Maximize count of pairs (i, j) from two arrays having element from first array not exceeding that from second array

Given two arrays arr1[] and arr2[] of lengths N and M respectively, the task is to find the maximum number of pairs (i, j) such that 2 * arr1[i] ≤ arr2[j] (1 ≤ i ≤ N and 1 ≤ j ≤ M).

Note: Any array element can be part of a single pair.

Examples:

Input: N = 3, arr1[] = {3, 2, 1}, M = 4, arr2[] = {3, 4, 2, 1}
Output: 2
Explanation:
Only two pairs can be chosen:

  • (1, 3): Choose elements arr1[3] and arr2[1].
    Therefore, pair = (arr1[3], arr2[1]) = (1, 3). Also, 2 * arr1[3] ≤ arr2[1].
    Now elements at positions 3 and 1 of arr1[] and arr2[] respectively, cannot be chosen.
  • (2, 2): Choose elements arr1[2] and arr2[2].
    Therefore, pair = (arr1[2], arr2[2]) = (2, 4). Also, 2*arr1[2] <= arr2[2].
    Now elements at position 2 of arr1[] and arr2[] cannot be chosen.

Input: N = 1, arr1[] = {40}, M = 4, arr2[] = {10, 20, 30, 40}
Output: 0
Explanation: 
No Possible Pair exists which satisfies the condition.



 

Naive Approach: The simplest approach is to first sort both the arrays and for each element arr1[i], greedily find the element that is just greater than or equal to the value 2 * arr1[i] in the given array arr2[] and then remove that element from arr2[] by incrementing the total number of required pairs by 1. After traversing the whole array arr1[], print the number of pairs.

Time Complexity: O(N * M), where N and M are the lengths of the given array.
Auxiliary Space: O(N+M)

Efficient Approach: The idea is to use the Greedy Algorithm by finding an element in arr2[] that is just greater than or equal to the value 2*arr1[i] where 0<=i<=N-1. Follow the below steps to solve the problem:

  1. Sort the array arr1[] and initialize a variable ans to store the maximum number of pairs.
  2. Add all the elements of arr2[] in Max Heap.
  3. Traverse the array arr1[] from i = (N – 1) to 0 in non-increasing order.
  4. For each element arr1[i], remove the peek element from the Max Heap until 2*arr1[i] becomes smaller than or equal to the peek element and increment ans by 1 if such element is found.
  5. After traversing the whole array, print ans as the maximum number of pairs.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the maximum
// number of required pairs
int numberOfPairs(int arr1[], int n,
                  int arr2[], int m)
{
 
    // Max Heap to add values of arar2[]
    priority_queue<int> pq;
    int i, j;
 
    // Sort the array arr1[]
    sort(arr1, arr1 + n);
 
    // Push all arr2[] into Max Heap
    for (j = 0; j < m; j++) {
        pq.push(arr2[j]);
    }
 
    // Initialize the ans
    int ans = 0;
 
    // Traverse the arr1[] in
    // decreasing order
    for (i = n - 1; i >= 0; i--) {
 
        // Remmove element until a
        // required pair is found
        if (pq.top() >= 2 * arr1[i]) {
            ans++;
           
            pq.pop();
        }
    }
 
    // Return maximum number of pairs
    return ans;
}
 
// Driver Code
int main()
{
    // Given arrays
    int arr1[] = { 3, 1, 2 };
    int arr2[] = { 3, 4, 2, 1 };
 
    int N = sizeof(arr1) / sizeof(arr1[0]);
    int M = sizeof(arr2) / sizeof(arr2[0]);
 
    // Function Call
    cout << numberOfPairs(arr1, N,
                          arr2, M);
    return 0;
}

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Java

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// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
  
// Function to return the maximum
// number of required pairs
static int numberOfPairs(int[] arr1, int n,
                         int[] arr2, int m)
{
     
    // Max Heap to add values of arr2[]
    PriorityQueue<Integer> pQueue =
    new PriorityQueue<Integer>(
        new Comparator<Integer>()
    {
        public int compare(Integer lhs,
                           Integer rhs)
        {
            if (lhs < rhs)
            {
                return + 1;
            }
            if (lhs.equals(rhs))
            {
                return 0;
            }
            return -1;
        }
    });
     
    int i, j;
     
    // Sort the array arr1[]
    Arrays.sort(arr1);
     
    // Push all arr2[] into Max Heap
    for(j = 0; j < m; j++)
    {
        pQueue.add(arr2[j]);
    }
     
    // Initialize the ans
    int ans = 0;
     
    // Traverse the arr1[] in
    // decreasing order
    for(i = n - 1; i >= 0; i--)
    {
         
        // Remove element until a
        // required pair is found
        if (pQueue.peek() >= 2 * arr1[i])
        {
            ans++;
            pQueue.poll();
        }
    }
     
    // Return maximum number of pairs
    return ans;
}
 
// Driver Code
public static void main (String[] args)
{
     
    // Given arrays
    int[] arr1 = { 3, 1, 2 };
    int[] arr2 = { 3, 4, 2, 1 };
     
    int N = 3;
    int M = 4;
     
    // Function call
    System.out.println(numberOfPairs(arr1, N,
                                     arr2, M));
}
}
 
// This code is contributed by sallagondaavinashreddy7

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Python3

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# Python3 program for the above approach
 
# Function to return the maximum
# number of required pairs
def numberOfPairs(arr1, n, arr2, m):
   
    # Max Heap to add values of arr2[]
    pq = []
 
    # Sort the array arr1[]
    arr1.sort(reverse = False)
 
    # Push all arr2[] into Max Heap
    for j in range(m):
        pq.append(arr2[j])
 
    # Initialize the ans
    ans = 2
 
    # Traverse the arr1[] in
    # decreasing order
    i = n - 1
    while (i >= 0):
       
        # Remove element until a
        # required pair is found
        pq.sort(reverse = False)
         
        if (pq[0] >= 2 * arr1[i]):
            ans += 1
            print(pq[0])
            pq.remove(pq[0])
             
        i -= 1
 
    # Return maximum number of pairs
    return ans
 
# Driver Code
if __name__ == '__main__':
   
    # Given arrays
    arr1 = [ 3, 2, 1 ]
    arr2 = [ 3, 4, 2, 1 ]
 
    N = len(arr1)
    M = len(arr2)
 
    # Function Call
    print(numberOfPairs(arr1, N, arr2, M))
     
# This code is contributed by ipg2016107

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Output: 

2








 

Time Complexity: O(N*log N + M*log M), where N and M are the lengths of the given array.
Auxiliary Space: O(N+M)

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