# Find the permutation p from the array q such that q[i] = p[i+1] – p[i]

Given an array Q[] of length N, the task is to find the permutation P[] of integers from the range [1, N + 1] such that Q[i] = P[i + 1] – P[i] for all valid i. If it is not possible then print -1.

Examples:

Input: Q[] = {-2, 1}
Output: 3 1 2
q[0] = p[1] – p[0] = 1 – 3 = -2
q[1] = p[2] – p[1] = 2 – 1 = 1

Input: Q[] = {1, 1, 1, 1}
Output: 1 2 3 4 5

Input: Q[] = {-1, 2, 2}
Output: -1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
Let,

P[0] = x then P[1] = P[0] + (P[1] – P[0]) = x + Q[0]

and, P[2] = P[0] + (P[1] – P[0]) + (P[2] – P[1]) = x + Q[0] + Q[1].

Similarly,

P[n] = x + Q[0] + Q[1] + Q[2 ] + ….. + Q[N – 1].

It means that the sequence p’ = 0, Q[1], Q[1] + Q[2], ….., + Q[1] + Q[2] + Q[3] + ….. + Q[N – 1] is the required permutation if we add x to each element.

To find the value of x, find an i such that p'[i] is minimum.

As, p'[i] + x is the minimum value in the series then it must be equal to 1 as series can have values from [1, N].
So x = 1 – p'[i].

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include    using namespace std;    // Function to return the minimum // value of x from the given array q int Get_Minimum(vector q) {     int minimum = 0;     int sum = 0;     for(int i = 0; i < q.size() - 1; i++)     {         sum += q[i];         if (sum < minimum)             minimum = sum;     }     return minimum; }    // Function to return the required permutation vector Find_Permutation(vector q, int n) {     vector p(n, 0);     int min_value = Get_Minimum(q);        // Set the value of p[0] i.e. x = p[0]     p[0] = 1 - min_value;        // Iterate over array q[]     for (int i = 0; i < n - 1; i++)         p[i + 1] = p[i] + q[i];        bool okay = true;        // Check if formed permutation     // is correct or not     for (int i = 0; i < n; i++)     {         if (p[i] < 1 or p[i] > n)             okay = false;         set w(p.begin(), p.end());         if (w.size() != n)             okay = false;     }        // Return the permutation p     if (okay)         return p;     else         return {-1}; }    // Driver code int main() {     vector q = {-2, 1};     int n = q.size() + 1;     cout << "[ ";     for (int i:Find_Permutation(q, n))         cout << i << " ";     cout << "]";      }    // This code is contributed by Mohit Kumar

## Java

 // Java implementation of the approach import java.util.*;    class GFG  {    // Function to return the minimum // value of x from the given array q static int Get_Minimum(int [] q) {     int minimum = 0;     int sum = 0;     for(int i = 0; i < q.length - 1; i++)     {         sum += q[i];         if (sum < minimum)             minimum = sum;     }     return minimum; }    // Function to return the required permutation static int [] Find_Permutation(int [] q, int n) {     int [] p = new int[n];     int min_value = Get_Minimum(q);        // Set the value of p[0] i.e. x = p[0]     p[0] = 1 - min_value;        // Iterate over array q[]     for (int i = 0; i < n - 1; i++)         p[i + 1] = p[i] + q[i];        boolean okay = true;        // Check if formed permutation     // is correct or not     for (int i = 0; i < n; i++)     {         if (p[i] < 1 || p[i] > n)             okay = false;         Set w = new HashSet<>();         if (w.size() != n)             okay = true;     }        // Return the permutation p     if (okay)         return p;     else         return new int []{-1}; }    // Driver code public static void main(String args[])  {     int []q = {-2, 1};     int n = q.length + 1;     System.out.print("[ ");     for (int i:Find_Permutation(q, n))         System.out.print(i + " ");     System.out.print("]"); } }    // This code is contributed by 29AjayKumar

## Python3

 # Python3 implementation of the approach    # Function to return the minimum  # value of x from the given array q def Get_Minimum(q):     minimum = 0     sum = 0     for i in range(n - 1):         sum += q[i]         if sum < minimum:             minimum = sum     return minimum    # Function to return the  # required permutation def Find_Permutation(q):     p = [0] * n     min_value = Get_Minimum(q)        # Set the value of p[0]      # i.e. x = p[0]     p[0]= 1 - min_value        # Iterate over array q[]     for i in range(n - 1):         p[i + 1] = p[i] + q[i]        okay = True        # Check if formed permutation      # is correct or not     for i in range(n):         if p[i] < 1 or p[i] > n:             okay = False     if len(set(p)) != n:         okay = False        # Return the permutation p     if okay:         return p     else:         return -1    # Driver code if __name__=="__main__":     q = [-2, 1]     n = len(q) + 1     print(Find_Permutation(q))

## C#

 // C# implementation of the approach using System; using System.Collections.Generic;    class GFG  {    // Function to return the minimum // value of x from the given array q static int Get_Minimum(int [] q) {     int minimum = 0;     int sum = 0;     for(int i = 0; i < q.Length - 1; i++)     {         sum += q[i];         if (sum < minimum)             minimum = sum;     }     return minimum; }    // Function to return the required permutation static int [] Find_Permutation(int [] q, int n) {     int [] p = new int[n];     int min_value = Get_Minimum(q);        // Set the value of p[0] i.e. x = p[0]     p[0] = 1 - min_value;        // Iterate over array q[]     for (int i = 0; i < n - 1; i++)         p[i + 1] = p[i] + q[i];        bool okay = true;        // Check if formed permutation     // is correct or not     for (int i = 0; i < n; i++)     {         if (p[i] < 1 || p[i] > n)             okay = false;         HashSet w = new HashSet();         if (w.Count != n)             okay = true;     }        // Return the permutation p     if (okay)         return p;     else         return new int []{-1}; }    // Driver code public static void Main(String []args)  {     int []q = {-2, 1};     int n = q.Length + 1;     Console.Write("[ ");     foreach (int i in Find_Permutation(q, n))         Console.Write(i + " ");     Console.Write("]"); } }    // This code is contributed by PrinciRaj1992

Output:

[3, 1, 2]

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