Minimum number of integers required such that each Segment contains at least one of them

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Given two arrays start[] and end[] consisting of positive integers denoting the starting and ending points of a segment respectively, the task is to find the minimum number of integers which lies in at least one of the given segments and each segment contains at least one of them.

Examples:

Input: start[] = {1, 2, 3}, end[] = { 3, 5, 6}
Output: 3
Explanation:
All three ranges ([1, 3], [2, 5], [3, 6]) contains the integer 3.

Input: start[] = {4, 1, 2, 5}, end[] = {7, 3, 5, 6}
Output: 3 6
Explanation:
Segments {1, 3} and {2, 5} are contains the integer 3.
Segments {4, 7} and {5, 6} contains the integer 6.

Mathematical formulation:
The mathematical way of describing the problem is to consider each given range of integers to be a line segment defined by two integer coordinates [a_i, b_i] on a line. Then the minimum number of integers required to cover each of the given range is the minimum number of points such that each segment contains at least one point.
The representation of Example 1 is shown below:

Naive approach:

The simplest way to solve the problem is to find the least value of all the starting points and maximum value of all ending points of all segments. Iterate over this range, and for each point in this range keep track of the number of segments which can be covered using this point. Use an array to store the number of segments as:

arr[point] = number of segments that can be covered using this point

Find the maximum value in the array arr[].
If this maximum value is equal to N, the index corresponding to this value is the point which covers all segments.
If this maximum value is less than N, then the index corresponding to this value is a point which covers some segments.
Repeat the steps 1 to 3 for array arr[] excluding this maximum value until the sum of all the maximum values found is equal to N.

Time Complexity: O((A-B+1)*N), where A is maximum of ending points of segments and B is the minimum of the starting points of the segments.
Auxiliary Space: O(1)

Efficient Approach:

Approach 1:

The problem can be solved efficiently by using the Greedy Technique. Follow the steps given below to solve the problem:

Sort the segments by their end points.
Select the point(or coordinate) corresponding to minimum end point of all segments.
Now, All the segments whose starting point are less than this selected point and whose ending points are greater than this selected point can be covered by this point.
Then print the minimum number of points.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// function to sort the 2D vector
// on basis of second element.
bool sortcol(const pair<int, int> p1,
             const pair<int, int> p2)
{
    return p1.second < p2.second;
}
 
// Function to compute minimum number
// of points which cover all segments
void minPoints(pair<int, int> points[], int n)
{
     
    // Sort the list of tuples by
    // their second element.
    sort(points, points + n, sortcol);
 
    // To store the solution
    vector<int> coordinates;
    int i = 0;
 
    // Iterate over all the segments
    while (i < n)
    {
        int seg = points[i].second;
        coordinates.push_back(seg);
        int p = i + 1;
 
        if (p >= n)
            break;
 
        // Get the start point of next segment
        int arrived = points[p].first;
 
        // Loop over all those segments whose
        // start point is less than the end
        // point of current segment
        while (seg >= arrived)
        {
            p += 1;
 
            if (p >= n)
                break;
 
            arrived = points[p].first;
        }
        i = p;
    }
     
    // Print the possible values of M
    for(auto point : coordinates)
        cout << point << " ";
}
 
// Driver code
int main()
{
    int n = 4;
 
    // Starting points of segments
    int start[] = { 4, 1, 2, 5 };
 
    // Ending points of segments
    int end[] = { 7, 3, 5, 6 };
 
    pair<int, int> points[n];
 
    // Insert ranges in points[]
    for(int i = 0; i < n; i++)
    {
        points[i] = { start[i], end[i] };
    }
 
    // Function call
    minPoints(points, n);
 
    return 0;
}
 
// This code is contributed by Kingash

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to compute minimum number
// of points which cover all segments
static void minPoints(int[][] points, int n)
{
     
    // Sort the list of tuples by
    // their second element.
    Arrays.sort(points, (a, b) -> a[1] - b[1]);
 
    // To store the solution
    ArrayList<Integer> coordinates = new ArrayList<>();
    int i = 0;
 
    // Iterate over all the segments
    while (i < n) 
    {
        int seg = points[i][1];
        coordinates.add(seg);
        int p = i + 1;
 
        if (p >= n)
            break;
 
        // Get the start point of next segment
        int arrived = points[p][0];
 
