Generate original array from difference between every two consecutive elements

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Given N – 1 differences between two consecutive elements of an array containing N numbers which are in range of 1 to N. The task is to determine the original array using the given differences. If possible print the array, else print -1.
Examples:

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

Approach: Since we want to generate permutations in the range [1, n] and each digit should occur only once. Take for example,

arr[] = {2, -3, 2}
Here, P2 – P1 = 2, P3 – P2 = -3, P4 – P3 = 2.
Let P1 = x then P2 = x + 2, P3 = P2 – 3 = x + 2 – 3 = x – 1, P4 = P3 + 2 = x – 1 + 2 = x + 1.
So, P1 = x, P2 = x + 2, P3 = x – 1, P4 = x + 1.
Now since we want a permutation from 1 to n, therefore the P[i]’s we get above must also satisfy the condition. Since the value must be satisfied for each and every x, hence for our simplicity we take x = 0.
Now, P1 = 0, P2 = 2, P3 = -1, P4 = 1.

We will sort the p[i]’s, after sorting the consecutive difference between each element must be equal to 1. If at any index we encounter an element p[i] such that p[i] – p[i – 1] != 1 then it is not possible to generate the permutation. To generate the final permutation we will keep track of indices, we can use map or unordered_map to do so.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the required permutation
void findPerm(int n, vector<int>& differences)
{
    vector<int> ans;
    ans.clear();
    ans.push_back(0);
 
    // Take x = 0 for simplicity
    int x = 0;
 
    // Calculate all the differences
    // and store it in a vector
    for (int i = 0; i <= n - 2; ++i) {
        int diff = differences[i];
        x = x + diff;
        ans.push_back(x);
    }
 
    // Preserve the original array
    vector<int> anss = ans;
    sort(ans.begin(), ans.end());
    int flag = -1;
 
    // Check if all the consecutive elements
    // have difference = 1
    for (int i = 1; i <= n - 1; ++i) {
        int res = ans[i] - ans[i - 1];
        if (res != 1) {
            flag = 0;
        }
    }
 
    // If consecutive elements don't have
    // difference 1 at any position then
    // the answer is impossible
    if (flag == 0) {
        cout << -1;
        return;
    }
 
    // Else store the indices and values
    // at those indices in a map
    // and finainty print them
    else {
        unordered_map<int, int> mpp;
        mpp.clear();
        int j = 1;
        vector<int> value_at_index;
        for (auto& x : ans) {
            mpp[x] = j;
            ++j;
        }
        for (auto& x : anss) {
            value_at_index.push_back(mpp[x]);
        }
        for (auto& x : value_at_index) {
            cout << x << " ";
        }
        cout << endl;
    }
}
 
// Driver code
int main()
{
    vector<int> differences;
    differences.push_back(2);
    differences.push_back(-3);
    differences.push_back(2);
    int n = differences.size() + 1;
    findPerm(n, differences);
 
    return 0;
}

Java

// Java program to implement the above approach
import java.util.*;
class GFG 
{
 
// Function to print the required permutation
static void findPerm(int n, Vector<Integer> differences)
{
    Vector<Integer> ans = new Vector<Integer>();
    ans.clear();
    ans.add(0);
 
    // Take x = 0 for simplicity
    int x = 0;
 
    // Calculate all the differences
    // and store it in a vector
    for (int i = 0; i <= n - 2; ++i) 
    {
        int diff = differences.get(i);
        x = x + diff;
        ans.add(x);
    }
 
    // Preserve the original array
    Vector<Integer> anss = new Vector<Integer>();
    for(Integer obj:ans)
        anss.add(obj);
         
    Collections.sort(ans);
    int flag = -1;
 
    // Check if all the consecutive elements
    // have difference = 1
    for (int i = 1; i <= n - 1; ++i)
    {
        int res = ans.get(i) - ans.get(i - 1);
        if (res != 1) 
        {
            flag = 0;
        }
    }
 
    // If consecutive elements don't have
    // difference 1 at any position then
    // the answer is impossible
    if (flag == 0) 
    {
        System.out.print(-1);
        return;
    }
 
    // Else store the indices and values
    // at those indices in a map
    // and finainty print them
    else
    {
        Map<Integer,Integer> mpp = new HashMap<>();
        mpp.clear();
        int j = 1;
        Vector<Integer> value_at_index = new Vector<Integer>();
        for (Integer x1 : ans) 
        {
            mpp.put(x1, j);
            ++j;
        }
        for (Integer x2 : anss) 
        {
            value_at_index.add(mpp.get(x2));
        }
        for (Integer x3 : value_at_index)
        {
            System.out.print(x3 + " ");
        }
        System.out.println();
    }
}
 
// Driver code
public static void main(String[] args)
{
    Vector<Integer> differences = new Vector<Integer>();
    differences.add(2);
    differences.add(-3);
    differences.add(2);
    int n = differences.size() + 1;
    findPerm(n, differences);
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
# Function to print the required permutation
def findPerm(n, differences):
    ans = []
    ans.append(0)
 
