Dados N números, encuentre el número de permutaciones en las que la suma de los elementos en el índice impar y la suma de los elementos en el índice par son iguales.
Ejemplos:
Entrada: 1 2 3
Salida: 2
Las permutaciones son:
1 3 2 suma en índice impar = 1+2 = 3, suma en índice par = 3
2 3 1 suma en índice impar = 2+1 = 3, suma en índice par = 3
Entrada: 1 2 1 2
Salida: 3
Las permutaciones son:
1 2 2 1
2 1 1 2
2 2 1 1
El enfoque del problema será usar next_permutation() en C++ STL, que ayuda a generar todas las permutaciones posibles de N números. Si la suma de los elementos del índice impar es igual a la suma de los elementos del índice par de la permutación generada, aumente el recuento. Cuando todas las permutaciones estén marcadas, imprima el conteo.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find number of permutations
// such that sum of elements at odd index
// and even index are equal
#include <bits/stdc++.h>
using namespace std;
// Function that returns the number of permutations
int numberOfPermutations(int a[], int n)
{
int sumEven, sumOdd, c = 0;
// iterate for all permutations
do {
// stores the sum of odd and even index elements
sumEven = sumOdd = 0;
// iterate for elements in permutation
for (int i = 0; i < n; i++) {
// if odd index
if (i % 2)
sumOdd += a[i];
else
sumEven += a[i];
}
// If condition holds
if (sumOdd == sumEven)
c++;
} while (next_permutation(a, a + n));
// return the number of permutations
return c;
}
// Driver Code
int main()
{
int a[] = { 1, 2, 3 };
int n = sizeof(a) / sizeof(a[0]);
// Calling Function
cout << numberOfPermutations(a, n);
return 0;
}
Java
// Java program to find number of permutations
// such that sum of elements at odd index
// and even index are equal
class GFG {
// Function that returns the number of permutations
static int numberOfPermutations(int a[], int n) {
int sumEven, sumOdd, c = 0;
// iterate for all permutations
do {
// stores the sum of odd and even index elements
sumEven = sumOdd = 0;
// iterate for elements in permutation
for (int i = 0; i < n; i++) {
// if odd index
if (i % 2 == 0) {
sumOdd += a[i];
} else {
sumEven += a[i];
}
}
// If condition holds
if (sumOdd == sumEven) {
c++;
}
} while (next_permutation(a));
// return the number of permutations
return c;
}
static boolean next_permutation(int[] p) {
for (int a = p.length - 2; a >= 0; --a) {
if (p[a] < p[a + 1]) {
for (int b = p.length - 1;; --b) {
if (p[b] > p[a]) {
int t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.length - 1; a < b; ++a, --b) {
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true;
}
}
}
}
return false;
}
// Driver Code
public static void main(String args[]) {
int a[] = {1, 2, 3};
int n = a.length;
System.out.println(numberOfPermutations(a, n));
}
}
/*This code is contributed by 29AjayKumar*/
Python3
# Python3 program to find number of permutations # such that sum of elements at odd index # and even index are equal def next_permutation(arr): arrCount = len(arr); # the head of the suffix i = arrCount - 1; # find longest suffix while (i > 0 and arr[i] <= arr[i - 1]): i-=1; # are we at the last permutation already? if (i <= 0): return [False,arr]; # get the pivot pivotIndex = i - 1; # find rightmost element that exceeds the pivot j = arrCount - 1; while (arr[j] <= arr[pivotIndex]): j-=1; # swap the pivot with j temp = arr[pivotIndex]; arr[pivotIndex] = arr[j]; arr[j] = temp; # reverse the suffix j = arrCount - 1; while (i < j): temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; i+=1; j-=1; return [True,arr]; # Function that returns the number # of permutations def numberOfPermutations(a, n): sumEven=0; sumOdd=0; c = 0; # iterate for all permutations while (True): # stores the sum of odd and # even index elements sumEven = 0; sumOdd = 0; # iterate for elements in permutation for i in range(n): # if odd index if (i % 2): sumOdd += a[i]; else: sumEven += a[i]; # If condition holds if (sumOdd == sumEven): c+=1; xx=next_permutation(a); if(xx[0]==False): break; a=xx[1]; # return the number of permutations return c; # Driver Code a = [1, 2, 3]; n = len(a); # Calling Function print(numberOfPermutations(a, n)); # This code is contributed by mits
C#