        // Loop over all those segments whose
        // start point is less than the end
        // point of current segment
        while (seg >= arrived)
        {
            p += 1;
             
            if (p >= n)
                break;
                 
            arrived = points[p][0];
        }
        i = p;
    }
 
    // Print the possible values of M
    for(Integer point : coordinates)
        System.out.print(point + " ");
}
 
// Driver code
public static void main(String[] args)
{
 
    int n = 4;
 
    // Starting points of segments
    int[] start = { 4, 1, 2, 5 };
 
    // Ending points of segments
    int[] end = { 7, 3, 5, 6 };
 
    int[][] points = new int[n][2];
 
    // Insert ranges in points[]
    for(int i = 0; i < n; i++) 
    {
        points[i][0] = start[i];
        points[i][1] = end[i];
    }
 
    // Function call
    minPoints(points, n);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
 
# Function to compute minimum number
# of points which cover all segments
def minPoints(points):
 
    # Sort the list of tuples by 
    # their second element.
    points.sort(key = lambda x: x[1])
 
    # To store the solution
    coordinates = []
    i = 0
 
    # Iterate over all the segments
    while i < n:
 
        seg = points[i][1]
        coordinates.append(seg)
        p = i + 1
 
        if p >= n:
            break
 
        # Get the start point of next segment
        arrived = points[p][0]
 
        # Loop over all those segments whose
        # start point is less than the end
        # point of current segment
        while seg >= arrived:
 
            p += 1
            if p >= n:
                break
            arrived = points[p][0]
        i = p
 
# Print the possible values of M
    for point in coordinates:
        print(point, end =" ")
 
 
# Driver Code
n = 4
 
# Starting points of segments
start = [4, 1, 2, 5]
 
# Ending points of segments
end = [7, 3, 5, 6] 
 
points = []
 
# Insert ranges in points[]
for i in range(n):
    tu = (start[i], end[i])
    points.append(tu)
 
# Function Call
minPoints(points)

C#

// C# program for the above approach
using System;
using System.Linq;
using System.Collections.Generic;
 
 
class GFG{
 
  // Function to compute minimum number
  // of points which cover all segments
  static void minPoints(List<int []> points, int n)
  {
 
    // Sort the list of tuples by
    // their second element.
    points = points.OrderBy(p => p[1]).ToList();
 
 
    // To store the solution
    List<int> coordinates = new List<int>();
    int i = 0;
 
    // Iterate over all the segments
    while (i < n) 
    {
      int seg = points[i][1];
      coordinates.Add(seg);
      int p = i + 1;
 
      if (p >= n)
        break;
 
      // Get the start point of next segment
      int arrived = points[p][0];
 
      // Loop over all those segments whose
      // start point is less than the end
      // point of current segment
      while (seg >= arrived)
      {
        p += 1;
 
        if (p >= n)
          break;
 
        arrived = points[p][0];
      }
      i = p;
    }
 
    // Print the possible values of M
    foreach (int point in coordinates)
      Console.Write(point + " ");
  }
 
  // Driver code
  public static void Main(string[] args)
  {
 
    int n = 4;
 
    // Starting points of segments
    int[] start = { 4, 1, 2, 5 };
 
    // Ending points of segments
    int[] end = { 7, 3, 5, 6 };
 
    List<int []> points = new List<int []>();
 
    // Insert ranges in points[]
    for(int i = 0; i < n; i++) 
    {
      points.Add( new [] {start[i], end[i]});
    }
 
    // Function call
    minPoints(points, n);
  }
}
 
// This code is contributed by phasing17

Javascript

<script>
 
// JavaScript program for the above approach
 
// Function to compute minimum number
// of points which cover all segments
function minPoints(points){
 
    // Sort the list of tuples by
    // their second element.
    points.sort((a,b)=>a[1]-b[1])
 
    // To store the solution
    let coordinates = []
    let i = 0
 
    // Iterate over all the segments
    while(i < n){
 
        let seg = points[i][1]
        coordinates.push(seg)
        let p = i + 1
 
        if(p >= n)
            break
 
        // Get the start point of next segment
        let arrived = points[p][0]
 
        // Loop over all those segments whose
        // start point is less than the end
        // point of current segment
        while(seg >= arrived){
 
            p += 1
            if(p >= n)
                break
            arrived = points[p][0]
        }
        i = p
    }
 
    // Print the possible values of M
    for(let point of coordinates)
        document.write(point," ")
}
 
// Driver Code
let n = 4
 
// Starting points of segments
let start = [4, 1, 2, 5]
 
// Ending points of segments
let end = [7, 3, 5, 6]
 
let points = []
 
// Insert ranges in points[]
for(let i = 0; i < n; i++){
    let tu = [start[i], end[i]]
    points.push(tu)
}
 
// Function Call
minPoints(points)
 
// This code is contributed by shinjanpatra
 
</script>

Output

3 6

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

Approach 2: Divide and Conquer Algorithm

Another approach to solve the problem of finding the minimum number of points needed to cover a set of intervals, which uses a divide and conquer algorithm.

