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

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++

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// C++ implementation of the approach
#include<bits/stdc++.h>
  
using namespace std;
  
// Function to return the minimum
// value of x from the given array q
int Get_Minimum(vector<int> 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<int> Find_Permutation(vector<int> q, int n)
{
    vector<int> 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<int> 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<int> 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

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Java

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// 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<Integer> 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

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Python3

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# 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))

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C#

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// 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<int> w = new HashSet<int>();
        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

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Output:

[3, 1, 2]


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