// C# program to find number of permutations
// such that sum of elements at odd index
// and even index are equal
using System;
public class GFG {
// Function that returns the number of permutations
static int numberOfPermutations(int []a, int n) {
int sumEven, sumOdd, c = 0;
// iterate for all permutations
do {
// stores the sum of odd and even index elements
sumEven = sumOdd = 0;
// iterate for elements in permutation
for (int i = 0; i < n; i++) {
// if odd index
if (i % 2 == 0) {
sumOdd += a[i];
} else {
sumEven += a[i];
}
}
// If condition holds
if (sumOdd == sumEven) {
c++;
}
} while (next_permutation(a));
// return the number of permutations
return c;
}
static bool next_permutation(int[] p) {
for (int a = p.Length - 2; a >= 0; --a) {
if (p[a] < p[a + 1]) {
for (int b = p.Length - 1;; --b) {
if (p[b] > p[a]) {
int t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.Length - 1; a < b; ++a, --b) {
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true;
}
}
}
}
return false;
}
// Driver Code
public static void Main() {
int []a = {1, 2, 3};
int n = a.Length;
Console.WriteLine(numberOfPermutations(a, n));
}
}
/*This code is contributed by 29AjayKumar*/
PHP
<?php
// PHP program to find number of permutations
// such that sum of elements at odd index
// and even index are equal
function next_permutation(&$input)
{
$inputCount = count($input);
// the head of the suffix
$i = $inputCount - 1;
// find longest suffix
while ($i > 0 && $input[$i] <= $input[$i - 1])
{
$i--;
}
// are we at the last permutation already?
if ($i <= 0)
{
return false;
}
// get the pivot
$pivotIndex = $i - 1;
// find rightmost element that exceeds the pivot
$j = $inputCount - 1;
while ($input[$j] <= $input[$pivotIndex])
{
$j--;
}
// swap the pivot with j
$temp = $input[$pivotIndex];
$input[$pivotIndex] = $input[$j];
$input[$j] = $temp;
// reverse the suffix
$j = $inputCount - 1;
while ($i < $j)
{
$temp = $input[$i];
$input[$i] = $input[$j];
$input[$j] = $temp;
$i++;
$j--;
}
return true;
}
// Function that returns the number
// of permutations
function numberOfPermutations($a, $n)
{
$sumEven;
$sumOdd;
$c = 0;
// iterate for all permutations
do {
// stores the sum of odd and
// even index elements
$sumEven = $sumOdd = 0;
// iterate for elements in permutation
for ($i = 0; $i < $n; $i++)
{
// if odd index
if ($i % 2)
$sumOdd += $a[$i];
else
$sumEven += $a[$i];
}
// If condition holds
if ($sumOdd == $sumEven)
$c++;
} while (next_permutation($a));
// return the number of permutations
return $c;
}
// Driver Code
$a = array(1, 2, 3);
$n = count($a);
// Calling Function
echo numberOfPermutations($a, $n);
// This code is contributed by
// Rajput-Ji
?>
Javascript
<script>
// javascript program to find number of permutations
// such that sum of elements at odd index
// and even index are equal // Function that returns the number of permutations
function numberOfPermutations( a , n) {
var sumEven, sumOdd, c = 0;
// iterate for all permutations
do {
// stores the sum of odd and even index elements
sumEven = sumOdd = 0;
// iterate for elements in permutation
for (var i = 0; i < n; i++) {
// if odd index
if (i % 2 == 0) {
sumOdd += a[i];
} else {
sumEven += a[i];
}
}
// If condition holds
if (sumOdd == sumEven) {
c++;
}
} while (next_permutation(a));
// return the number of permutations
return c;
}
function next_permutation(p) {
for (var a = p.length - 2; a >= 0; --a) {
if (p[a] < p[a + 1]) {
for (var b = p.length - 1;; --b) {
if (p[b] > p[a]) {
var t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.length - 1; a < b; ++a, --b) {
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true;
}
}
}
}
return false;
}
// Driver Code
var a = [ 1, 2, 3 ];
var n = a.length;
document.write(numberOfPermutations(a, n));
// This code contributed by Princi Singh
</script>
Producción:
2
Complejidad temporal: O(N! * N)
Publicación traducida automáticamente
Artículo escrito por RohanDeoli y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA