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

Given an 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:

Initialize a variable count = 0 which stores the number pairs that lie between the current pair.

Traverse the array of pairs and for each pair, check if the count is equal to the total number of pairs or not.

If found to be true, it means that the pair can cover all other pairs. Print the pair.

Otherwise, set count = 0 and repeat the above steps for the next pair.

If no such pair exists, then print -1.

Below is the implementation of the above approach:

// 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 occurences 

    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); 

} 

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:

Iterate over the given array of pairs and find the minimum first pair and maximum second pair from all the pairs in arr[][]. 

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.

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.

If there exists any position in the above step, then print that position.

Otherwise, print -1.

Below is the implementation of the above approach:

// 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) 

{ 

    // Postion 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); 

} 

Output:

2

Time Complexity: O(N)

Auxiliary Space: O(1)