Find permutation array from the cumulative sum array

Last Updated : 01 Mar, 2022

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# 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)

Find the sum of array after performing every query

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Find permutation array from the cumulative sum array

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