Check if any K ranges overlap at any point

Given N ranges [L, R] and an integer K, the task is to check if there are any K ranges which overlap at any point.

Examples:

Input: ranges[][] = {{1, 3}, {2, 4}, {3, 4}, {7, 10}}, K = 3 
Output: Yes 
3 is a common point among the 
ranges {1, 3}, {2, 4} and {3, 4}.

Input: ranges[][] = {{1, 2}, {3, 4}, {5, 6}, {7, 8}}, K = 2 
Output: No 

Approach: The idea is to make a vector of pair and store the starting point for every range as pair in this vector as (starting point, -1) and the ending point as (ending point, 1). Now, sort the vector then traverse the vector and if the current element is a starting point then push it in a stack and if it is an ending point then pop an element from the stack. If at any instance of time, the size of the stack is greater than or equal K then print Yes else print No in the end.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Comparator to sort the vector of pairs
bool sortby(const pair<int, int>& a,
            const pair<int, int>& b)
{
    if (a.first != b.first)
        return a.first < b.first;
    return (a.second < b.second);
}

// Function that returns true if any k
// segments overlap at any point
bool kOverlap(vector<pair<int, int> > pairs, int k)
{
    // Vector to store the starting point
    // and the ending point
    vector<pair<int, int> > vec;
    for (int i = 0; i < pairs.size(); i++) {

        // Starting points are marked by -1
        // and ending points by +1
        vec.push_back({ pairs[i].first, -1 });
        vec.push_back({ pairs[i].second, +1 });
    }

    // Sort the vector by first element
    sort(vec.begin(), vec.end());

    // Stack to store the overlaps
    stack<pair<int, int> > st;

    for (int i = 0; i < vec.size(); i++) {
        // Get the current element
        pair<int, int> cur = vec[i];

        // If it is the starting point
        if (cur.second == -1) {
            // Push it in the stack
            st.push(cur);
        }

        // It is the ending point
        else {
            // Pop an element from stack
            st.pop();
        }

        // If more than k ranges overlap
        if (st.size() >= k) {
            return true;
        }
    }

    return false;
}

// Driver code
int main()
{
    vector<pair<int, int> > pairs;
    pairs.push_back(make_pair(1, 3));
    pairs.push_back(make_pair(2, 4));
    pairs.push_back(make_pair(3, 5));
    pairs.push_back(make_pair(7, 10));

    int n = pairs.size(), k = 3;

    if (kOverlap(pairs, k))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

Java

// Java implementation of the approach
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Stack;

class GFG{

static class Pair
{
    int first, second;

    public Pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function that returns true if any k
// segments overlap at any point
static boolean kOverlap(ArrayList<Pair> pairs,
                        int k)
{

    // Vector to store the starting point
    // and the ending point
    ArrayList<Pair> vec = new ArrayList<>();
    for(int i = 0; i < pairs.size(); i++)
    {

        // Starting points are marked by -1
        // and ending points by +1
        vec.add(new Pair(pairs.get(i).first, -1));
        vec.add(new Pair(pairs.get(i).second, +1));
    }

    // Sort the vector by first element
    Collections.sort(vec, new Comparator<Pair>()
    {

        // Comparator to sort the vector of pairs
        public int compare(Pair a, Pair b)
        {
            if (a.first != b.first)
                return a.first - b.first;

            return (a.second - b.second);
        }
    });

    // Stack to store the overlaps
    Stack<Pair> st = new Stack<>();

    for(int i = 0; i < vec.size(); i++)
    {

        // Get the current element
        Pair cur = vec.get(i);

        // If it is the starting point
        if (cur.second == -1)
        {

            // Push it in the stack
            st.push(cur);
        }

        // It is the ending point
        else
        {

            // Pop an element from stack
            st.pop();
        }

        // If more than k ranges overlap
        if (st.size() >= k)
        {
            return true;
        }
    }
    return false;
}

// Driver code
public static void main(String[] args)
{
    ArrayList<Pair> pairs = new ArrayList<>();
    pairs.add(new Pair(1, 3));
    pairs.add(new Pair(2, 4));
    pairs.add(new Pair(3, 5));
    pairs.add(new Pair(7, 10));

    int n = pairs.size(), k = 3;

    if (kOverlap(pairs, k))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}

// This code is contributed by sanjeev2552

Python3

# Python3 implementation of the approach

# Function that returns true if any k
# segments overlap at any point
def kOverlap(pairs: list, k):

    # Vector to store the starting point
    # and the ending point
    vec = list()
    for i in range(len(pairs)):

        # Starting points are marked by -1
        # and ending points by +1
        vec.append((pairs[0], -1))
        vec.append((pairs[1], 1))

