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 that 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 pairs 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 into 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 to 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

Time Complexity: O(N*logN), as we sort an array of size N. Where N is the number of pairs in the array.
Auxiliary Space: O(N), as we are using extra space for the array vec and stack st. Where N is the number of pairs in the array.

Approach(Using brute force):

1. The function kOverlap takes a vector of pairs representing N ranges, and an integer K as input.
2. The function first sorts the input vector of ranges in ascending order of the start point of each range.
3. It then iterates through the sorted ranges, keeping track of the maximum endpoint of the K ranges seen so far and the number of overlapping ranges.
4. If the current range overlaps with the K ranges seen so far, the function updates the maximum endpoint and increments the overlap count.
5. If the current range doesn’t overlap with the K ranges seen so far, the function checks if K overlapping ranges have been found so far. If yes, it returns true; otherwise, it resets the overlap count and maximum endpoint for the next set of ranges.
6. The function returns false if no K overlapping ranges are found in the entire vector of ranges.

C++

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

// Function that returns true if any k
// segments overlap at any point
bool kOverlap(vector<pair<int, int> > pairs, int k)
{
    // Sort the vector by start point of each
    // range
    sort(pairs.begin(), pairs.end());

    // Keep track of the maximum endpoint of=
    //  k ranges seen so far
    int maxEndpoint = pairs[0].second;
    int overlapCount = 1;

    for (int i = 1; i < pairs.size(); i++) {
        // If the current range overlaps with
        // the k ranges seen so far
        if (pairs[i].first <= maxEndpoint) {
            overlapCount++;
            maxEndpoint = max(maxEndpoint, pairs[i].second);
        }
        // If the current range doesn't overlap
        // with the k ranges seen so far
        else {
            // If k overlapping ranges
            /// have been found
            if (overlapCount >= k) {
                return true;
            }
            // Reset the counters for the
            // next set of ranges
            overlapCount = 1;
            maxEndpoint = pairs[i].second;
        }
    }

    // Check if the last set of ranges
    // overlap k times or more
    if (overlapCount >= k) {
        return true;
    }

    // No k overlapping ranges found
    return false;
}

// Driver code
int main()
{
    vector<pair<int, int> > pairs
        = { { 1, 3 }, { 2, 4 }, { 3, 5 }, { 7, 10 } };
    int k = 3;

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

    return 0;
}

Java

import java.util.Arrays;

public class Main {
    public static void main(String[] args) {
        int[][] pairs = {
            {1, 3},
            {2, 4},
            {3, 5},
            {7, 10}
        };
        int k = 3;

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

    public static boolean kOverlap(int[][] pairs, int k) {
        // Sort the array of pairs by the start point of each range
        Arrays.sort(pairs, (a, b) -> Integer.compare(a[0], b[0]));

        // Keep track of the maximum endpoint of k ranges seen so far
        int maxEndpoint = pairs[0][1];
        int overlapCount = 1;

        for (int i = 1; i < pairs.length; i++) {
            // If the current range overlaps with the k ranges seen so far
            if (pairs[i][0] <= maxEndpoint) {
                overlapCount++;
                maxEndpoint = Math.max(maxEndpoint, pairs[i][1]);
            } else {
                // If k overlapping ranges have been found
                if (overlapCount >= k) {
                    return true;
                }
                // Reset the counters for the next set of ranges
                overlapCount = 1;
                maxEndpoint = pairs[i][1];
            }
        }

        // Check if the last set of ranges overlap k times or more
        return overlapCount >= k;
    }
}

Python3

def k_overlap(pairs, k):
    pairs.sort(key=lambda x: x[0])  # Sort the pairs by the start point

    max_endpoint = pairs[0][1]
    overlap_count = 1

    for i in range(1, len(pairs)):
        if pairs[i][0] <= max_endpoint:  # If the current range overlaps
            overlap_count += 1
            max_endpoint = max(max_endpoint, pairs[i][1])
        else:  # If the current range doesn't overlap
            if overlap_count >= k:  # If k overlapping ranges have been found
                return True
            overlap_count = 1
            max_endpoint = pairs[i][1]

    if overlap_count >= k:  # Check if the last set of ranges overlap k times or more
        return True

    return False  # No k overlapping ranges found

# Driver code
pairs = [(1, 3), (2, 4), (3, 5), (7, 10)]
k = 3

# Function call
if k_overlap(pairs, k):
    print("Yes")
else:
    print("No")

C#

using System;
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)
    {
        // Sort the list of tuples by start point of each range
        pairs.Sort((a, b) => a.Item1.CompareTo(b.Item1));

        // Keep track of the maximum endpoint of k ranges seen so far
        int maxEndpoint = pairs[0].Item2;
        int overlapCount = 1;

        for (int i = 1; i < pairs.Count; i++)
        {
            // If the current range overlaps with the k ranges seen so far
            if (pairs[i].Item1 <= maxEndpoint)
            {
                overlapCount++;
                maxEndpoint = Math.Max(maxEndpoint, pairs[i].Item2);
            }
            // If the current range doesn't overlap with the k ranges seen so far
            else
            {
                // If k overlapping ranges have been found
                if (overlapCount >= k)
                {
                    return true;
                }
                // Reset the counters for the next set of ranges
                overlapCount = 1;
                maxEndpoint = pairs[i].Item2;
            }
        }

        // Check if the last set of ranges overlap k times or more
        if (overlapCount >= k)
        {
            return true;
        }

        // No k overlapping ranges found
        return false;
    }

    // Driver code
    static void Main()
    {
        List<Tuple<int, int>> pairs = new List<Tuple<int, int>>()
        {
            new Tuple<int, int>(1, 3),
            new Tuple<int, int>(2, 4),
            new Tuple<int, int>(3, 5),
            new Tuple<int, int>(7, 10)
        };
        int k = 3;

        // Function call
        if (KOverlap(pairs, k))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

Javascript

// Function that returns true if any k segments overlap at any point
function kOverlap(pairs, k) {
    // Sort the array of pairs by the start point of each range
    pairs.sort((a, b) => a[0] - b[0]);

    // Keep track of the maximum endpoint of k ranges seen so far
    let maxEndpoint = pairs[0][1];
    let overlapCount = 1;

    for (let i = 1; i < pairs.length; i++) {
        // If the current range overlaps with the k ranges seen so far
        if (pairs[i][0] <= maxEndpoint) {
            overlapCount++;
            maxEndpoint = Math.max(maxEndpoint, pairs[i][1]);
        } else {
            // If k overlapping ranges have been found
            if (overlapCount >= k) {
                return true;
            }
            // Reset the counters for the next set of ranges
            overlapCount = 1;
            maxEndpoint = pairs[i][1];
        }
    }

    // Check if the last set of ranges overlap k times or more
    if (overlapCount >= k) {
        return true;
    }

    // No k overlapping ranges found
    return false;
}

// Driver code
const pairs = [
    [1, 3],
    [2, 4],
    [3, 5],
    [7, 10]
];
const k = 3;

// Function call
if (kOverlap(pairs, k)) {
    console.log("Yes");
} else {
    console.log("No");
}

Output

Yes

Time Complexity: O(NlogN)
Auxiliary Space:  O(1)