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]ifL1 ≥ L2andR1 ≤ R2.

**Examples:**

Input:N = 5, ranges[][] = {{1, 5}, {2, 10}, {3, 10}, {2, 2}, {2, 15}}Output:4 1Explanation: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:-1Explanation: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(N^{2})* 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 indices` `vector<pair<pair<` `int` `, ` `int` `>, ` `int` `> > tup;` `// Function to find a pair of intersecting ranges` `void` `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 Code` `int` `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);` `}` |

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## Python3

`# Python3 program for the above approach` `# Store ranges and their` `# corresponding array indices` `# Function to find a pair of intersecting ranges` `def` `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 Code` `if` `__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` |

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**Output:**

4 1

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

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