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

0 0 6 minutes read

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++

// 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

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<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

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<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

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function to return the minimum
// value of x from the given array q
function Get_Minimum(q)
{
    let minimum = 0;
    let sum = 0;
    for(let i = 0; i < q.length - 1; i++)
    {
        sum += q[i];
        if (sum < minimum)
            minimum = sum;
    }
    return minimum;
}
 
// Function to return the required permutation    
function Find_Permutation(q, n)
{
    let p = new Array(n);
    let 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(let i = 0; i < n - 1; i++)
        p[i + 1] = p[i] + q[i];
   
    let okay = true;
   
    // Check if formed permutation
    // is correct or not
    for(let i = 0; i < n; i++)
    {
        if (p[i] < 1 || p[i] > n)
            okay = false;
             
        let w = new Set();
        if (w.size != n)
            okay = true;
    }
   
    // Return the permutation p
    if (okay)
        return p;
    else
        return new [-1];
}
 
// Driver code
let q = [ -2, 1 ];
let n = q.length + 1;
document.write("[ ");
 
let temp = Find_Permutation(q, n);
for(let i = 0; i < temp.length; i++)
    document.write(temp[i] + " ");
     
document.write("]");
 
// This code is contributed by patel2127
 
</script>

Output:

[3, 1, 2]

Time Complexity: O(n²)
Auxiliary Space: O(n)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!