Skip to content
Related Articles

Related Articles

Count of pairs from arrays A and B such that element in A is greater than element in B at that index
  • Last Updated : 19 Jan, 2021

Given two arrays A[] and B[] of size N, the task is to count the maximum number of pairs, where each pair contains one from each array, such that A[i] > B[i]. Also the array A can be rearranged any number of times.

Examples:  

Input: A[] = {20, 30, 50}, B[]= {60, 40, 25} 
Output:
Explanation: 
Initially: 
A[0] = 20 < B[0] = 60 
A[1] = 30 < B[1] = 40 
A[2] = 50 > B[2] = 25 
Clearly, this arrangement has only 1 value such that A[i] > B[i]. 
This array A[] when rearranged to {20, 50, 30}: 
A[0] = 20 < B[0] = 60 
A[1] = 50 > B[1] = 40 
A[2] = 30 > B[2] = 25 
2 values follow the condition A[i] > B[i] which is the maximum for these set of arrays. 

Input: A[] = {10, 3, 7, 5, 8}, B[] = {8, 6, 2, 5, 9} 
Output:
Explanation: 
Initially: 
A[0] = 10 > B[0] = 8 
A[1] = 3 < B[1] = 6 
A[2] = 7 > B[2] = 2 
A[3] = 5 = B[3] = 5 
A[4] = 8 < B[4] = 9 
Clearly, this arrangement has only 2 values such that A[i] > B[i]. 
This array A[] when rearranged to {10, 8, 5, 7, 3}: 
A[0] = 10 > B[0] = 8 
A[1] = 8 > B[1] = 6 
A[2] = 5 > B[2] = 2 
A[3] = 7 > B[3] = 5 
A[4] = 3 < B[4] = 9 
4 values follow the condition A[i] > B[i] which is the maximum for these set of arrays. 
 

Approach: The idea is to use the concept of heap. Since the arrangement of B[] doesn’t matter in the question, we can perform max heap on both the arrays. After performing max heap and storing the values in two different heaps, iterate through the heap corresponding to A[] and B[] to count the number of indices satisfying the given condition A[i] > B[i]



Below is the implementation of the above approach:  

C++14




// C++ program to find the maximum count of
// values that follow the given condition
#include<bits/stdc++.h>
using namespace std;
 
// Function to find the maximum count of
// values that follow the given condition
int check(int A[], int B[], int N)
{
 
    // Initializing the max-heap for the array A[]
    priority_queue <int> pq1,pq2;
 
    // Adding the values of A[] into max heap
    for (int i = 0; i < N; i++) {
        pq1.push(A[i]);
    }
 
    // Adding the values of B[] into max heap
    for (int i = 0; i < N; i++) {
        pq2.push(B[i]);
    }
 
    // Counter variable
    int c = 0;
 
    // Loop to iterate through the heap
    for (int i = 0; i < N; i++) {
 
        // Comparing the values at the top.
        // If the value of heap A[] is greater,
        // then counter is incremented
        if (pq1.top()>pq2.top()) {
            c++;
            pq1.pop();
            pq2.pop();
        }
        else {
            if (pq2.size() == 0) {
                break;
            }
            pq2.pop();
        }
    }
    return (c);
}
 
// Driver code
int main()
{
    int A[] = { 10, 3, 7, 5, 8 };
    int B[] = { 8, 6, 2, 5, 9 };
    int N = sizeof(A)/sizeof(A[0]);
 
    cout<<(check(A, B, N));
}
 
// This code is contributed by mohit kumar 29   

Java




// Java program to find the maximum count of
// values that follow the given condition
 
import java.util.*;
public class GFG {
 
    // Function to find the maximum count of
    // values that follow the given condition
    static int check(int A[], int B[], int N)
    {
 
        // Initializing the max-heap for the array A[]
        PriorityQueue<Integer> pq1
            = new PriorityQueue<Integer>(
Collections.reverseOrder());
 
