ending point            end = Math.min(ranges[i][1], end);             } else {            end = ranges[i][1];        }    }         return rem;}Â
// Driver Codevar ranges = [[19, 25], [10, 20], [16, 20]];console.log(minRemovels(ranges));Â
// This code is contributed by phasing17 |
Output
2
Time Complexity: O(n log n)
Auxiliary Space: O(1) as no extra space has been used.
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