Number of permutations such that sum of elements at odd index and even index are equal

Given N numbers, find the number of permutations in which the sum of elements at odd index and sum of elements at even index are equal. 

Examples:

Input: 1 2 3 
Output: 2 
The permutations are: 
1 3 2 sum at odd index = 1+2 = 3, sum at even index = 3 
2 3 1 sum at odd index = 2+1 = 3, sum at even index = 3

Input: 1 2 1 2 
Output: 3 
The permutations are: 
1 2 2 1 
2 1 1 2 
2 2 1 1 

The approach to the problem will be to use next_permutation() in C++ STL which helps to generate all the possible permutation of N numbers. If the sum of the odd index elements is equal to the sum of even index elements of the generated permutation, then increase the count. When all permutations are checked, print the count. 

Below is the implementation of the above approach: 

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>

Output

2

Time Complexity: O(N! * N)  

Auxiliary Space: O(1) because it is using constant space for variables