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Find a pair of overlapping intervals from a given Set
  • Difficulty Level : Easy
  • Last Updated : 25 Feb, 2021

Given a 2D array arr[][] with each row of the form {l, r}, the task is to find a pair (i, j) such that the ith interval lies within the jth interval. If multiple solutions exist, then print anyone of them. Otherwise, print -1.

Examples:

Input: N = 5, arr[][] = { { 1, 5 }, { 2, 10 }, { 3, 10}, {2, 2}, {2, 15}}
Output: 3 0
Explanation: [2, 2] lies inside [1, 5].

Input: N = 4, arr[][] = { { 2, 10 }, { 1, 9 }, { 1, 8 }, { 1, 7 } }
Output: -1
Explanation: No such pair of intervals exist.

Native Approach: The simplest approach to solve this problem is to generate all possible pairs of the array. For every pair (i, j), check if the ith interval lies within the jth interval or not. If found to be true, then print the pairs. Otherwise, print -1
Time Complexity: O(N2)
Auxiliary Space:O(1)

Efficient Approach: The idea is to sort the segments firstly by their left border in increasing order and in case of equal left borders, sort them by their right borders in decreasing order. Then, just find the intersecting intervals by keeping track of the maximum right border.



Follow the steps below to solve the problem:

  1. Sort the given array of intervals according to their left border and if any two left borders are equal, sort them with their right border in decreasing order.
  2. Now, traverse from left to right, keep the maximum right border of processed segments and compare it to the current segment.
  3. If the segments are overlapping, print their indices.
  4. Otherwise, after traversing, if no overlapping segments are found, print -1.

Below is the implementation of the above approach:

C++




// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find a pair(i, j) such that
// i-th interval lies within the j-th interval
void findOverlapSegement(int N, int a[], int b[])
{
 
    // Store interval and index of the interval
    // in the form of { {l, r}, index }
    vector<pair<pair<int, int>, int> > tup;
 
    // Traverse the array, arr[][]
    for (int i = 0; i < N; i++) {
 
        int x, y;
 
        // Stores l-value of
        // the interval
        x = a[i];
 
        // Stores r-value of
        // the interval
        y = b[i];
 
        // Push current interval and index into tup
        tup.push_back(pair<pair<int, int>, int>(
            pair<int, int>(x, y), i));
    }
 
    // Sort the vector based on l-value
    // of the intervals
    sort(tup.begin(), tup.end());
 
    // Stores r-value of current interval
    int curr = tup[0].first.second;
 
    // Stores index of current interval
    int currPos = tup[0].second;
 
    // Traverse the vector, tup[]
    for (int i = 1; i < N; i++) {
 
        // Stores l-value of previous interval
        int Q = tup[i - 1].first.first;
 
        // Stores l-value of current interval
        int R = tup[i].first.first;
 
        // If Q and R are equal
        if (Q == R) {
 
            // Print the index of interval
            if (tup[i - 1].first.second
                < tup[i].first.second)
                cout << tup[i - 1].second << ' '
                     << tup[i].second;
 
            else
                cout << tup[i].second << ' '
                     << tup[i - 1].second;
 
            return;
        }
 
        // Stores r-value of current interval
        int T = tup[i].first.second;
 
        // If T is less than or equal to curr
        if (T <= curr) {
            cout << tup[i].second << ' ' << currPos;
            return;
        }
        else {
 
            // Update curr
            curr = T;
 
            // Update currPos
            currPos = tup[i].second;
        }
    }
 
    // If such intervals found
    cout << "-1 -1";
}
 
// Driver Code
int main()
{
 
    // Given l-value of segments
    int a[] = { 1, 2, 3, 2, 2 };
 
    // Given r-value of segments
    int b[] = { 5, 10, 10, 2, 15 };
 
    // Given size
    int N = sizeof(a) / sizeof(int);
 
    // Function Call
    findOverlapSegement(N, a, b);
}


Java




// Java program to implement
// the above approach
import java.util.*;
import java.lang.*;
 
class pair{
  int l,r,index;
 
  pair(int l, int r, int index){
    this.l = l;
    this.r = r;
    this.index=index;
  }
}
class GFG {
 
  // Function to find a pair(i, j) such that
  // i-th interval lies within the j-th interval
  static void findOverlapSegement(int N, int[] a, int[] b)
  {
 
    // Store interval and index of the interval
    // in the form of { {l, r}, index }
    ArrayList<pair> tup = new ArrayList<>();
 
    // Traverse the array, arr[][]
    for (int i = 0; i < N; i++) {
 
      int x, y;
 
      // Stores l-value of
      // the interval
      x = a[i];
 
      // Stores r-value of
      // the interval
      y = b[i];
 
      // Push current interval and index into tup
      tup.add(new pair(x, y, i));
    }
 
    // Sort the vector based on l-value
    // of the intervals
    Collections.sort(tup,(aa,bb)->(aa.l!=bb.l)?aa.l-bb.l:aa.r-bb.r);
 
    // Stores r-value of current interval
    int curr = tup.get(0).r;
 
    // Stores index of current interval
    int currPos = tup.get(0).index;
 
    // Traverse the vector, tup[]
    for (int i = 1; i < N; i++) {
 
      // Stores l-value of previous interval
      int Q = tup.get(i - 1).l;
 
      // Stores l-value of current interval
      int R = tup.get(i).l;
 
