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# 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 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 Codeint 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 approachimport 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 Codepublic 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 arrdef 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 Codeif __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 approachusing 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 Codepublic 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



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 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 Codeint 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 approachimport 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 Codepublic 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 approachimport 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 Codeif __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 approachusing System;class GFG{ // Function to find the position// of the pair that covers// every pair in the array [,]arrstatic 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 Codepublic 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



Output

2

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