Find permutation array from the cumulative sum array

Given an array arr[] of N elements where each arr[i] is the cumulative sum of the subarray P[0…i] of another array P[] where P is the permutation of the integers from 1 to N. The task is to find the array P[], if no such P exists then print -1.
Examples:

Input: arr[] = {2, 3, 6} 
Output: 2 1 3
Input: arr[] = {1, 2, 2, 4} 
Output: -1 

Approach:

• The first element of the cumulative array must be the first element of permutation array and the element at the i^th position will be arr[i] – arr[i – 1] as arr[] is the cumulative sum array of the permutation array.
• So, find the array from the cumulative sum array and then mark the occurrence of every element from 1 to N that is present in the generated array.
• If any element is appearing more than once then the permutation is invalid else print the permutation.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the valid permutation
void getPermutation(int a[], int n)
{

    // Find the array from the cumulative sum
    vector<int> ans(n);
    ans[0] = a[0];
    for (int i = 1; i < n; i++)
        ans[i] = a[i] - a[i - 1];

    // To mark the occurrence of an element
    bool present[n + 1] = { false };
    for (int i = 0; i < ans.size(); i++) {

        // If current element has already
        // been seen previously
        if (present[ans[i]]) {
            cout << "-1";
            return;
        }

        // Mark the current element's occurrence
        else
            present[ans[i]] = true;
    }

    // Print the required permutation
    for (int i = 0; i < n; i++)
        cout << ans[i] << " ";
}

// Driver code
int main()
{
    int a[] = { 2, 3, 6 };
    int n = sizeof(a) / sizeof(a[0]);

    getPermutation(a, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to find the valid permutation
static void getPermutation(int a[], int n)
{

    // Find the array from the cumulative sum
    int []ans = new int[n];
    ans[0] = a[0];
    for (int i = 1; i < n; i++)
        ans[i] = a[i] - a[i - 1];

    // To mark the occurrence of an element
    boolean []present = new boolean[n + 1];
    for (int i = 0; i < ans.length; i++)
    {

        // If current element has already
        // been seen previously
        if (present[ans[i]])
        {
            System.out.print("-1");
            return;
        }

        // Mark the current element's occurrence
        else
            present[ans[i]] = true;
    }

    // Print the required permutation
    for (int i = 0; i < n; i++)
        System.out.print(ans[i] + " ");
}

// Driver code
public static void main(String[] args)
{
    int a[] = { 2, 3, 6 };
    int n = a.length;

    getPermutation(a, n);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach

# Function to find the valid permutation
def getPermutation(a, n) :

    # Find the array from the cumulative sum
    ans = [0] * n;
    ans[0] = a[0];
    for i in range(1, n) :
        ans[i] = a[i] - a[i - 1];

    # To mark the occurrence of an element
    present = [0] * (n + 1);

    for i in range(n) :

        # If current element has already
        # been seen previously
        if (present[ans[i]]) :
            print("-1", end = "");
            return;

        # Mark the current element's occurrence
        else :
            present[ans[i]] = True;

    # Print the required permutation
    for i in range(n) :
        print(ans[i], end = " ");

# Driver code
if __name__ == "__main__" :

    a = [ 2, 3, 6 ];
    n = len(a);

    getPermutation(a, n);

# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;
using System.Collections.Generic;

class GFG
{

// Function to find the valid permutation
static void getPermutation(int [] a, int n)
{

    // Find the array from the cumulative sum
    List<int> ans = new List<int>();
    ans.Add(a[0]);
    for (int i = 1; i < n; i++)
        ans.Add(a[i] - a[i - 1]);

    // To mark the occurrence of an element
    List<int> present = new List<int>();

    for (int i = 0; i < n+1; i++)
        present.Add(0);

    for (int i = 0; i < ans.Count; i++)
    {

        // If current element has already
        // been seen previously
        if (present[ans[i]] == 1)
        {
            Console.Write("-1");
            return;
        }

        // Mark the current element's occurrence
        else
            present[ans[i]] = 1;
    }

    // Print the required permutation
    for (int i = 0; i < n; i++)
        Console.Write(ans[i] + " ");
}

// Driver code
static public void Main()
{
    int[] a = { 2,3,6};
    int n = a.Length;
    getPermutation(a, n);
}
}

// This code is contributed by mohit kumar 29

Javascript

<script>

// Javascript implementation of the approach

// Function to find the valid permutation
function getPermutation(a, n)
{

    // Find the array from the cumulative sum
    var ans = Array(n);
    ans[0] = a[0];
    for (var i = 1; i < n; i++)
        ans[i] = a[i] - a[i - 1];

    // To mark the occurrence of an element
    var present = Array(n+1).fill(false);
    for (var i = 0; i < ans.length; i++) {

        // If current element has already
        // been seen previously
        if (present[ans[i]]) {
            document.write( "-1");
            return;
        }

        // Mark the current element's occurrence
        else
            present[ans[i]] = true;
    }

    // Print the required permutation
    for (var i = 0; i < n; i++)
        document.write( ans[i] + " ");
}

// Driver code
var a = [2, 3, 6];
var n = a.length;
getPermutation(a, n);

// This code is contributed by rrrtnx.
</script>

Output:

2 1 3

Time Complexity: O(n)

Auxiliary Space: O(n)