# C++ program to find permutation for which sum of adjacent elements sort is same as given array

Suppose we have an array A with n elements. A function F(p) is a sorted array of sums of adjacent elements in p. So F(p) = sort([p1 + p2, p2 + p3, ... pn-1 + pn]). We have a permutation represented in A. We have to find the different permutation of A where F(A) is same.

So, if the input is like A = [2, 1, 6, 5, 4, 3], then the output will be [1, 2, 5, 6, 3, 4], because F(A)=sort([2+1, 1+6, 6+5, 5+4, 4+3]) = sort([3, 7, 11, 9, 7]) = [3,7,7,9,11]. And sort([1+2, 2+5, 5+6, 6+3, 3+4]) = sort([3, 7, 11, 9, 7]) = [3, 7, 7, 9, 11]. (There can be other answers too)

## Steps

To solve this, we will follow these steps −

n := size of A
for initialize i := n - 1, when i >= 0, update (decrease i by 1), do:
print A[i]

## Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
void solve(vector<int> A) {
int n = A.size();
for (int i = n - 1; i >= 0; i--)
cout << A[i] << ", ";
}
int main() {
vector<int> A = { 2, 1, 6, 5, 4, 3 };
solve(A);
}

## Input

{ 2, 1, 6, 5, 4, 3 }

## Output

3, 4, 5, 6, 1, 2,