Maximum length intersection of all K ranges among all given ranges

Last Updated : 28 Mar, 2023

Given an array arr[] consisting of N ranges of the form [L, R], the task is to select K ranges such that the intersection length of the K ranges is maximum.

Examples:

Input: arr[] = {{5, 15}, {1, 10}, {14, 50}, {30, 70}, {99, 100}}, K = 2
Output: 21
Explanation: Selecting ranges (14, 50) and (30, 70) would give the maximum answer. Therefore, the maximum length intersection of above 2 ranges is (50 - 30 + 1) = 21.

Input: arr[] = {{-8, 6}, {7, 9}, {-10, -5}, {-4, 5}}, K = 3
Output: 0

Input: arr[] = {{1, 10}, {2, 5}, {3, 7}, {3, 4}, {3, 5}, {3, 10}, {4, 7}}, K = 3

Fig.1

Output: 5
Explanation: Selecting ranges (1, 10), (3, 7), (3, 10), which will give us the maximum length possible.

Approach: The problem can be solved based on the following idea:

Sorting ranges according to { L, R } and considering each given range as a potential starting point L to our required answer as [L, Kth largest ending point of ranges seen so far].

Follow the steps mentioned below to implement the idea:

Use vector pair asc to store ranges in non-decreasing order of starting point l followed by ending point r.
Traverse asc from the beginning and maintain min-heap (priority_queue) pq to store K largest ending point r of ranges seen so far.
Discard the values from the heap if it is less than the starting point l[i] of the current range i.
Update the answer as the maximum of previous answers we got and pq.top() - l[i] + 1.

Below is the implementation of the above approach:

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to get maximum
// intersection length
int maxKRanges(vector<vector<int> > range, int k)
{
    int n = range.size();
    vector<pair<int, int> > asc;
    for (int i = 0; i < n; i++) {
        asc.push_back({ range[i][0], range[i][1] });
    }

    // Sorting ranges in
    // non-descending order
    sort(asc.begin(), asc.end());

    // Min-heap to store k largest ending
    // point of ranges seen so far
    priority_queue<int, vector<int>, greater<int> > pq;

    int ans = 0;
    for (int i = 0; i < n; i++) {

        // Ending point of ith range
        pq.push(asc[i].second);

        // Ranges having zero intersection
        while (!pq.empty() && pq.top() < asc[i].first)
            pq.pop();

        // Size upto k
        while (pq.size() > k)
            pq.pop();

        // Update answer
        if (pq.size() == k)
            ans = max(ans, pq.top() - asc[i].first + 1);
    }

    return ans;
}

// Driver Code
int main()
{
    // Number of ranges
    int N = 5;

    // Number of ranges selected at a time
    int K = 2;

    vector<vector<int> > range = { { 5, 15 },
                                   { 1, 10 },
                                   { 14, 50 },
                                   { 30, 70 },
                                   { 99, 100 } };

    // Function call
    cout << maxKRanges(range, K) << endl;

    return 0;
}

Python3

# Python code to implement the approach
import heapq

# Function to get maximum
# intersection length
def maxKRanges(range_, k):
    n = len(range_)
    asc = []
    for i in range(n):
        asc.append((range_[i][0], range_[i][1]))

    # Sorting ranges in
    # non-descending order
    asc.sort()

    # Min-heap to store k largest ending
    # point of ranges seen so far
    pq = []

    ans = 0
    for i in range(n):

        # Ending point of ith range
        heapq.heappush(pq, asc[i][1])

        # Ranges having zero intersection
        while pq and pq[0] < asc[i][0]:
            heapq.heappop(pq)

        # Size upto k
        while len(pq) > k:
            heapq.heappop(pq)

        # Update answer
        if len(pq) == k:
            ans = max(ans, pq[0] - asc[i][0] + 1)

    return ans


# Driver Code
# Number of ranges
N = 5

# Number of ranges selected at a time
K = 2

range_ = [[5, 15],
          [1, 10],
          [14, 50],
          [30, 70],
          [99, 100]]

# Function call
print(maxKRanges(range_, K))

using System;
using System.Collections.Generic;
using System.Linq;

class Program
{
  static int MaxKRanges(List<List<int>> range, int k)
  {
    int n = range.Count;
    List<Tuple<int, int>> asc = new List<Tuple<int, int>>();
    for (int i = 0; i < n; i++)
    {
      asc.Add(new Tuple<int, int>(range[i][0], range[i][1]));
    }

