Dados N segmentos de línea donde la longitud del i-ésimo segmento de línea es 
. La tarea es encontrar el número máximo de paralelogramos que se pueden hacer con esos segmentos de línea si cada segmento de línea se usa como máximo en un paralelogramo.
Ejemplos:
Input: arr[] = {1, 2, 1, 2}
Output: 1
Only one parallelogram can be made with sides 1, 2, 1, 2
Input: arr[] = {1, 3, 5, 7, 1, 3, 5, 7, 1, 3, 5, 7}
Output: 2
Enfoque: Haga una array de frecuencias de la longitud de los segmentos de línea. Entonces la respuesta será el número total de paralelogramos que se pueden hacer con 4 lados iguales + paralelogramos que se pueden hacer con 2 lados iguales. Como un paralelogramo debe tener lados opuestos iguales.
C++
// Function to find the maximum number
// of parallelograms can be made
#include <bits/stdc++.h>
using namespace std;
void convert(int n, int a[])
{
// Finding the length of the frequency array
int z = a[0];
for (int i = 1; i < n; i++) {
if (a[i] > z)
z = a[i];
}
z = z + 1;
int ff[z] = { 0 };
for (int i = 0; i < n; i++) {
// Increasing the occurrence of each segment
ff[a[i]] += 1;
}
// To store the count of parallelograms
int cc = 0;
for (int i = 0; i < z; i++) {
// Counting parallelograms that can
// be made using 4 similar sides
cc += int(ff[i] / 4);
ff[i] = ff[i] % 4;
}
int vv = 0;
for (int i = 0; i < z; i++) {
// counting segments which have occurrence left >= 2
if (ff[i] >= 2)
vv += 1;
}
// Adding parallelograms that can be
// made using 2 similar sides
cc += int(vv / 2);
cout << (cc);
}
// Driver function
int main()
{
int n = 4;
int a[] = { 1, 2, 1, 2 };
convert(n, a);
}
// This code is contributed by
// Surendra_Gangwar
Java
// Function to find the maximum number
// of parallelograms can be made
import java.util.*;
class GFG
{
static void convert(int n, int a[])
{
// Finding the length of the frequency array
int z = a[0];
for (int i = 1; i < n; i++)
{
if (a[i] > z)
z = a[i];
}
z = z + 1;
int ff[] = new int[z];
for (int i = 0; i < n; i++)
{
// Increasing the occurrence of each segment
ff[a[i]] += 1;
}
// To store the count of parallelograms
int cc = 0;
for (int i = 0; i < z; i++)
{
// Counting parallelograms that can
// be made using 4 similar sides
cc += (ff[i] / 4);
ff[i] = ff[i] % 4;
}
int vv = 0;
for (int i = 0; i < z; i++)
{
// counting segments which have occurrence left >= 2
if (ff[i] >= 2)
vv += 1;
}
// Adding parallelograms that can be
// made using 2 similar sides
cc += (vv / 2);
System.out.println(cc);
}
// Driver code
public static void main(String[] args)
{
int n = 4;
int a[] = { 1, 2, 1, 2 };
convert(n, a);
}
}
/* This code is contributed by PrinciRaj1992 */
Python
# Function to find the maximum number # of parallelograms can be made def convert(n, a): # Finding the length of the frequency array z = max(a)+1 ff =[0]*z for i in range(n): # Increasing the occurrence of each segment ff[a[i]]+= 1 # To store the count of parallelograms cc = 0 for i in range(z): # Counting parallelograms that can # be made using 4 similar sides cc+= ff[i]//4 ff[i]= ff[i]% 4 vv = 0 for i in range(z): # counting segments which have occurrence left >= 2 if(ff[i]>= 2): vv+= 1 # Adding parallelograms that can be # made using 2 similar sides cc+= vv//2 print(cc) n = 4 a =[1, 2, 1, 2] convert(n, a)
C#
// Function to find the maximum number
// of parallelograms can be made
using System;
class GFG
{
static void convert(int n, int []a)
{
// Finding the length of the frequency array
int z = a[0];
for (int i = 1; i < n; i++)
{
if (a[i] > z)
z = a[i];
}
z = z + 1;
int []ff = new int[z];
for (int i = 0; i < n; i++)
{
// Increasing the occurrence of each segment
ff[a[i]] += 1;
}
// To store the count of parallelograms
int cc = 0;
for (int i = 0; i < z; i++)
{
// Counting parallelograms that can
// be made using 4 similar sides
cc += (ff[i] / 4);
ff[i] = ff[i] % 4;
}
int vv = 0;
for (int i = 0; i < z; i++)
{
// counting segments which have occurrence left >= 2
if (ff[i] >= 2)
vv += 1;
}
// Adding parallelograms that can be
// made using 2 similar sides
cc += (vv / 2);
Console.WriteLine(cc);
}
// Driver code
static public void Main ()
{
int n = 4;
int []a = { 1, 2, 1, 2 };
convert(n, a);
}
}
/* This code is contributed by ajit_23*/
Javascript
<script>
// Function to find the maximum number
// of parallelograms can be made
function convert(n, a)
{
// Finding the length of the frequency array
let z = a[0];
for (let i = 1; i < n; i++)
{
if (a[i] > z)
z = a[i];
}
z = z + 1;
let ff = new Array(z);
ff.fill(0);
for (let i = 0; i < n; i++)
{
// Increasing the occurrence of each segment
ff[a[i]] += 1;
}
// To store the count of parallelograms
let cc = 0;
for (let i = 0; i < z; i++)
{
// Counting parallelograms that can
// be made using 4 similar sides
cc += parseInt(ff[i] / 4, 10);
ff[i] = ff[i] % 4;
}
let vv = 0;
for (let i = 0; i < z; i++)
{
// counting segments which have occurrence left >= 2
if (ff[i] >= 2)
vv += 1;
}
// Adding parallelograms that can be
// made using 2 similar sides
cc += parseInt(vv / 2, 10);
document.write(cc);
}
let n = 4;
let a = [ 1, 2, 1, 2 ];
convert(n, a);
</script>
Producción:
1