    # Sort the vector by first element
    vec.sort(key = lambda a: a[0])

    # Stack to store the overlaps
    st = list()

    for i in range(len(vec)):

        # Get the current element
        cur = vec[i]

        # If it is the starting point
        if cur[1] == -1:

            # Push it in the stack
            st.append(cur)

        # It is the ending point
        else:

            # Pop an element from stack
            st.pop()

        # If more than k ranges overlap
        if len(st) >= k:
            return True
    return False


# Driver Code
if __name__ == "__main__":
    pairs = list()
    pairs.append((1, 3))
    pairs.append((2, 4))
    pairs.append((3, 5))
    pairs.append((7, 10))

    n = len(pairs)
    k = 3

    if kOverlap(pairs, k):
        print("Yes")
    else:
        print("No")

# This code is contributed by
# sanjeev2552

C#

// C# implementation of the approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{

// Function that returns true if any k
// segments overlap at any point
static bool kOverlap(List<Tuple<int,int>> pairs,
                        int k)
{

    // Vector to store the starting point
    // and the ending point
    List<Tuple<int,int>> vec = new List<Tuple<int,int>>();
    for(int i = 0; i < pairs.Count; i++)
    {

        // Starting points are marked by -1
        // and ending points by +1
        vec.Add(new Tuple<int,int>(pairs[i].Item1,-1));
        vec.Add(new Tuple<int,int>(pairs[i].Item2,1));
    }
    vec.Sort();

    // Stack to store the overlaps
    Stack st = new Stack();
    for(int i = 0; i < vec.Count; i++)
    {

        // Get the current element
        Tuple<int,int> cur = vec[i];

        // If it is the starting point
        if (cur.Item2 == -1)
        {

            // Push it in the stack
            st.Push(cur);
        }

        // It is the ending point
        else
        {

            // Pop an element from stack
            st.Pop();
        }

        // If more than k ranges overlap
        if (st.Count >= k)
        {
            return true;
        }
    }
    return false;
}

// Driver code
public static void Main(params string[] args)
{
    List<Tuple<int,int>> pairs = new List<Tuple<int,int>>();
    pairs.Add(new Tuple<int,int>(1, 3));
    pairs.Add(new Tuple<int,int>(2, 4));
    pairs.Add(new Tuple<int,int>(3, 5));
    pairs.Add(new Tuple<int,int>(7, 10));

    int n = pairs.Count, k = 3;
    if (kOverlap(pairs, k))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}

// This code is contributed by rutvik_56/

Javascript

<script>

// JavaScript implementation of the approach

// Function that returns true if any k
// segments overlap at any point
function kOverlap(pairs, k)
{
    // Vector to store the starting point
    // and the ending point
    var vec = [];
    for (var i = 0; i < pairs.length; i++) {

        // Starting points are marked by -1
        // and ending points by +1
        vec.push([pairs[i][0], -1 ]);
        vec.push([pairs[i][1], +1 ]);
    }

    // Sort the vector by first element
    vec.sort((a,b)=>{
        if(a[0]!=b[0])
            return a[0]-b[0]
        return a[1]-b[1]
    });

    // Stack to store the overlaps
    var st = [];

    for (var i = 0; i < vec.length; i++) {
        // Get the current element
        var cur = vec[i];

        // If it is the starting point
        if (cur[1] == -1) {
            // Push it in the stack
            st.push(cur);
        }

        // It is the ending point
        else {
            // Pop an element from stack
            st.pop();
        }

        // If more than k ranges overlap
        if (st.length >= k) {
            return true;
        }
    }

    return false;
}

// Driver code
var pairs = [];
pairs.push([1, 3]);
pairs.push([2, 4]);
pairs.push([3, 5]);
pairs.push([7, 10]);
var n = pairs.length, k = 3;
if (kOverlap(pairs, k))
    document.write( "Yes");
else
    document.write( "No");


</script>

Output:

Yes