        // Initializing the max-heap for the array B[]
        PriorityQueue<Integer> pq2
            = new PriorityQueue<Integer>(
Collections.reverseOrder());
 
        // Adding the values of A[] into max heap
        for (int i = 0; i < N; i++) {
            pq1.add(A[i]);
        }
 
        // Adding the values of B[] into max heap
        for (int i = 0; i < N; i++) {
            pq2.add(B[i]);
        }
 
        // Counter variable
        int c = 0;
 
        // Loop to iterate through the heap
        for (int i = 0; i < N; i++) {
 
            // Comparing the values at the top.
            // If the value of heap A[] is greater,
            // then counter is incremented
            if (pq1.peek().compareTo(pq2.peek()) == 1) {
                c++;
                pq1.poll();
                pq2.poll();
            }
            else {
                if (pq2.size() == 0) {
                    break;
                }
                pq2.poll();
            }
        }
        return (c);
    }
 
    // Driver code
    public static void main(String args[])
    {
        int A[] = { 10, 3, 7, 5, 8 };
        int B[] = { 8, 6, 2, 5, 9 };
        int N = A.length;
 
        System.out.println(check(A, B, N));
    }
}

Python3




# Python3 program to find the maximum count of
# values that follow the given condition
import heapq
 
# Function to find the maximum count of
# values that follow the given condition
def check(A, B,N):
 
    # Initializing the max-heap for the array A[]
    pq1 = []
    pq2 = []
 
    # Adding the values of A[] into max heap
    for i in range(N):
        heapq.heappush(pq1,-A[i])
 
    # Adding the values of B[] into max heap
    for i in range(N):
        heapq.heappush(pq2,-B[i])
 
    # Counter variable
    c = 0
 
    # Loop to iterate through the heap
    for i in range(N):
 
        # Comparing the values at the top.
        # If the value of heap A[] is greater,
        # then counter is incremented
        if -pq1[0] > -pq2[0]:
            c += 1
            heapq.heappop(pq1)
            heapq.heappop(pq2)
 
        else:
            if len(pq2) == 0:
                break
            heapq.heappop(pq2)
    return (c)
 
# Driver code
A = [ 10, 3, 7, 5, 8 ]
B = [ 8, 6, 2, 5, 9 ]
N = len(A)
 
print(check(A, B, N))
 
# This code is contributed by apurva raj

C#




// C# program to find the maximum count of
// values that follow the given condition
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find the maximum count of
// values that follow the given condition
static int check(int[] A, int[] B, int N)
{
     
    // Initializing the max-heap for the array A[]
    List<int> pq1 = new List<int>();
     
    // Initializing the max-heap for the array B[]
    List<int> pq2 = new List<int>();
     
    // Adding the values of A[] into max heap
    for(int i = 0; i < N; i++)
    {
        pq1.Add(A[i]);
    }
     
    // Adding the values of B[] into max heap
    for(int i = 0; i < N; i++)
    {
        pq2.Add(B[i]);
    }
    pq1.Sort();
    pq1.Reverse();
    pq2.Sort();
    pq2.Reverse();
     
    // Counter variable
    int c = 0;
     
    // Loop to iterate through the heap
    for(int i = 0; i < N; i++)
    {
         
        // Comparing the values at the top.
        // If the value of heap A[] is greater,
        // then counter is incremented
        if (pq1[0] > pq2[0])
        {
            c++;
            pq1.RemoveAt(0);
            pq2.RemoveAt(0);
        }
        else
        {
            if (pq2.Count == 0)
            {
                break;
            }
            pq2.RemoveAt(0);
        }
    }
    return c;
}
 
// Driver code
static public void Main()
{
    int[] A = { 10, 3, 7, 5, 8 };
    int[] B = { 8, 6, 2, 5, 9 };
    int N = A.Length;
     
    Console.WriteLine(check(A, B, N));
}
}
 
// This code is contributed by avanitrachhadiya2155
Output: 
4

 

Time Complexity: O(N * log(N))
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :