Pair having all other given pairs lying between its minimum and maximum

Given a 2D array arr[][] consisting of N pairs of integers, the task is to find the pair which covers all other pairs of the given array. If it is impossible to find such a pair, then print -1.

A pair {a, ?b} will cover another pair {c, ?d}, if the condition (a???c???d???b) holds true. 

Examples:

Input: arr[][2] = {{2, 2}, {3, 3}, {3, 5}, {4, 5}, {1, 1}, {1, 5}}
Output: 6
Explanation:
There exist a pair (1, 5) which cover all other pair because all other pair lies in the pair {1, 5}.
Therefore, the position of the pair {1, 5} is 6. So, the output is 6.

Input: arr[][] = {{1, 20}, {2, 22}, {3, 18}}
Output: -1
Explanation:
No such pair exists which covers all the remaining pairs.
Therefore, the output is -1 

Naive Approach: The simplest approach is to compare each pair with all other pairs and check if any pair covers all the pairs or not. Below are the steps:

1. Initialize a variable count = 0 which stores the number of pairs that lie between the current pair.
2. Traverse the array of pairs and for each pair, check if the count is equal to the total number of pairs or not.
3. If found to be true, it means that the pair can cover all other pairs. Print the pair.
4. Otherwise, set count = 0 and repeat the above steps for the next pair.
5. If no such pair exists, then print -1.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach

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

// Function to find the position of
// the pair that covers every pair
// in the array arr[][]
void position(int arr[][2], int N)
{
    // Stores the index of the
    // resultant pair
    int pos = -1;

    // To count the occurrences
    int count;

    // Iterate to check every pair
    for (int i = 0; i < N; i++) {

        // Set count to 0
        count = 0;

        for (int j = 0; j < N; j++) {

            // Condition to checked for
            // overlapping of pairs
            if (arr[i][0] <= arr[j][0]
                && arr[i][1] >= arr[j][1]) {
                count++;
            }
        }

        // If that pair can cover all other
        // pairs then store its position
        if (count == N) {
            pos = i;
        }
    }

    // If position not found
    if (pos == -1) {

        cout << pos;
    }

    // Otherwise
    else {

        cout << pos + 1;
    }
}

// Driver Code
int main()
{
    // Given array of pairs
    int arr[][2] = {{ 3, 3 }, { 1, 3 },
                    { 2, 2 }, { 2, 3 },
                    { 1, 2 }};

    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    position(arr, N);
}

Java

// Java program for
// the above approach
import java.util.*;
class GFG{

// Function to find the position of
// the pair that covers every pair
// in the array arr[][]
static void position(int arr[][],
                     int N)
{
  // Stores the index of the
  // resultant pair
  int pos = -1;

  // To count the occurrences
  int count;

  // Iterate to check every pair
  for (int i = 0; i < N; i++)
  {
    // Set count to 0
    count = 0;

    for (int j = 0; j < N; j++)
    {
      // Condition to checked for
      // overlapping of pairs
      if (arr[i][0] <= arr[j][0] &&
          arr[i][1] >= arr[j][1])
      {
        count++;
      }
    }

    // If that pair can cover all other
    // pairs then store its position
    if (count == N)
    {
      pos = i;
    }
  }

  // If position not found
  if (pos == -1)
  {
    System.out.print(pos);
  }

  // Otherwise
  else
  {
    System.out.print(pos + 1);
  }
}

// Driver Code
public static void main(String[] args)
{
  // Given array of pairs
  int arr[][] = {{3, 3}, {1, 3},
                 {2, 2}, {2, 3},
                 {1, 2}};

  int N = arr.length;

  // Function Call
  position(arr, N);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for
# the above approach

# Function to find the position of
# the pair that covers every pair
# in the array arr
def position(arr, N):

    # Stores the index of the
    # resultant pair
    pos = -1;

    # To count the occurrences
    count = 0;

    # Iterate to check every pair
    for i in range(N):

        # Set count to 0
        count = 0;

        for j in range(N):
            # Condition to checked for
            # overlapping of pairs
            if (arr[i][0] <= arr[j][0] and
                arr[i][1] >= arr[j][1]):
                count += 1;

        # If that pair can cover
        # all other pairs then
        # store its position
        if (count == N):
            pos = i;

    # If position not found
    if (pos == -1):
        print(pos);

    # Otherwise
    else:
        print(pos + 1);

# Driver Code
if __name__ == '__main__':

    # Given array of pairs
    arr = [[3, 3], [1, 3],
           [2, 2], [2, 3],
           [1, 2]];

    N = len(arr);

