Generate a permutation of first N natural numbers from an array of differences between adjacent elements

Given an array arr[] consisting of (N – 1), the task is to construct a permutation array P[] consisting of the first N Natural Numbers such that arr[i] = (P[i +1] – P[i]). If there exists no such permutation, then print “-1”.

Examples:

Input: arr[] = {-1, 2, -3, -1}
Output: 4 3 5 2 1
Explanation:
For the array {4, 3, 5, 2, 1}, the adjacent difference array of consecutive elements is {4 – 3, 5 – 3, 2 – 5, 1 – 2} = {-1, 2, -3, -1} which is the same as the array arr[].

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

Approach: The given problem can be solved by considering the first element of the permutation as 0 and then constructing a new permutation array by using the given array arr[]. After this, add the minimum element of the new array to each element to make the array elements over the range [1, N]. Follow the steps below to solve the problem:

• Initialize an array, say perm[] of size N to store the resultant permutation.
• Initialize perm[0] as 0, and also initialize a variable, say lastEle as 0.
• Iterate over the range [1, N] using the variable i, and add the value of arr[i – 1] to the element lastEle and update the value of perm[i] as lastEle.
• Initialize a variable, say minimumElement to the minimum element of the array perm[].
• Initialize a HashSet of integers st, to store all elements of the permutation. Also, initialize a variable mx as 0 to store the maximum element in the perm[] array.
• Traverse through the perm[] array and add the value of (-sm) + 1 to the value perm[i], update the value of mx as max(mx, perm[i]) and add perm[i] to st.
• After completing the above steps, if the value of mx and the size of HashSet st is N, then print the array perm[] as the resultant array. Otherwise, print -1.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the permutation of
// N integers from the given difference
// array A[]
void findPermutation(int A[], int N)
{
    int lasEle = 0;

    // Stores the resultant permutation
    int perm[N];
    perm[0] = 0;

    for (int i = 1; i < N; i++) {

        // Update the value of lastEle
        lasEle += A[i - 1];

        // Initialize the value of
        // perm[i]
        perm[i] = lasEle;
    }

    // Stores the minimum element of
    // the array perm[]
    int sm = *min_element(perm, perm + N);

    // Stores the elements of the
    // permutation array perm[]
    unordered_set<int> st;
    int mx = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        // Update the value of perm[i]
        perm[i] += (-sm) + 1;

        // Update the value of mx
        mx = max(mx, perm[i]);

        // Insert the current element
        // in the hashset
        st.insert(perm[i]);
    }

    // Check if the maximum element and
    // the size of hashset is N or not
    if (mx == N and st.size() == N) {

        // Print the permutation
        for (int i = 0; i < N; i++) {
            cout << perm[i] << " ";
        }
    }

    // Otherwise print -1
    else {
        cout << -1 << " ";
    }
}

// Driver Code
int main()
{
    int arr[] = { -1, 2, -3, -1 };
    int N = sizeof(arr) / sizeof(arr[0]);
    findPermutation(arr, N + 1);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the permutation of
// N integers from the given difference
// array A[]
static void findPermutation(int []A, int N)
{
    int lasEle = 0;

    // Stores the resultant permutation
    int []perm = new int[N];
    perm[0] = 0;

    for (int i = 1; i < N; i++) {

        // Update the value of lastEle
        lasEle += A[i - 1];

        // Initialize the value of
        // perm[i]
        perm[i] = lasEle;
    }

    // Stores the minimum element of
    // the array perm[]
     int sm = perm[0];
        //Loop through the array
        for (int i = 0; i < perm.length; i++) {
            //Compare elements of array with min
           if(perm[i] <sm)
               sm = perm[i];
        }

    // Stores the elements of the
    // permutation array perm[]
    Set<Integer> st = new HashSet<Integer>();
    int mx = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        // Update the value of perm[i]
        perm[i] += (-sm) + 1;

        // Update the value of mx
        mx = Math.max(mx, perm[i]);

        // Insert the current element
        // in the hashset
        st.add(perm[i]);
    }

    // Check if the maximum element and
    // the size of hashset is N or not
    if (mx == N && st.size() == N) {

        // Print the permutation
        for (int i = 0; i < N; i++) {
            System.out.print(perm[i]+" ");
        }
    }

    // Otherwise print -1
    else {
        System.out.print(-1);
    }
}

// Driver Code
public static void main(String args[])
{
    int arr[] = { -1, 2, -3, -1 };
    int N = arr.length;
    findPermutation(arr, N + 1);
}
}

// This code is contributed by SURENDRA_GANGWAR.

