Count of pairs from arrays A and B such that element in A is greater than element in B at that index

• Difficulty Level : Hard
• Last Updated : 15 Jun, 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 = 20 < B = 60
A = 30 < B = 40
A = 50 > B = 25
Clearly, this arrangement has only 1 value such that A[i] > B[i].
This array A[] when rearranged to {20, 50, 30}:
A = 20 < B = 60
A = 50 > B = 40
A = 30 > B = 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 = 10 > B = 8
A = 3 < B = 6
A = 7 > B = 2
A = 5 = B = 5
A = 8 < B = 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 = 10 > B = 8
A = 8 > B = 6
A = 5 > B = 2
A = 7 > B = 5
A = 3 < B = 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#includeusing namespace std; // Function to find the maximum count of// values that follow the given conditionint check(int A[], int B[], int N){     // Initializing the max-heap for the array A[]    priority_queue 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 codeint main(){    int A[] = { 10, 3, 7, 5, 8 };    int B[] = { 8, 6, 2, 5, 9 };    int N = sizeof(A)/sizeof(A);     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 pq1            = new PriorityQueue(Collections.reverseOrder());         // Initializing the max-heap for the array B[]        PriorityQueue pq2            = new PriorityQueue(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 conditionimport heapq # Function to find the maximum count of# values that follow the given conditiondef 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 > -pq2:            c += 1            heapq.heappop(pq1)            heapq.heappop(pq2)         else:            if len(pq2) == 0:                break            heapq.heappop(pq2)    return (c) # Driver codeA = [ 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 conditionusing System;using System.Collections.Generic; class GFG{     // Function to find the maximum count of// values that follow the given conditionstatic int check(int[] A, int[] B, int N){         // Initializing the max-heap for the array A[]    List pq1 = new List();         // Initializing the max-heap for the array B[]    List pq2 = new List();         // 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 > pq2)        {            c++;            pq1.RemoveAt(0);            pq2.RemoveAt(0);        }        else        {            if (pq2.Count == 0)            {                break;            }            pq2.RemoveAt(0);        }    }    return c;} // Driver codestatic 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

Javascript


Output:
4

Time Complexity: O(N * log(N))

My Personal Notes arrow_drop_up