    # Function Call
    position(arr, N);

# This code is contributed by shikhasingrajput

C#

// C# program for
// the above approach
using System;
class GFG{

// Function to find the position of
// the pair that covers every pair
// in the array arr[][]
static void position(int[,] arr,
                     int N)
{
  // Stores the index of the
  // resultant pair
  int pos = -1;

  // To count the occurrences
  int count;

  // Iterate to check every pair
  for(int i = 0; i < N; i++)
  {
    // Set count to 0
    count = 0;

    for(int j = 0; j < N; j++)
    {
      // Condition to checked for
      // overlapping of pairs
      if (arr[i, 0] <= arr[j, 0] &&
          arr[i, 1] >= arr[j, 1])
      {
        count++;
      }
    }

    // If that pair can cover
    // all other pairs then
    // store its position
    if (count == N)
    {
      pos = i;
    }
  }

  // If position not found
  if (pos == -1)
  {
    Console.Write(pos);
  }

  // Otherwise
  else
  {
    Console.Write(pos + 1);
  }
}

// Driver Code
public static void Main()
{
  // Give array of pairs
  int[,] arr = {{3, 3}, {1, 3},
                {2, 2}, {2, 3},
                {1, 2}};

  int N = arr.GetLength(0);

  // Function Call
  position(arr, N);
}
}

// This code is contributed by sanjoy_62

Javascript

<script>

// Javascript program for
// the above approach

// Function to find the position of
// the pair that covers every pair
// in the array arr[][]
function position(arr, N)
{
  // Stores the index of the
  // resultant pair
  let pos = -1;

  // To count the occurrences
  let count;

  // Iterate to check every pair
  for (let i = 0; i < N; i++)
  {
    // Set count to 0
    count = 0;

    for (let j = 0; j < N; j++)
    {
      // Condition to checked for
      // overlapping of pairs
      if (arr[i][0] <= arr[j][0] &&
          arr[i][1] >= arr[j][1])
      {
        count++;
      }
    }

    // If that pair can cover all other
    // pairs then store its position
    if (count == N)
    {
      pos = i;
    }
  }

  // If position not found
  if (pos == -1)
  {
    document.write(pos);
  }

  // Otherwise
  else
  {
    document.write(pos + 1);
  }
}

// Driver Code

    // Given array of pairs
  let arr = [[3, 3], [1, 3],
                 [2, 2], [2, 3],
                 [1, 2]];

  let N = arr.length;

  // Function Call
  position(arr, N);

</script>

Output

2

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to observe that the answer is always unique because there is always a unique pair that contains both minimum and maximum value. Below are the steps: 

1. Iterate over the given array of pairs and find the minimum first pair and maximum second pair from all the pairs in arr[][].
2. After finding the maximum and minimum in the above step, there must exist any pair with arr[i][0] = minimum and arr[i][1] = maximum.
3. Iterate through every array of pairs and check if there exists a pair whose arr[i][0] is equal to the minimum and arr[i][1] is equal to the maximum.
4. If there exists any position in the above step, then print that position.
5. Otherwise, print -1.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the position of
// the pair that covers every pair
// in the array arr[][]
void position(int arr[][2], int N)
{
    // Position to store the index
    int pos = -1;

    // Stores the maximum second value
    int right = INT_MIN;

    // Stores the minimum first value
    int left = INT_MAX;

    // Iterate over the array of pairs
    for (int i = 0; i < N; i++) {

        // Update right maximum
        if (arr[i][1] > right) {
            right = arr[i][1];
        }

        // Update left minimum
        if (arr[i][0] < left) {

            left = arr[i][0];
        }
    }

    // Iterate over the array of pairs
    for (int i = 0; i < N; i++) {

        // If any pair exists with value
        // {left, right} then store it
        if (arr[i][0] == left
            && arr[i][1] == right) {

            pos = i + 1;
        }
    }

    // Print the answer
    cout << pos << endl;
}

// Driver Code
int main()
{
    // Given array of pairs
    int arr[][2] = {{ 3, 3 }, { 1, 3 },
                    { 2, 2 }, { 2, 3 },
                    { 1, 2 }};

    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    position(arr, N);
}

Java

// Java program for the
// above approach
import java.util.*;
class GFG{

// Function to find the position
// of the pair that covers
// every pair in the array arr[][]
static void position(int arr[][],
                     int N)
{
  // Position to store
  // the index
  int pos = -1;

  // Stores the maximum
  // second value
  int right = Integer.MIN_VALUE;

  // Stores the minimum
  // first value
  int left = Integer.MAX_VALUE;

  // Iterate over the array
  // of pairs
  for (int i = 0; i < N; i++)
  {
    // Update right maximum
    if (arr[i][1] > right)
    {
      right = arr[i][1];
    }

    // Update left minimum
    if (arr[i][0] < left)
    {
      left = arr[i][0];
    }
  }

  // Iterate over the array
  // of pairs
  for (int i = 0; i < N; i++)
  {
    // If any pair exists
    // with value {left,
    // right} then store it
    if (arr[i][0] == left &&
        arr[i][1] == right)
    {
      pos = i + 1;
    }
  }

  // Print the answer
  System.out.print(pos + "\n");
}

// Driver Code
public static void main(String[] args)
{
  // Given array of pairs
  int arr[][] = {{3, 3}, {1, 3},
                 {2, 2}, {2, 3},
                 {1, 2}};

  int N = arr.length;

  // Function Call
  position(arr, N);
}
}

// This code is contributed by Princi Singh

Python3

# Python3 program for the above approach
import sys

# Function to find the position of
# the pair that covers every pair
# in the array arr[][]
def position(arr, N):

    # Position to store the index
    pos = -1

    # Stores the minimum second value
    right = -sys.maxsize - 1

    # Stores the maximum first value
    left = sys.maxsize

    # Iterate over the array of pairs
    for i in range(N):

        # Update right maximum
        if (arr[i][1] > right):
            right = arr[i][1]

        # Update left minimum
        if (arr[i][0] < left):
            left = arr[i][0]

    # Iterate over the array of pairs
    for i in range(N):

        # If any pair exists with value
        # {left, right then store it
        if (arr[i][0] == left and
            arr[i][1] == right):
            pos = i + 1

    # Print the answer
    print(pos)

# Driver Code
if __name__ == '__main__':

    # Given array of pairs
    arr = [ [ 3, 3 ], [ 1, 3 ],
            [ 2, 2 ], [ 2, 3 ],
            [ 1, 2 ] ]

    N = len(arr)

    # Function call
    position(arr, N)

# This code is contributed by mohit kumar 29

C#

// C# program for the
// above approach
using System;
class GFG{

// Function to find the position
// of the pair that covers
// every pair in the array [,]arr
static void position(int [,]arr,
                     int N)
{
  // Position to store
  // the index
  int pos = -1;

  // Stores the maximum
  // second value
  int right = int.MinValue;

  // Stores the minimum
  // first value
  int left = int.MaxValue;

  // Iterate over the array
  // of pairs
  for (int i = 0; i < N; i++)
  {
    // Update right maximum
    if (arr[i, 1] > right)
    {
      right = arr[i, 1];
    }

    // Update left minimum
    if (arr[i, 0] < left)
    {
      left = arr[i, 0];
    }
  }

  // Iterate over the array
  // of pairs
  for (int i = 0; i < N; i++)
  {
    // If any pair exists
    // with value {left,
    // right} then store it
    if (arr[i, 0] == left &&
        arr[i, 1] == right)
    {
      pos = i + 1;
    }
  }

  // Print the answer
  Console.Write(pos + "\n");
}

// Driver Code
public static void Main(String[] args)
{
  // Given array of pairs
  int [,]arr = {{3, 3}, {1, 3},
                {2, 2}, {2, 3},
                {1, 2}};

  int N = arr.GetLength(0);

  // Function Call
  position(arr, N);
}
}

// This code is contributed by shikhasingrajput

Javascript

<script>

// JavaScript program for the above approach

// Function to find the position of
// the pair that covers every pair
// in the array arr[][]
function position(arr, N){

    // Position to store the index
    let pos = -1

    // Stores the minimum second value
    let right = Number.MIN_VALUE

    // Stores the maximum first value
    let left = Number.MAX_VALUE

    // Iterate over the array of pairs
    for(let i=0;i<N;i++){

        // Update right maximum
        if (arr[i][1] > right)
            right = arr[i][1]

        // Update left minimum
        if (arr[i][0] < left)
            left = arr[i][0]

    }

    // Iterate over the array of pairs
    for(let i=0;i<N;i++){

        // If any pair exists with value
        // {left, right then store it
        if (arr[i][0] == left && arr[i][1] == right)
            pos = i + 1

    }

    // Print the answer
    document.write(pos,"</br>")

}

// Driver Code


// Given array of pairs
let arr = [ [ 3, 3 ], [ 1, 3 ], [ 2, 2 ], [ 2, 3 ], [ 1, 2 ] ]

let N = arr.length

// Function call
position(arr, N)

// This code is contributed by shinjanpatra

</script>

Output

2

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