Given a 2D array ranges[][] of size N * 2, with each row representing a range of the form [L, R], the task is to find two ranges such that the first range completely lies ins the second range and print their indices. If no such pair of ranges can be obtained, print -1. If multiple such ranegs exist, print any one of them.
Segment [L1, R1] lies within segment [L2, R2] if L1 ≥ L2 and R1 ≤ R2.
Examples:
Input: N = 5, ranges[][] = {{1, 5}, {2, 10}, {3, 10}, {2, 2}, {2, 15}}
Output: 4 1
Explanation: Segment [2, 2] lies completely within the segment [1, 5], as 1 ≤ 2 and 2 ≤ 5.Input: N = 4, ranges[][] = {{2, 10}, {1, 9}, {1, 8}, {1, 7}}
Output: -1
Explanation: No such pair of segments exist.
Naive Approach: The simplest approach to solve the problem is to iterate over the array and for each range, traverse over the remaining array to check if any range exists or not which lies completely inside the current range or vice versa.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to sort the array of ranges using a comparator function and check whether any segment lies inside any other segment or not. Follow the steps given below to solve this problem:
- Sort the segments in increasing order of their starting points. In the case of a pair having equal starting points, sort in decreasing order of ending points.
- Now, traverse the array and maintain the maximum ending point obtained so and compare it with that of the current segment.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Store ranges and their// corresponding array indicesvector<pair<pair<int, int>, int> > tup;// Function to find a pair of intersecting rangesvoid findIntersectingRange(int N, int ranges[][2]){ // Stores ending point // of every range int curr; // Stores the maximum // ending point obtained int currPos; // Iterate from 0 to N - 1 for (int i = 0; i < N; i++) { int x, y; // Starting point of // the current range x = ranges[i][0]; // End point of // the current range y = ranges[i][1]; // Push pairs into tup tup.push_back({ { x, y }, i + 1 }); } // Sort the tup vector sort(tup.begin(), tup.end()); curr = tup[0].first.second; currPos = tup[0].second; // Iterate over the ranges for (int i = 1; i < N; i++) { int Q = tup[i - 1].first.first; int R = tup[i].first.first; // If starting points are equal if (Q == R) { 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; } int T = tup[i].first.second; // Print the indices of the // intersecting ranges if (T <= curr) { cout << tup[i].second << ' ' << currPos; return; } else { curr = T; currPos = tup[i].second; } } // If no such pair of segments exist cout << "-1 -1";}// Driver Codeint main(){ // Given N int N = 5; // Given 2d ranges[][] array int ranges[][2] = { { 1, 5 }, { 2, 10 }, { 3, 10 }, { 2, 2 }, { 2, 15 }}; // Function call findIntersectingRange(N, ranges);} |
Python3
# Python3 program for the above approach# Store ranges and their# corresponding array indices# Function to find a pair of intersecting rangesdef findIntersectingRange(tup, N, ranges): # Stores ending po # of every range curr = 0 # Stores the maximum # ending poobtained currPos = 0 # Iterate from 0 to N - 1 for i in range(N): # Starting poof # the current range x = ranges[i][0] # End poof # the current range y = ranges[i][1] # Push pairs into tup tup.append([ [ x, y ], i + 1 ]) # Sort the tup vector tup = sorted(tup) curr = tup[0][0][1] currPos = tup[0][1] # Iterate over the ranges for i in range(1, N): Q = tup[i - 1][0][0] R = tup[i][0][0] #If starting points are equal if (Q == R): 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 T = tup[i][0][1] # Prthe indices of the # intersecting ranges if (T <= curr): print(tup[i][1], currPos) return else: curr = T currPos = tup[i][1] # If no such pair of segments exist print("-1 -1")# Driver Codeif __name__ == '__main__': # Given N N = 5 # Given 2d ranges[][] array ranges= [[ 1, 5 ], [ 2, 10 ], [ 3, 10 ], [ 2, 2 ], [ 2, 15 ]] # Function call findIntersectingRange([], N, ranges) # This code is contributed by mohit kumar 29 |
4 1
Time Complexity: O(N LogN)
Auxiliary Space: O(N)
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