The idea behind the divide and conquer algorithm is to recursively divide the intervals into two halves and find the minimum number of points needed to cover each half.

Then, we can combine the solutions for the two halves by using a point to cover the intervals that overlap.

Follow the Below steps to solve the above approach:

If there is only one interval, return 1.
Divide the intervals into two halves.
Recursively find the minimum number of points needed to cover the left and right halves.
Initialize the minimum number of points to cover the overlapping intervals to the maximum of the two halves.
Iterate through the left and right halves.
If the intervals overlap, update the minimum number of points to cover the overlapping intervals.
Return the minimum number of points to cover the overlapping intervals.

Below is the implementation of the above approach:

C++

// CPP program to find the minimum number of points needed
// to cover all of the intervals using Divide Algorithm
#include <bits/stdc++.h>
using namespace std;
 
int minimumPoints(vector<pair<int, int> > intervals)
{
    sort(intervals.begin(), intervals.end(),
         [](pair<int, int> a, pair<int, int> b) {
             return a.second < b.second;
         });
    vector<int> points;
    points.push_back(intervals[0].second);
    for (int i = 1; i < intervals.size(); i++) {
        if (intervals[i].first > points.back()) {
            points.push_back(intervals[i].second);
        }
    }
    return points.size();
}
 
int main()
{
    // Define the intervals
    vector<pair<int, int> > intervals
        = { { 1, 3 }, { 2, 4 }, { 3, 5 }, { 6, 8 } };
    // Compute the minimum number of points needed to cover
    // the intervals and print it
    cout << minimumPoints(intervals) << endl; // Output: 2
 
    intervals = { { 1, 2 }, { 3, 4 }, { 2, 5 }, { 5, 8 } };
    cout << minimumPoints(intervals) << endl; // Output: 3
}
 
// This code is contributed by Susobhan Akhuli

Java

// Java program to find the minimum number of points needed
// to cover all of the intervals using Divide Algorithm
import java.util.Arrays;
import java.util.Comparator;
 
public class GFG {
    public static int minimumPoints(int[][] intervals)
    {
        // sort intervals by their ending point
        Arrays.sort(intervals, new Comparator<int[]>() {
            public int compare(int[] a, int[] b)
            {
                return a[1] - b[1];
            }
        });
 
        int[] points = new int[intervals.length];
        int index = 0;
        points[index] = intervals[0][1];
        // iterate through the intervals
        for (int i = 1; i < intervals.length; i++) {
            // if the starting point of the current interval
            // is greater than the ending point of the last
            // selected interval, add its ending point
            if (intervals[i][0] > points[index]) {
                points[++index] = intervals[i][1];
            }
        }
        // return the number of points
        return index + 1;
    }
 
    public static void main(String[] args)
    {
        int[][] intervals
            = { { 1, 3 }, { 2, 4 }, { 3, 5 }, { 6, 8 } };
        System.out.println(
            minimumPoints(intervals)); // Output: 2
        int[][] intervals1
            = { { 1, 2 }, { 3, 4 }, { 2, 5 }, { 5, 8 } };
        System.out.println(
            minimumPoints(intervals1)); // Output: 3
    }
}
 
// This code is contributed by Susobhan Akhuli

Python3

# Python3 program to find the minimum number of points needed
# to cover all of the intervals using Divide Algorithm
from typing import List, Tuple
 
 
def minimumPoints(intervals: List[Tuple[int, int]]) -> int:
    intervals.sort(key=lambda x: x[1])
    points = [intervals[0][1]]
    for i in range(1, len(intervals)):
        if intervals[i][0] > points[-1]:
            points.append(intervals[i][1])
    return len(points)
 
 
if __name__ == "__main__":
    # Define the intervals
    intervals = [(1, 3), (2, 4), (3, 5), (6, 8)]
    # Compute the minimum number of points needed to cover the intervals
    minPoints = minimumPoints(intervals)
    # Print the result
    print(minPoints)  # Output: 2
 
    intervals = [(1, 2), (3, 4), (2, 5), (5, 8)]
    minPoints = minimumPoints(intervals)
    print(minPoints)  # Output: 3
 