    # Take x = 0 for simplicity
    x = 0
 
    # Calculate all the differences
    # and store it in a vector
    for i in range(n - 1):
        diff = differences[i]
        x = x + diff
        ans.append(x)
 
    # Preserve the original array
    anss = ans
    ans = sorted(ans)
    flag = -1
 
    # Check if all the consecutive elements
    # have difference = 1
    for i in range(1, n):
        res = ans[i] - ans[i - 1]
        if (res != 1):
            flag = 0
 
    # If consecutive elements don't have
    # difference 1 at any position then
    # the answer is impossible
    if (flag == 0):
        print("-1")
        return
 
    # Else store the indices and values
    # at those indices in a map
    # and infinity print them
    else:
        mpp = dict()
        j = 1
        value_at_index = []
        for x in ans:
            mpp[x] = j
            j += 1
 
        for x in anss:
            value_at_index.append(mpp[x])
 
        for x in value_at_index:
            print(x, end = " ")
        print()
 
# Driver code
differences=[]
differences.append(2)
differences.append(-3)
differences.append(2)
n = len(differences) + 1
findPerm(n, differences)
 
# This code is contributed by mohit kumar

C#

// C# program to implement the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to print the required permutation
static void findPerm(int n, List<int> differences)
{
    List<int> ans = new List<int>();
    ans.Clear();
    ans.Add(0);
 
    // Take x = 0 for simplicity
    int x = 0;
 
    // Calculate all the differences
    // and store it in a vector
    for(int i = 0; i <= n - 2; ++i) 
    {
        int diff = differences[i];
        x = x + diff;
        ans.Add(x);
    }
 
    // Preserve the original array
    List<int> anss = new List<int>();
    foreach(int obj in ans)
        anss.Add(obj);
         
    ans.Sort();
    int flag = -1;
 
    // Check if all the consecutive elements
    // have difference = 1
    for(int i = 1; i <= n - 1; ++i)
    {
        int res = ans[i] - ans[i - 1];
        if (res != 1) 
        {
            flag = 0;
        }
    }
 
    // If consecutive elements don't have
    // difference 1 at any position then
    // the answer is impossible
    if (flag == 0) 
    {
        Console.Write(-1);
        return;
    }
 
    // Else store the indices and values
    // at those indices in a map
    // and finainty print them
    else
    {
        Dictionary<int,
                   int> mpp = new Dictionary<int,
                                             int>();
        mpp.Clear();
        int j = 1;
         
        List<int> value_at_index = new List<int>();
        foreach (int x1 in ans) 
        {
            mpp.Add(x1, j);
            ++j;
        }
         
        foreach (int x2 in anss) 
        {
            value_at_index.Add(mpp[x2]);
        }
         
        foreach (int x3 in value_at_index)
        {
            Console.Write(x3 + " ");
        }
        Console.WriteLine();
    }
}
 
// Driver code
public static void Main(String[] args)
{
    List<int> differences = new List<int>();
    differences.Add(2);
    differences.Add(-3);
    differences.Add(2);
     
    int n = differences.Count + 1;
     
    findPerm(n, differences);
}
}
 
// This code is contributed by Amit Katiyar 

Javascript

<script>
// Javascript program to implement the above approach
     
    // Function to print the required permutation
function findPerm(n,differences)
{
    let ans = [];    
    ans.push(0);
   
    // Take x = 0 for simplicity
    let x = 0;
   
    // Calculate all the differences
    // and store it in a vector
    for (let i = 0; i <= n - 2; ++i) 
    {
        let diff = differences[i];
        x = x + diff;
        ans.push(x);
    }
   
    // Preserve the original array
    let anss = [];
    for(let obj = 0; obj < ans.length; obj++)
        anss.push(ans[obj]);
           
    ans.sort(function(a,b){return a-b;});
    let flag = -1;
   
    // Check if all the consecutive elements
    // have difference = 1
    for (let i = 1; i <= n - 1; ++i)
    {
        let res = ans[i] - ans[i - 1];
        if (res != 1) 
        {
            flag = 0;
        }
    }
   
    // If consecutive elements don't have
    // difference 1 at any position then
    // the answer is impossible
    if (flag == 0) 
    {
        document.write(-1);
        return;
    }
   
    // Else store the indices and values
    // at those indices in a map
    // and finainty print them
    else
    {
        let mpp = new Map();
         
        let j = 1;
        let value_at_index = [];
        for (let x1 = 0; x1 < ans.length; x1++) 
        {
            mpp.set(ans[x1], j);
            ++j;
        }
        for (let x2 = 0; x2 < anss.length; x2++) 
        {
            value_at_index.push(mpp.get(anss[x2]));
        }
        for (let x3 = 0; x3 < value_at_index.length; x3++)
        {
            document.write(value_at_index[x3] + " ");
        }
        document.write("<br>");
    }
}
 
    // Driver code
    let differences =[];
    differences.push(2);
    differences.push(-3);
    differences.push(2);
    let n = differences.length + 1;
    findPerm(n, differences);
     
// This code is contributed by unknown2108.
</script>

Output:

2 4 1 3

Time Complexity: O(n*log(n))
Auxiliary Space: O(n)

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