      // If Q and R are equal
      if (Q == R) {
 
        // Print the index of interval
        if (tup.get(i - 1).r < tup.get(i).r)
          System.out.print(tup.get(i - 1).index + " " + tup.get(i).index);
 
        else
          System.out.print(tup.get(i).index + " " + tup.get(i - 1).index);
 
        return;
      }
 
      // Stores r-value of current interval
      int T = tup.get(i).r;
 
      // If T is less than or equal to curr
      if (T <= curr) {
        System.out.print(tup.get(i).index + " " + currPos);
        return;
      }
      else {
 
        // Update curr
        curr = T;
 
        // Update currPos
        currPos = tup.get(i).index;
      }
    }
 
    // If such intervals found
    System.out.print("-1 -1");
  }    
 
  // Driver code
  public static void main (String[] args)
  {
 
    // Given l-value of segments
    int[] a = { 1, 2, 3, 2, 2 };
 
    // Given r-value of segments
    int[] b = { 5, 10, 10, 2, 15 };
 
    // Given size
    int N = a.length;
 
    // Function Call
    findOverlapSegement(N, a, b);
  }
}
 
// This code is contributed by offbeat.


Python3




# Python3 program to implement
# the above approach
 
# Function to find a pair(i, j) such that
# i-th interval lies within the j-th interval
def findOverlapSegement(N, a, b) :
     
    # Store interval and index of the interval
    # in the form of { {l, r}, index }
    tup = []
     
    # Traverse the array, arr[][]
    for i in range(N) :
         
        # Stores l-value of
        # the interval
        x = a[i]
         
        # Stores r-value of
        # the interval
        y = b[i]
         
        # Push current interval and index into tup
        tup.append(((x,y),i))
         
    # Sort the vector based on l-value
    # of the intervals
    tup.sort()
     
    # Stores r-value of current interval
    curr = tup[0][0][1]
     
    # Stores index of current interval
    currPos = tup[0][1]
     
    # Traverse the vector, tup[]
    for i in range(1,N) :
         
        # Stores l-value of previous interval
        Q = tup[i - 1][0][0]
         
        # Stores l-value of current interval
        R = tup[i][0][0]
         
        # If Q and R are equal
        if Q == R :
             
            # Print the index of interval
            if tup[i - 1][0][1] < tup[i][0][1] :
                 
                print(tup[i - 1][1], tup[i][1])
                 
            else :
                 
                print(tup[i][1], tup[i - 1][1])
                 
            return
         
        # Stores r-value of current interval
        T = tup[i][0][1]
         
        # If T is less than or equal to curr
        if (T <= curr) :
             
            print(tup[i][1], currPos)
             
            return
        else :
             
            # Update curr
            curr = T
             
            # Update currPos
            currPos = tup[i][1]
             
    # If such intervals found
    print("-1", "-1", end = "")
     
# Given l-value of segments
a = [ 1, 2, 3, 2, 2 ]
 
# Given r-value of segments
b = [ 5, 10, 10, 2, 15 ]
 
# Given size
N = len(a)
 
# Function Call
findOverlapSegement(N, a, b)
 
# This code is contributed by divyesh072019


C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG {
 
  // Function to find a pair(i, j) such that
  // i-th interval lies within the j-th interval
  static void findOverlapSegement(int N, int[] a, int[] b)
  {
 
    // Store interval and index of the interval
    // in the form of { {l, r}, index }
    List<Tuple<Tuple<int,int>, int>> tup = new List<Tuple<Tuple<int,int>, int>>();
 
    // Traverse the array, arr[][]
    for (int i = 0; i < N; i++) {
 
      int x, y;
 
      // Stores l-value of
      // the interval
      x = a[i];
 
      // Stores r-value of
      // the interval
      y = b[i];
 
      // Push current interval and index into tup
      tup.Add(new Tuple<Tuple<int,int>, int>(new Tuple<int, int>(x, y), i));
    }
 
    // Sort the vector based on l-value
    // of the intervals
    tup.Sort();
 
    // Stores r-value of current interval
    int curr = tup[0].Item1.Item2;
 
    // Stores index of current interval
    int currPos = tup[0].Item2;
 
    // Traverse the vector, tup[]
    for (int i = 1; i < N; i++) {
 
      // Stores l-value of previous interval
      int Q = tup[i - 1].Item1.Item1;
 
      // Stores l-value of current interval
      int R = tup[i].Item1.Item1;
 
      // If Q and R are equal
      if (Q == R) {
 
        // Print the index of interval
        if (tup[i - 1].Item1.Item2 < tup[i].Item1.Item2)
          Console.Write(tup[i - 1].Item2 + " " + tup[i].Item2);
 
        else
          Console.Write(tup[i].Item2 + " " + tup[i - 1].Item2);
 
        return;
      }
 
      // Stores r-value of current interval
      int T = tup[i].Item1.Item2;
 
      // If T is less than or equal to curr
      if (T <= curr) {
        Console.Write(tup[i].Item2 + " " + currPos);
        return;
      }
      else {
 
        // Update curr
        curr = T;
 
        // Update currPos
        currPos = tup[i].Item2;
      }
    }
 
    // If such intervals found
    Console.Write("-1 -1");
  }    
 
  // Driver code
  static void Main()
  {
 
    // Given l-value of segments
    int[] a = { 1, 2, 3, 2, 2 };
 
    // Given r-value of segments
    int[] b = { 5, 10, 10, 2, 15 };
 
    // Given size
    int N = a.Length;
 
    // Function Call
    findOverlapSegement(N, a, b);
  }
}
 
// This code is contributed by divyeshrabadiya07


Output: 

3 0

 

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

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