Python3

# Python program for the above approach

# Function to find the permutation of
# N integers from the given difference
# array A[]
def findPermutation(A, N):
    lasEle = 0

    # Stores the resultant permutation
    perm = [0]*N
    perm[0] = 0

    for i in range(1,N):
        # Update the value of lastEle
        lasEle += A[i - 1]

        # Initialize the value of
        # perm[i]
        perm[i] = lasEle

    # Stores the minimum element of
    # the array perm[]
    sm = min(perm)

    # Stores the elements of the
    # permutation array perm[]
    st = {}
    mx = 0

    # Traverse the array
    for i in range(N):
        # Update the value of perm[i]
        perm[i] += (-sm) + 1

        # Update the value of mx
        mx = max(mx, perm[i])

        # Insert the current element
        # in the hashset
        st[perm[i]] = 1

    # Check if the maximum element and
    # the size of hashset is N or not
    if (mx == N and len(st) == N):

        # Print the permutation
        for i in range(N):
            print(perm[i],end=" ")
    # Otherwise print -1
    else:
        print(-1,end=" ")

# Driver Code
if __name__ == '__main__':
    arr = [-1, 2, -3, -1]
    N = len(arr)
    findPermutation(arr, N + 1)

# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
using System.Linq;
using System.Collections.Generic;

class GFG {

// Function to find the permutation of
// N integers from the given difference
// array A[]
static void findPermutation(int[] A, int N)
{
    int lasEle = 0;

    // Stores the resultant permutation
    int[] perm = new int[N];
    perm[0] = 0;

    for (int i = 1; i < N; i++) {

        // Update the value of lastEle
        lasEle += A[i - 1];

        // Initialize the value of
        // perm[i]
        perm[i] = lasEle;
    }

    // Stores the minimum element of
    // the array perm[]
    int sm = perm.Min();

    // Stores the elements of the
    // permutation array perm[]
   List<int> st = new List<int>();
    int mx = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        // Update the value of perm[i]
        perm[i] += (-sm) + 1;

        // Update the value of mx
        mx = Math.Max(mx, perm[i]);

        // Insert the current element
        // in the hashset
        st.Add(perm[i]);
    }

    // Check if the maximum element and
    // the size of hashset is N or not
    if (mx == N && st.Count == N) {

        // Print the permutation
        for (int i = 0; i < N; i++) {
           Console.Write(perm[i] + " ");
        }
    }

    // Otherwise print -1
    else {
        Console.Write(-1 + " ");
    }
}

    // Driver Code
    static void Main()
    {
        int[] arr= { -1, 2, -3, -1 };
    int N = arr.Length;
    findPermutation(arr, N + 1);
    }
}

// This code is contributed by sanjoy_62.

Javascript

<script>

// JavaScript program for the above approach


// Function to find the permutation of
// N integers from the given difference
// array A[]
function findPermutation(A, N) {
    let lasEle = 0;

    // Stores the resultant permutation
    let perm = new Array(N);
    perm[0] = 0;

    for (let i = 1; i < N; i++) {

        // Update the value of lastEle
        lasEle += A[i - 1];

        // Initialize the value of
        // perm[i]
        perm[i] = lasEle;
    }

    // Stores the minimum element of
    // the array perm[]

    let temp = [...perm];
    let sm = temp.sort((a, b) => a - b)[0]

    // Stores the elements of the
    // permutation array perm[]
    let st = new Set();
    let mx = 0;

    // Traverse the array
    for (let i = 0; i < N; i++) {

        // Update the value of perm[i]
        perm[i] += (-sm) + 1;

        // Update the value of mx
        mx = Math.max(mx, perm[i]);

        // Insert the current element
        // in the hashset
        st.add(perm[i]);
    }

    // Check if the maximum element and
    // the size of hashset is N or not
    if (mx == N && st.size == N) {

        // Print the permutation
        for (let i of perm) {
            document.write(i + " ")
        }
    }

    // Otherwise print -1
    else {
        document.write(-1 + " ");
    }
}

// Driver Code

let arr = [-1, 2, -3, -1];
let N = arr.length
findPermutation(arr, N + 1);

</script>

Output:

4 3 5 2 1

Time Complexity: O(N)
Auxiliary Space: O(N)