# This code is contributed by Susobhan Akhuli

C#

// C# program to find the minimum number of points needed
// to cover all of the intervals using Divide Algorithm
 
using System;
using System.Collections.Generic;
using System.Linq;
 
class GFG {
    public static int MinimumPoints(int[][] intervals)
    {
        // sort intervals by their ending point
        Array.Sort(intervals,
                   new Comparison<int[]>(
                       (a, b) => a[1].CompareTo(b[1])));
 
        int[] points = new int[intervals.Length];
        int index = 0;
        points[index] = intervals[0][1];
 
        // iterate through the intervals
        for (int i = 1; i < intervals.Length; i++) {
            // if the starting point of the current interval
            // is greater than the ending point of the last
            // selected interval, add its ending point
            if (intervals[i][0] > points[index]) {
                points[++index] = intervals[i][1];
            }
        }
 
        // return the number of points
        return index + 1;
    }
 
    public static void Main(string[] args)
    {
        int[][] intervals
            = { new[] { 1, 3 }, new[] { 2, 4 },
                new[] { 3, 5 }, new[] { 6, 8 } };
        Console.WriteLine(
            MinimumPoints(intervals)); // Output: 2
 
        int[][] intervals1
            = { new[] { 1, 2 }, new[] { 3, 4 },
                new[] { 2, 5 }, new[] { 5, 8 } };
        Console.WriteLine(
            MinimumPoints(intervals1)); // Output: 3
    }
}
 
// This code is contributed by phasing17

Javascript

function minimumPoints(intervals) {
  intervals.sort((a, b) => a[1] - b[1]);
  const points = [intervals[0][1]];
  for (let i = 1; i < intervals.length; i++) {
    if (intervals[i][0] > points[points.length - 1]) {
      points.push(intervals[i][1]);
    }
  }
  return points.length;
}
 
// Define the intervals
let intervals = [[1, 3], [2, 4], [3, 5], [6, 8]];
 
 // Compute the minimum number of points needed to cover
    // the intervals and print it
console.log(minimumPoints(intervals)); // Output: 2
intervals = [[1, 2], [3, 4], [2, 5], [5, 8]]
console.log(minimumPoints(intervals)); // Output: 3
 
 // This code is contributed by imruhrbf8.

Output

2
3

Complexity Analysis:

Time complexity: O(N*logN), due to the recursive nature of the algorithm, where N is the number of intervals.
Auxiliary Space: O(N), since we need to store the intervals in each recursive call.

Approach 3: Dynamic Programming

Another approach to solve the problem of finding the minimum number of integers required such that each Segment contains at least one of them, which uses a dynamic programming algorithm.

The idea behind the dynamic programming algorithm is to iterate through the intervals and, for each interval, compute the minimum number of integers required such that each Segment contains at least one of them.

To do this, we can use a two-dimensional array, where the first dimension represents the index of the interval and the second dimension represents the number of points.

At each step, we have two options: We can either use a new point to cover the current interval, or we can use one of the points that we have already used to cover a previous interval. If we use a new point, we need to increment the number of points by 1. If we use an existing point, we do not need to increment the number of points.

We can compute the minimum number of points needed to cover the intervals using the following recurrence:
dp[i][j] = min(dp[i-1][j], dp[i-1][j-1] + 1)
The base case is when i is equal to 0, in which case dp[i][j] is equal to j.
Using this recurrence, we can compute the minimum number of points needed to cover the intervals in O(nk) time, where n is the number of intervals and k is the maximum number of points needed to cover the intervals.

Follow the Below steps to solve the above approach:

Sort the intervals by their starting point.
Initialize a list minPoints with the length of the intervals, and set all elements to 1.
Iterate through the intervals from the second to the last.
For each interval, iterate through the minPoints list from the first element to the current interval.
If the ending point of the previous interval is greater than or equal to the starting point of the current interval, update the current element of minPoints as the minimum of itself and the previous element plus 1.
Return the last element of minPoints.

