Restore a permutation from the given helper array

Last Updated : 01 Mar, 2022

Given an array Q[] of size N – 1 such that each Q[i] = P[i + 1] – P[i] where P[] is the permutation of the first N natural numbers, the task is to find this permutation. If no valid permutation P[] can be found then print -1.

Examples:

Input: Q[] = {-2, 1}
Output: 3 1 2

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

Approach: This is a mathematical algorithmic question. Lets denote P[i] = x. Therefore P[i + 1] = P[i] + (P[i + 1] – P[i]) = x + Q[i] (Since Q[i] = P[i + 1] – P[i]).
Therefore, P[i + 2]= P[i] + (P[i + 1] – P[i]) + (P[i + 2] – P[i + 1]) = x + Q[i] + Q[i + 1]. Observe, the pattern forming here. P is nothing but [x, x + Q[1], x + Q[1] + Q[2] + … + x + Q[1] + Q[2] + … + Q[n – 1]] where x = P[i] which is still unknown.
Lets have a permutation P’ where P'[i] = P[i] – x. Therefore, P’ = [0, Q[1], Q[1] + Q[2], Q[1] + Q[2] + Q[3], …, Q[1] + Q[2] + … + Q[n – 1]].

To find x, lets find the smallest element in P’. Let it be P'[k]. Therefore, x = 1 – P'[k]. This is because, the original permutation P has integers from 1 to n and so 1 can be the minimum element in P. After finding x, add x to each P’ to get the original permutation P.

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 required permutation
void findPerm(int Q[], int n)
{
 
    int minval = 0, qsum = 0;
    for (int i = 0; i < n - 1; i++) {
 
        // Each element in P' is like a
        // cumulative sum in Q
        qsum += Q[i];
 
        // minval is the minimum
        // value in P'
        if (qsum < minval)
            minval = qsum;
    }
    vector<int> P(n);
    P[0] = 1 - minval;
 
    // To check if each entry in P
    // is from the range [1, n]
    bool permFound = true;
    for (int i = 0; i < n - 1; i++) {
        P[i + 1] = P[i] + Q[i];
 
        // Invalid permutation
        if (P[i + 1] > n || P[i + 1] < 1) {
            permFound = false;
            break;
        }
    }
 
    // If a valid permutation exists
    if (permFound) {
 
        // Print the permutation
        for (int i = 0; i < n; i++) {
            cout << P[i] << " ";
        }
    }
    else {
 
        // No valid permutation
        cout << -1;
    }
}
 
// Driver code
int main()
{
    int Q[] = { -2, 1 };
    int n = 1 + (sizeof(Q) / sizeof(int));
 
    findPerm(Q, n);
 
    return 0;
}

Java

// Java implementation of the approach
 
class GFG
{
 
// Function to find the required permutation
static void findPerm(int Q[], int n)
{
 
    int minval = 0, qsum = 0;
    for (int i = 0; i < n - 1; i++) 
    {
 
        // Each element in P' is like a
        // cumulative sum in Q
        qsum += Q[i];
 
        // minval is the minimum
        // value in P'
        if (qsum < minval)
            minval = qsum;
    }
    int []P = new int[n];
    P[0] = 1 - minval;
 
    // To check if each entry in P
    // is from the range [1, n]
    boolean permFound = true;
    for (int i = 0; i < n - 1; i++) 
    {
        P[i + 1] = P[i] + Q[i];
 
        // Invalid permutation
        if (P[i + 1] > n || P[i + 1] < 1)
        {
            permFound = false;
            break;
        }
    }
 
    // If a valid permutation exists
    if (permFound)
    {
 
        // Print the permutation
        for (int i = 0; i < n; i++) 
        {
            System.out.print(P[i]+ " ");
        }
    }
    else
    {
 
        // No valid permutation
        System.out.print(-1);
    }
}
 
// Driver code
public static void main(String[] args)
{
    int Q[] = { -2, 1 };
    int n = 1 + Q.length;
 
    findPerm(Q, n);
 
