Given an array Q[] of length N, the task is to find the permutation P[] of integers from the range [1, N + 1] such that Q[i] = P[i + 1] – P[i] for all valid i. If it is not possible then print -1.
Examples:
Input: Q[] = {-2, 1}
Output: 3 1 2
q[0] = p[1] – p[0] = 1 – 3 = -2
q[1] = p[2] – p[1] = 2 – 1 = 1Input: Q[] = {1, 1, 1, 1}
Output: 1 2 3 4 5Input: Q[] = {-1, 2, 2}
Output: -1
Approach:
Let,
P[0] = x then P[1] = P[0] + (P[1] – P[0]) = x + Q[0]
and, P[2] = P[0] + (P[1] – P[0]) + (P[2] – P[1]) = x + Q[0] + Q[1].
Similarly,
P[n] = x + Q[0] + Q[1] + Q[2 ] + ….. + Q[N – 1].
It means that the sequence p’ = 0, Q[1], Q[1] + Q[2], ….., + Q[1] + Q[2] + Q[3] + ….. + Q[N – 1] is the required permutation if we add x to each element.
To find the value of x, find an i such that p'[i] is minimum.
As, p'[i] + x is the minimum value in the series then it must be equal to 1 as series can have values from [1, N].
So x = 1 – p'[i].
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include<bits/stdc++.h>using namespace std;// Function to return the minimum// value of x from the given array qint Get_Minimum(vector<int> q){ int minimum = 0; int sum = 0; for(int i = 0; i < q.size() - 1; i++) { sum += q[i]; if (sum < minimum) minimum = sum; } return minimum;}// Function to return the required permutationvector<int> Find_Permutation(vector<int> q, int n){ vector<int> p(n, 0); int min_value = Get_Minimum(q); // Set the value of p[0] i.e. x = p[0] p[0] = 1 - min_value; // Iterate over array q[] for (int i = 0; i < n - 1; i++) p[i + 1] = p[i] + q[i]; bool okay = true; // Check if formed permutation // is correct or not for (int i = 0; i < n; i++) { if (p[i] < 1 or p[i] > n) okay = false; set<int> w(p.begin(), p.end()); if (w.size() != n) okay = false; } // Return the permutation p if (okay) return p; else return {-1};}// Driver codeint main(){ vector<int> q = {-2, 1}; int n = q.size() + 1; cout << "[ "; for (int i:Find_Permutation(q, n)) cout << i << " "; cout << "]"; }// This code is contributed by Mohit Kumar |
Java
// Java implementation of the approachimport java.util.*;class GFG {// Function to return the minimum// value of x from the given array qstatic int Get_Minimum(int [] q){ int minimum = 0; int sum = 0; for(int i = 0; i < q.length - 1; i++) { sum += q[i]; if (sum < minimum) minimum = sum; } return minimum;}// Function to return the required permutationstatic int [] Find_Permutation(int [] q, int n){ int [] p = new int[n]; int min_value = Get_Minimum(q); // Set the value of p[0] i.e. x = p[0] p[0] = 1 - min_value; // Iterate over array q[] for (int i = 0; i < n - 1; i++) p[i + 1] = p[i] + q[i]; boolean okay = true; // Check if formed permutation // is correct or not for (int i = 0; i < n; i++) { if (p[i] < 1 || p[i] > n) okay = false; Set<Integer> w = new HashSet<>(); if (w.size() != n) okay = true; } // Return the permutation p if (okay) return p; else return new int []{-1};}// Driver codepublic static void main(String args[]) { int []q = {-2, 1}; int n = q.length + 1; System.out.print("[ "); for (int i:Find_Permutation(q, n)) System.out.print(i + " "); System.out.print("]");}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach# Function to return the minimum # value of x from the given array qdef Get_Minimum(q): minimum = 0 sum = 0 for i in range(n - 1): sum += q[i] if sum < minimum: minimum = sum return minimum# Function to return the # required permutationdef Find_Permutation(q): p = [0] * n min_value = Get_Minimum(q) # Set the value of p[0] # i.e. x = p[0] p[0]= 1 - min_value # Iterate over array q[] for i in range(n - 1): p[i + 1] = p[i] + q[i] okay = True # Check if formed permutation # is correct or not for i in range(n): if p[i] < 1 or p[i] > n: okay = False if len(set(p)) != n: okay = False # Return the permutation p if okay: return p else: return -1# Driver codeif __name__=="__main__": q = [-2, 1] n = len(q) + 1 print(Find_Permutation(q)) |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG {// Function to return the minimum// value of x from the given array qstatic int Get_Minimum(int [] q){ int minimum = 0; int sum = 0; for(int i = 0; i < q.Length - 1; i++) { sum += q[i]; if (sum < minimum) minimum = sum; } return minimum;}// Function to return the required permutationstatic int [] Find_Permutation(int [] q, int n){ int [] p = new int[n]; int min_value = Get_Minimum(q); // Set the value of p[0] i.e. x = p[0] p[0] = 1 - min_value; // Iterate over array q[] for (int i = 0; i < n - 1; i++) p[i + 1] = p[i] + q[i]; bool okay = true; // Check if formed permutation // is correct or not for (int i = 0; i < n; i++) { if (p[i] < 1 || p[i] > n) okay = false; HashSet<int> w = new HashSet<int>(); if (w.Count != n) okay = true; } // Return the permutation p if (okay) return p; else return new int []{-1};}// Driver codepublic static void Main(String []args) { int []q = {-2, 1}; int n = q.Length + 1; Console.Write("[ "); foreach (int i in Find_Permutation(q, n)) Console.Write(i + " "); Console.Write("]");}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript implementation of the approach// Function to return the minimum// value of x from the given array qfunction Get_Minimum(q){ let minimum = 0; let sum = 0; for(let i = 0; i < q.length - 1; i++) { sum += q[i]; if (sum < minimum) minimum = sum; } return minimum;}// Function to return the required permutation function Find_Permutation(q, n){ let p = new Array(n); let min_value = Get_Minimum(q); // Set the value of p[0] i.e. x = p[0] p[0] = 1 - min_value; // Iterate over array q[] for(let i = 0; i < n - 1; i++) p[i + 1] = p[i] + q[i]; let okay = true; // Check if formed permutation // is correct or not for(let i = 0; i < n; i++) { if (p[i] < 1 || p[i] > n) okay = false; let w = new Set(); if (w.size != n) okay = true; } // Return the permutation p if (okay) return p; else return new [-1];}// Driver codelet q = [ -2, 1 ];let n = q.length + 1;document.write("[ ");let temp = Find_Permutation(q, n);for(let i = 0; i < temp.length; i++) document.write(temp[i] + " "); document.write("]");// This code is contributed by patel2127</script> |
Output:
[3, 1, 2]
Time Complexity: O(n2)
Auxiliary Space: O(n)
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