Given an array arr[] of size N. The task is to reorder arr[] and find the maximum number of GCD pairs which follows the conditions given below.
- Choose any two elements of array Ai and Aj of array where 0 <= i < j < N.
- Calculate GCD after multiplying Aj with 2 like (Ai, 2 * Aj) which is greater than 1.
Examples
Input: arr[] = { 3, 6 . 5, 3}
Output: 4
Explanation: Reorder array like this : { 6, 5, 3, 3 } and below are the pairs formed from arr[].
P1 = GCD( 6, 5 * 2) => (6, 10) => 2 > 1
P2 = GCD( 6, 3 * 2) => (6, 6) => 6 > 1
P3 = GCD( 6, 3 * 2) => (6, 6) => 6 > 1
P4 = GCD( 3, 3 * 2) => (3, 6) => 3 > 1
Input: arr[] = { 1, 7 }
Output: 0
Explanation: If array is order like { 7, 1 } no pair can be formed
GCD(7, 1 * 2) = > (7, 2 ), GCD(1, 7 * 2) => (1, 14) == 1
Approach: If we observe that if even elements are in starting position then the pairs of (GCD > 1) is maximum because there is a condition to multiply the arr[j] * 2, and the ordering of odd elements does not matter its number of pair always same. Follow the steps below to solve the given problem.
- Initialize the variable idx with value 0.
- Traverse the array.
- If arr[i] is even then swap with arr[idx] and increment idx by 1.
- After traversing all the elements.
- Initialize the variable ans with 0.
- Use two loops first with i and second with j = i + 1.
- Now if gcd(arr[i], 2 * arr[j] * 2) > 1 increment the ans by 1.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
int maximumpairs(int arr[], int n)
{
int idx = 0;
for (int i = 0; i < n; i++) {
if (arr[i] % 2 == 0) {
swap(arr[i], arr[idx]);
idx++;
}
}
int ans = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (__gcd(arr[i], 2 * arr[j]) > 1) {
ans++;
}
}
}
return ans;
}
int main()
{
int arr[] = { 5, 3, 6, 3 };
int N = sizeof(arr) / sizeof(int);
int ans = maximumpairs(arr, N);
cout << ans;
return 0;
}
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Java
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static int maximumpairs(int arr[], int n)
{
int idx = 0;
for (int i = 0; i < n; i++) {
if (arr[i] % 2 == 0) {
int temp = arr[i];
arr[i] = arr[idx];
arr[idx] = temp;
idx++;
}
}
int ans = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
int a = arr[i];
int b = 2 * arr[j];
if (gcd(a, b) > 1) {
ans++;
}
}
}
return ans;
}
public static void main (String[] args)
{
int arr[] = { 5, 3, 6, 3 };
int N = arr.length;
int ans = maximumpairs(arr, N);
System.out.println(ans);
}
}
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Python3
def gcd(a, b):
if(b == 0):
return a
else:
return gcd(b, a % b)
def maximumpairs(arr,n):
idx = 0
for i in range(0, n):
if (arr[i] % 2 == 0):
arr[i], arr[idx] = arr[idx], arr[i]
idx = idx + 1
ans = 0
for i in range(0,n):
for j in range(i + 1,n):
if (gcd(arr[i], 2*arr[j]) > 1):
ans = ans + 1
return ans
arr = [ 5, 3, 6, 3 ]
N = len(arr)
ans = maximumpairs(arr, N)
print(ans)
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C#
using System;
class GFG {
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static int maximumpairs(int []arr, int n)
{
int idx = 0;
for (int i = 0; i < n; i++) {
if (arr[i] % 2 == 0) {
int temp = arr[i];
arr[i] = arr[idx];
arr[idx] = temp;
idx++;
}
}
int ans = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
int a = arr[i];
int b = 2 * arr[j];
if (gcd(a, b) > 1) {
ans++;
}
}
}
return ans;
}
public static void Main ()
{
int []arr = { 5, 3, 6, 3 };
int N = arr.Length;
int ans = maximumpairs(arr, N);
Console.Write(ans);
}
}
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Javascript
<script>
function gcd(a, b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
function maximumpairs(arr, n)
{
let idx = 0;
for (let i = 0; i < n; i++) {
if (arr[i] % 2 == 0) {
let temp = arr[i];
arr[i] = arr[idx];
arr[idx] = temp;
idx++;
}
}
let ans = 0;
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
let a = arr[i];
let b = 2 * arr[j];
if (gcd(a, b) > 1) {
ans++;
}
}
}
return ans;
}
let arr = [ 5, 3, 6, 3 ];
let N = arr.length;
let ans = maximumpairs(arr, N);
document.write(ans);
</script>
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Time Complexity: O(N * N * logN)
Auxiliary Space: O(1)