Below is the implementation of the above approach:

C++

// CPP program to find the minimum number of points needed
// to cover all of the intervals using DP
 
#include <bits/stdc++.h>
using namespace std;
 
int minimumPoints(vector<pair<int, int> > intervals)
{
 
    // Sort the intervals by their starting point
    sort(intervals.begin(), intervals.end());
 
    // Number of intervals
    int n = intervals.size();
 
    // Initialize a list minPoints with value 1
    // and size with length of the intervals
    vector<int> minPoints(n, 1);
 
    // Iterate through the intervals from
    // the second to the last
    for (int i = 1; i < n; i++)
 
        // For each interval, iterate through
        // the minPoints list from the first
        // element to the current interval
        for (int j = 0; j < i; j++) {
 
            // If the ending point of the
            // previous interval is greater
            // than or equal to the starting
            // point of the current interval,
            // update the current element
            // of minPoints
            if (intervals[j].second <= intervals[i].first)
                minPoints[i]
                    = max(minPoints[i], minPoints[j] + 1);
            else
                minPoints[i] = 1;
        }
 
    // Return the last element of minPoints
    return minPoints[n - 1];
}
 
int main()
{
    // Define the intervals
    vector<pair<int, int> > intervals
        = { { 1, 3 }, { 2, 4 }, { 3, 5 }, { 6, 8 } };
    // Compute the minimum number of points needed to cover
    // the intervals and print it
    cout << minimumPoints(intervals) << endl; // Output: 2
 
    intervals = { { 1, 2 }, { 3, 4 }, { 2, 5 }, { 5, 8 } };
    cout << minimumPoints(intervals) << endl; // Output: 3
}
 
// This code is contributed by Susobhan Akhuli

Java

// Java program to find the minimum number of points needed
// to cover all of the intervals using DP
import java.util.Arrays;
import java.util.Comparator;
 
public class GFG {
    public static int minimumPoints(int[][] intervals)
    {
 
        // sort intervals by their starting point
        Arrays.sort(intervals, new Comparator<int[]>() {
            public int compare(int[] a, int[] b)
            {
                return a[0] - b[0];
            }
        });
 
        // Number of intervals
        int n = intervals.length;
 
        // Initialize a list minPoints with value 1
        // and size with length of the intervals
        int[] minPoints = new int[n];
        Arrays.fill(minPoints, 1);
 
        // Iterate through the intervals from
        // the second to the last
        for (int i = 1; i < n; i++) {
 
            // For each interval, iterate through
            // the minPoints list from the first
            // element to the current interval
            for (int j = 0; j < i; j++) {
 
                // If the ending point of the
                // previous interval is greater
                // than or equal to the starting
                // point of the current interval,
                // update the current element
                // of minPoints
                if (intervals[j][1] <= intervals[i][0])
                    minPoints[i] = Math.max(
                        minPoints[i], minPoints[j] + 1);
                else
                    minPoints[i] = 1;
            }
        }
        // Return the last element of minPoints
        return minPoints[n - 1];
    }
 
    public static void main(String[] args)
    {
        int[][] intervals
            = { { 1, 3 }, { 2, 4 }, { 3, 5 }, { 6, 8 } };
          // Compute the minimum number of points needed to cover
        // the intervals and print it
        System.out.println(
            minimumPoints(intervals)); // Output: 2
        int[][] intervals1
            = { { 1, 2 }, { 3, 4 }, { 2, 5 }, { 5, 8 } };
        System.out.println(
            minimumPoints(intervals1)); // Output: 3
    }
}
 
// This code is contributed by Susobhan Akhuli

Python3

# Python3 program to find the minimum number of points needed
# to cover all of the intervals using DP
 
from typing import List, Tuple
 
 
def minimumPoints(intervals: List[Tuple[int, int]]) -> int:
 
    # Sort the intervals by their starting point
    intervals.sort(key=lambda x: x[0])
 
    # Number of intervals
    n = len(intervals)
 
    # Initialize a list minPoints with value 1
    # and size with length of the intervals
    minPoints = [1] * n
 
    # Iterate through the intervals from
    # the second to the last
    for i in range(1, n):
 
        # For each interval, iterate through
        # the minPoints list from the first
        # element to the current interval
        for j in range(i):
 
            # If the ending point of the
            # previous interval is greater
            # than or equal to the starting
            # point of the current interval,
            # update the current element
            # of minPoints
            if intervals[j][1] <= intervals[i][0]:
                minPoints[i] = max(minPoints[i], minPoints[j] + 1)
            else:
                minPoints[i] = 1
 