    // Sorting ranges in non-descending order
    asc.Sort();

    // SortedSet to store k largest ending point of ranges seen so far
    SortedSet<int> pq = new SortedSet<int>();

    int ans = 0;
    for (int i = 0; i < n; i++)
    {
      // Ending point of ith range
      pq.Add(asc[i].Item2);

      // Ranges having zero intersection
      while (pq.Count > 0 && pq.Min < asc[i].Item1)
        pq.Remove(pq.Min);

      // Size upto k
      while (pq.Count > k)
        pq.Remove(pq.Max);

      // Update answer
      if (pq.Count == k)
        ans = Math.Max(ans, pq.Min - asc[i].Item1 + 1);
    }

    return ans;
  }

  // Driver Code
  static void Main()
  {
    // Number of ranges
    int N = 5;

    // Number of ranges selected at a time
    int K = 2;

    List<List<int>> range = new List<List<int>>() {
      new List<int>(){ 5, 15 },
      new List<int>(){ 1, 10 },
      new List<int>(){ 14, 50 },
      new List<int>(){ 30, 70 },
      new List<int>(){ 99, 100 }
    };

    // Function call
    Console.WriteLine(MaxKRanges(range, K));
  }
}

JavaScript

// Function to get maximum intersection length
function maxKRanges(range, k) {
    const n = range.length;
    const asc = [];

    for (let i = 0; i < n; i++) {
        asc.push([range[i][0], range[i][1]]);
    }

    // Sorting ranges in non-descending order
    asc.sort((a, b) => a[0] - b[0]);

    // Min-heap to store k largest ending
    // point of ranges seen so far
    const pq = [];

    let ans = 0;
    for (let i = 0; i < n; i++) {
        // Ending point of ith range
        pq.push(asc[i][1]);

        // Ranges having zero intersection
        while (pq.length && pq[0] < asc[i][0]) {
            pq.shift();
        }

        // Size upto k
        while (pq.length > k) {
            pq.shift();
        }

        // Update answer
        if (pq.length === k) {
            ans = Math.max(ans, pq[0] - asc[i][0] + 1);
        }
    }

    return ans;
}

// Driver code
// Number of ranges
const N = 5;

// Number of ranges selected at a time
const K = 2;

const range = [[5, 15],
               [1, 10],
               [14, 50],
               [30, 70],
               [99, 100]];

// Function call
console.log(maxKRanges(range, K));

// This code is contributed by Prajwal Kandekar

Java

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.PriorityQueue;

class GFG {
  public static int maxKRanges(ArrayList<ArrayList<Integer>> range, int k) {
    int n = range.size();
    int[][] asc = new int[n][2];
    for (int i = 0; i < n; i++) {
      asc[i][0] = range.get(i).get(0);
      asc[i][1] = range.get(i).get(1);
    }

    // Sorting ranges in non-descending order
    Arrays.sort(asc, (a, b) -> a[0] - b[0]);

    // Min-heap to store k largest ending point of ranges seen so far
    PriorityQueue<Integer> pq = new PriorityQueue<>(k);

    int ans = 0;
    for (int i = 0; i < n; i++) {
      // Ending point of ith range
      pq.offer(asc[i][1]);

      // Ranges having zero intersection
      while (!pq.isEmpty() && pq.peek() < asc[i][0])
        pq.poll();

      // Size up to k
      while (pq.size() > k)
        pq.poll();

      // Update answer
      if (pq.size() == k)
        ans = Math.max(ans, pq.peek() - asc[i][0] + 1);
    }

    return ans;
  }

  public static void main(String[] args) {
    // Number of ranges
    int N = 5;

    // Number of ranges selected at a time
    int K = 2;

    ArrayList<ArrayList<Integer>> range = new ArrayList<>();
    range.add(new ArrayList<Integer>(Arrays.asList(5, 15)));
    range.add(new ArrayList<Integer>(Arrays.asList(1, 10)));
    range.add(new ArrayList<Integer>(Arrays.asList(14, 50)));
    range.add(new ArrayList<Integer>(Arrays.asList(30, 70)));
    range.add(new ArrayList<Integer>(Arrays.asList(99, 100)));

    // Function call
    System.out.println(maxKRanges(range, K));
  }
}

Output

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

Maximum length intersection of all K ranges among all given ranges

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