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach 
 
# Function to find the required permutation 
def findPerm(Q, n) : 
 
    minval = 0; qsum = 0; 
    for i in range(n - 1) :
 
        # Each element in P' is like a 
        # cumulative sum in Q 
        qsum += Q[i]; 
 
        # minval is the minimum 
        # value in P' 
        if (qsum < minval) :
            minval = qsum; 
 
    P = [0]*n; 
    P[0] = 1 - minval; 
 
    # To check if each entry in P 
    # is from the range [1, n] 
    permFound = True; 
     
    for i in range(n - 1) :
        P[i + 1] = P[i] + Q[i]; 
 
        # Invalid permutation 
        if (P[i + 1] > n or P[i + 1] < 1) :
            permFound = False; 
            break; 
 
    # If a valid permutation exists 
    if (permFound) :
 
        # Print the permutation 
        for i in range(n) :
            print(P[i],end=" "); 
    else :
 
        # No valid permutation 
        print(-1); 
 
# Driver code 
if __name__ == "__main__" : 
 
    Q = [ -2, 1 ]; 
    n = 1 + len(Q) ;
 
    findPerm(Q, n); 
     
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Function to find the required permutation
static void findPerm(int []Q, int n)
{
 
    int minval = 0, qsum = 0;
    for (int i = 0; i < n - 1; i++) 
    {
 
        // Each element in P' is like a
        // cumulative sum in Q
        qsum += Q[i];
 
        // minval is the minimum
        // value in P'
        if (qsum < minval)
            minval = qsum;
    }
    int []P = new int[n];
    P[0] = 1 - minval;
 
    // To check if each entry in P
    // is from the range [1, n]
    bool permFound = true;
    for (int i = 0; i < n - 1; i++) 
    {
        P[i + 1] = P[i] + Q[i];
 
        // Invalid permutation
        if (P[i + 1] > n || P[i + 1] < 1)
        {
            permFound = false;
            break;
        }
    }
 
    // If a valid permutation exists
    if (permFound)
    {
 
        // Print the permutation
        for (int i = 0; i < n; i++) 
        {
            Console.Write(P[i]+ " ");
        }
    }
    else
    {
 
        // No valid permutation
        Console.Write(-1);
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int []Q = { -2, 1 };
    int n = 1 + Q.Length;
 
    findPerm(Q, n);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function to find the required permutation
function findPerm(Q, n)
{
    var minval = 0, qsum = 0;
    for(var i = 0; i < n - 1; i++) 
    {
         
        // Each element in P' is like a
        // cumulative sum in Q
        qsum += Q[i];
 
        // minval is the minimum
        // value in P'
        if (qsum < minval)
            minval = qsum;
    }
    var P = Array(n);
    P[0] = 1 - minval;
 
    // To check if each entry in P
    // is from the range [1, n]
    var permFound = true;
    for(var i = 0; i < n - 1; i++)
    {
        P[i + 1] = P[i] + Q[i];
 
        // Invalid permutation
        if (P[i + 1] > n || P[i + 1] < 1) 
        {
            permFound = false;
            break;
        }
    }
 
    // If a valid permutation exists
    if (permFound)
    {
         
        // Print the permutation
        for(var i = 0; i < n; i++) 
        {
            document.write( P[i] + " ");
        }
    }
    else
    {
         
        // No valid permutation
        document.write( -1);
    }
}
 
// Driver code
var Q = [ -2, 1 ];
var n = 1 + Q.length;
 
findPerm(Q, n);
 
// This code is contributed by famously
 
</script>

Output:

3 1 2

Time Complexity: O(n)

Auxiliary Space: O(n)

Longest permutation subsequence in a given array

Ripunjoy Medhi

Improve

Article Tags :

Practice Tags :

Restore a permutation from the given helper array

C++

Java

Python3

C#

Javascript

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