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 permutationvoid 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 codeint 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 approachimport java.util.*;class GFG {// Function to print the required permutationstatic 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 codepublic 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 permutationdef 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 codedifferences=[]differences.append(2)differences.append(-3)differences.append(2)n = len(differences) + 1findPerm(n, differences)# This code is contributed by mohit kumar |
C#
// C# program to implement the above approachusing System;using System.Collections.Generic;class GFG{// Function to print the required permutationstatic 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 codepublic 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 permutationfunction 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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