    # Return the last element of minPoints
    return minPoints[-1]
 
 
if __name__ == "__main__":
    # Define the intervals
    intervals = [(1, 3), (2, 4), (3, 5), (6, 8)]
    # Compute the minimum number of points needed to cover the intervals
    minPoints = minimumPoints(intervals)
    print(minPoints)  # Output: 2
 
    intervals = [(1, 2), (3, 4), (2, 5), (5, 8)]
    minPoints = minimumPoints(intervals)
    print(minPoints)  # Output: 3
 
# This code is contributed by Susobhan Akhuli

C#

// C# program to find the minimum number of points needed
// to cover all of the intervals using DP
 
using System;
using System.Collections.Generic;
using System.Linq;
 
class GFG {
    static int
    MinimumPoints(List<Tuple<int, int> > intervals)
    {
        // Sort the intervals by their starting point
        intervals
            = intervals.OrderBy(t => t.Item1).ToList();
        // Number of intervals
        int n = intervals.Count;
 
        // Initialize a list minPoints with value 1
        // and size with length of the intervals
        List<int> minPoints
            = Enumerable.Repeat(1, n).ToList();
 
        // Iterate through the intervals from
        // the second to the last
        for (int i = 1; i < n; i++) {
            // For each interval, iterate through
            // the minPoints list from the first
            // element to the current interval
            for (int j = 0; j < i; j++) {
                // If the ending point of the
                // previous interval is greater
                // than or equal to the starting
                // point of the current interval,
                // update the current element
                // of minPoints
                if (intervals[j].Item2
                    <= intervals[i].Item1)
                    minPoints[i] = Math.Max(
                        minPoints[i], minPoints[j] + 1);
                else
                    minPoints[i] = 1;
            }
        }
 
        // Return the last element of minPoints
        return minPoints[n - 1];
    }
 
    static void Main(string[] args)
    {
        // Define the intervals
        List<Tuple<int, int> > intervals
            = new List<Tuple<int, int> >() {
                  Tuple.Create(1, 3), Tuple.Create(2, 4),
                      Tuple.Create(3, 5), Tuple.Create(6, 8)
              };
 
        // Compute the minimum number of points needed to
        // cover the intervals and print it
        Console.WriteLine(
            MinimumPoints(intervals)); // Output: 2
 
        intervals = new List<Tuple<int, int> >() {
            Tuple.Create(1, 2), Tuple.Create(3, 4),
                Tuple.Create(2, 5), Tuple.Create(5, 8)
        };
 
        Console.WriteLine(
            MinimumPoints(intervals)); // Output: 3
    }
}

Javascript

// JavaScript program to find the minimum number of points
// needed to cover all of the intervals using DP
 
// Function to find the minimum number of points
// needed to cover all of the intervals
function minimumPoints(intervals)
{
    // Sort the intervals by their starting point
    intervals.sort((a, b) => a[0] - b[0]);
 
    // Number of intervals
    const n = intervals.length;
 
    // Initialize an array minPoints with value 1
    // and length with the number of intervals
    const minPoints = new Array(n).fill(1);
 
    // Iterate through the intervals from
    // the second to the last
    for (let i = 1; i < n; i++) {
        // For each interval, iterate through
        // the minPoints array from the first
        // element to the current interval
        for (let j = 0; j < i; j++) {
            // If the ending point of the
            // previous interval is greater
            // than or equal to the starting
            // point of the current interval,
            // update the current element
            // of minPoints
            if (intervals[j][1] <= intervals[i][0]) {
                minPoints[i] = Math.max(minPoints[i],
                                        minPoints[j] + 1);
            }
            else {
                minPoints[i] = 1;
            }
        }
    }
 
    // Return the last element of minPoints
    return minPoints[n - 1];
}
 
// Define the intervals
const intervals = [
    [ 1, 3 ],
    [ 2, 4 ],
    [ 3, 5 ],
    [ 6, 8 ],
];
 
// Compute the minimum number of points needed to cover
// the intervals and print it
console.log(minimumPoints(intervals)); // Output: 2
 
const intervals2 = [
    [ 1, 2 ],
    [ 3, 4 ],
    [ 2, 5 ],
    [ 5, 8 ],
];
 
// Compute the minimum number of points needed to cover
// the intervals and print it
console.log(minimumPoints(intervals2)); // Output: 3

Output

2
3

Complexity Analysis:

Time complexity: O(N²), due to the nested loops, where N is the number of intervals.
Auxiliary Space: O(N), since we need to store the minPoints list.

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