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: 2
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: 4
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 conditionint 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 codeint 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 conditionimport 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 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[0] > -pq2[0]: 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 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 |
Javascript
<script>// Javascript program to find the maximum count of// values that follow the given condition// Function to find the maximum count of// values that follow the given conditionfunction check(A,B,N){ let pq1=[]; let pq2=[]; // Adding the values of A[] into max heap for (let i = 0; i < N; i++) { pq1.push(A[i]); } // Adding the values of B[] into max heap for (let i = 0; i < N; i++) { pq2.push(B[i]); } pq1.sort(function(a,b){return a-b;}); pq1.reverse(); pq2.sort(function(a,b){return a-b;}); pq2.reverse(); // Counter variable let c = 0; // Loop to iterate through the heap for (let 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.shift(); pq2.shift(); } else { if (pq2.length == 0) { break; } pq2.shift(); } } return (c);}// Driver codelet A=[ 10, 3, 7, 5, 8];let B=[8, 6, 2, 5, 9 ];let N = A.length;document.write(check(A, B, N));// This code is contributed by patel2127</script> |
4
Time Complexity: O(N * log(N))
Auxiliary Space: O(N)
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