Given two given arrays A and B of equal length, the task is to find the maximum number of distinct pairs of elements that can be chosen such that the element from A is strictly greater than the element from B.
Examples:
Input:
A[]={20, 30, 50} , B[]={25, 60, 40}
Output: 2
Explanation:
(30, 25) and (50, 40) are the two possible pairs.Input:
A[]={20, 25, 60} , B[]={25, 60, 40}
Output: 1
Explanation:
(60, 25) or (60, 40) can be the required pair.
Approach: In order to solve this problem, we need to adopt the following approach:
- Sort both the arrays.
- Traverse the entire length of array A and check if the current element in A is greater than the element in B currently pointed at B.
- If so, point to the next element in both the arrays. Otherwise, move to the next element in A and check if it is greater than the element currently pointed at B.
- The number of elements traversed in array B after the entire traversal of array A gives the required answer.
Illustration:
Let us consider the two following arrays:
A[] = {30, 28, 45, 22} , B[] = {35, 25, 22, 48}
After sorting, the arrays appear to be
A[] = {22, 28, 30, 45} , B[] = {22, 25, 35, 48}
After the first iteration, since A[0] is not greater than B[0], we move to A[1].
After the second iteration, we move to B[1] as A[1] is greater than B[0].
After the third iteration, we move to B[2] as A[2] is greater than B[1].
Similarly, A[3] is greater than B[2] and we move to B[3].
Hence, the number of elements traversed in B, i.e. 3, is the final answer.
The possible pairs are (28,22), (30,25), (45, 35).
Below is the implementation of the above approach:
C++
// C++ Program to count number of distinct // pairs possible from the two arrays// such that element selected from one array is // always greater than the one selected from // the other array#include <bits/stdc++.h>using namespace std;// Function to return the count // of pairsint countPairs(vector<int> A, vector<int> B){ int n = A.size(); sort(A.begin(),A.end()); sort(B.begin(),B.end()); int ans = 0, i; for (int i = 0; i < n; i++) { if (A[i] > B[ans]) { ans++; } } return ans;}// Driver codeint main(){ vector<int> A = { 30, 28, 45, 22 }; vector<int> B = { 35, 25, 22, 48 }; cout << countPairs(A,B); return 0;} |
Java
// Java program to count number of distinct // pairs possible from the two arrays such // that element selected from one array is // always greater than the one selected from // the other arrayimport java.util.*;class GFG{// Function to return the count // of pairsstatic int countPairs(int [] A, int [] B){ int n = A.length; int ans = 0; Arrays.sort(A); Arrays.sort(B); for(int i = 0; i < n; i++) { if (A[i] > B[ans]) { ans++; } } return ans;}// Driver codepublic static void main(String[] args){ int [] A = { 30, 28, 45, 22 }; int [] B = { 35, 25, 22, 48 }; System.out.print(countPairs(A, B));}}// This code is contributed by sapnasingh4991 |
Python3
# Python3 program to count number of distinct# pairs possible from the two arrays# such that element selected from one array is# always greater than the one selected from# the other array# Function to return the count# of pairsdef countPairs(A, B): n = len(A) A.sort() B.sort() ans = 0 for i in range(n): if(A[i] > B[ans]): ans += 1 return ans# Driver codeif __name__ == '__main__': A = [30, 28, 45, 22] B = [35, 25, 22, 48] print(countPairs(A, B))# This code is contributed by Shivam Singh |
C#
// C# program to count number of distinct // pairs possible from the two arrays such // that element selected from one array is // always greater than the one selected from // the other arrayusing System;class GFG{// Function to return the count // of pairsstatic int countPairs(int [] A, int [] B){ int n = A.Length; int ans = 0; Array.Sort(A); Array.Sort(B); for(int i = 0; i < n; i++) { if (A[i] > B[ans]) { ans++; } } return ans;}// Driver codepublic static void Main(){ int []A = { 30, 28, 45, 22 }; int []B = { 35, 25, 22, 48 }; Console.Write(countPairs(A, B));}}// This code is contributed by Code_Mech |
Javascript
<script>// Javascript program to count number of distinct // pairs possible from the two arrays such // that element selected from one array is // always greater than the one selected from // the other array// Function to return the count // of pairsfunction countPairs(A, B){ let n = A.length; let ans = 0; A.sort(); B.sort(); for(let i = 0; i < n; i++) { if (A[i] > B[ans]) { ans++; } } return ans;}// Driver Code let A = [ 30, 28, 45, 22 ]; let B = [ 35, 25, 22, 48 ]; document.write(countPairs(A, B)); </script> |
3
Time Complexity: O(n * log n)
Auxiliary Space: O(1)
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