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 2Input: 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 permutationvoid 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 codeint main(){ int Q[] = { -2, 1 }; int n = 1 + (sizeof(Q) / sizeof(int)); findPerm(Q, n); return 0;} |
Java
// Java implementation of the approachclass GFG{// Function to find the required permutationstatic 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 codepublic 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 approachusing System;using System.Collections.Generic;class GFG{// Function to find the required permutationstatic 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 codepublic 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 permutationfunction 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 codevar 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)
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