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:
- Sort the array arr1[] and initialize a variable ans to store the maximum number of pairs.
- Add all the elements of arr2[] in Max Heap.
- Traverse the array arr1[] from i = (N – 1) to 0 in non-increasing order.
- 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.
- After traversing the whole array, print ans as the maximum number of pairs.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int numberOfPairs( int arr1[], int n,
int arr2[], int m)
{
priority_queue< int > pq;
int i, j;
sort(arr1, arr1 + n);
for (j = 0; j < m; j++) {
pq.push(arr2[j]);
}
int ans = 0;
for (i = n - 1; i >= 0; i--) {
if (pq.top() >= 2 * arr1[i]) {
ans++;
pq.pop();
}
}
return ans;
}
int main()
{
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]);
cout << numberOfPairs(arr1, N,
arr2, M);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG{
static int numberOfPairs( int [] arr1, int n,
int [] arr2, int m)
{
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;
Arrays.sort(arr1);
for (j = 0 ; j < m; j++)
{
pQueue.add(arr2[j]);
}
int ans = 0 ;
for (i = n - 1 ; i >= 0 ; i--)
{
if (pQueue.peek() >= 2 * arr1[i])
{
ans++;
pQueue.poll();
}
}
return ans;
}
public static void main (String[] args)
{
int [] arr1 = { 3 , 1 , 2 };
int [] arr2 = { 3 , 4 , 2 , 1 };
int N = 3 ;
int M = 4 ;
System.out.println(numberOfPairs(arr1, N,
arr2, M));
}
}
|
Python3
def numberOfPairs(arr1, n, arr2, m):
pq = []
arr1.sort(reverse = False )
for j in range (m):
pq.append(arr2[j])
ans = 2
i = n - 1
while (i > = 0 ):
pq.sort(reverse = False )
if (pq[ 0 ] > = 2 * arr1[i]):
ans + = 1
print (pq[ 0 ])
pq.remove(pq[ 0 ])
i - = 1
return ans
if __name__ = = '__main__' :
arr1 = [ 3 , 2 , 1 ]
arr2 = [ 3 , 4 , 2 , 1 ]
N = len (arr1)
M = len (arr2)
print (numberOfPairs(arr1, N, arr2, M))
|
C#
using System;
using System.Collections.Generic;
class Program
{
static int numberOfPairs( int [] arr1, int n, int [] arr2, int m) {
List< int > pq = new List< int >();
Array.Sort(arr1);
for ( int j = 0; j < m; j++) {
pq.Add(arr2[j]);
}
int ans = 2;
int i = n - 1;
while (i >= 0) {
pq.Sort();
if (pq[0] >= 2 * arr1[i]) {
ans += 1;
Console.WriteLine(pq[0]);
pq.Remove(pq[0]);
}
i -= 1;
}
return ans;
}
static void Main( string [] args) {
int [] arr1 = new int [] { 3, 2, 1 };
int [] arr2 = new int [] { 3, 4, 2, 1 };
int N = arr1.Length;
int M = arr2.Length;
Console.WriteLine(numberOfPairs(arr1, N, arr2, M));
}
}
|
Javascript
<script>
function numberOfPairs(arr1, n, arr2, m){
let pq = []
arr1.sort((a,b)=>a-b)
for (let j=0;j<m;j++)
pq.push(arr2[j])
let ans = 2
let i = n - 1
while (i >= 0){
pq.sort((a,b)=>a-b)
if (pq[0] >= 2 * arr1[i]){
ans += 1
pq.shift()
}
i -= 1
}
return ans
}
let arr1 = [ 3, 2, 1 ]
let arr2 = [ 3, 4, 2, 1 ]
let N = arr1.length
let M = arr2.length
document.write(numberOfPairs(arr1, N, arr2, M))
</script>
|
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)
Last Updated :
20 Feb, 2023
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