Given an array arr[] of N elements where each arr[i] is the cumulative sum of the subarray P[0…i] of another array P[] where P is the permutation of the integers from 1 to N. The task is to find the array P[], if no such P exists then print -1.
Examples:
Input: arr[] = {2, 3, 6}
Output: 2 1 3
Input: arr[] = {1, 2, 2, 4}
Output: -1
Approach:
- The first element of the cumulative array must be the first element of permutation array and the element at the ith position will be arr[i] – arr[i – 1] as arr[] is the cumulative sum array of the permutation array.
- So, find the array from the cumulative sum array and then mark the occurrence of every element from 1 to N that is present in the generated array.
- If any element is appearing more than once then the permutation is invalid else print the permutation.
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 valid permutationvoid getPermutation(int a[], int n){ // Find the array from the cumulative sum vector<int> ans(n); ans[0] = a[0]; for (int i = 1; i < n; i++) ans[i] = a[i] - a[i - 1]; // To mark the occurrence of an element bool present[n + 1] = { false }; for (int i = 0; i < ans.size(); i++) { // If current element has already // been seen previously if (present[ans[i]]) { cout << "-1"; return; } // Mark the current element's occurrence else present[ans[i]] = true; } // Print the required permutation for (int i = 0; i < n; i++) cout << ans[i] << " ";}// Driver codeint main(){ int a[] = { 2, 3, 6 }; int n = sizeof(a) / sizeof(a[0]); getPermutation(a, n); return 0;} |
Java
// Java implementation of the approachclass GFG{// Function to find the valid permutationstatic void getPermutation(int a[], int n){ // Find the array from the cumulative sum int []ans = new int[n]; ans[0] = a[0]; for (int i = 1; i < n; i++) ans[i] = a[i] - a[i - 1]; // To mark the occurrence of an element boolean []present = new boolean[n + 1]; for (int i = 0; i < ans.length; i++) { // If current element has already // been seen previously if (present[ans[i]]) { System.out.print("-1"); return; } // Mark the current element's occurrence else present[ans[i]] = true; } // Print the required permutation for (int i = 0; i < n; i++) System.out.print(ans[i] + " ");}// Driver codepublic static void main(String[] args){ int a[] = { 2, 3, 6 }; int n = a.length; getPermutation(a, n);}}// This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach # Function to find the valid permutation def getPermutation(a, n) : # Find the array from the cumulative sum ans = [0] * n; ans[0] = a[0]; for i in range(1, n) : ans[i] = a[i] - a[i - 1]; # To mark the occurrence of an element present = [0] * (n + 1); for i in range(n) : # If current element has already # been seen previously if (present[ans[i]]) : print("-1", end = ""); return; # Mark the current element's occurrence else : present[ans[i]] = True; # Print the required permutation for i in range(n) : print(ans[i], end = " "); # Driver code if __name__ == "__main__" : a = [ 2, 3, 6 ]; n = len(a); getPermutation(a, n); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the above approach using System; using System.Collections.Generic; class GFG { // Function to find the valid permutationstatic void getPermutation(int [] a, int n){ // Find the array from the cumulative sum List<int> ans = new List<int>(); ans.Add(a[0]); for (int i = 1; i < n; i++) ans.Add(a[i] - a[i - 1]); // To mark the occurrence of an element List<int> present = new List<int>(); for (int i = 0; i < n+1; i++) present.Add(0); for (int i = 0; i < ans.Count; i++) { // If current element has already // been seen previously if (present[ans[i]] == 1) { Console.Write("-1"); return; } // Mark the current element's occurrence else present[ans[i]] = 1; } // Print the required permutation for (int i = 0; i < n; i++) Console.Write(ans[i] + " ");}// Driver codestatic public void Main() { int[] a = { 2,3,6}; int n = a.Length; getPermutation(a, n); } }// This code is contributed by mohit kumar 29 |
Javascript
<script>// Javascript implementation of the approach// Function to find the valid permutationfunction getPermutation(a, n){ // Find the array from the cumulative sum var ans = Array(n); ans[0] = a[0]; for (var i = 1; i < n; i++) ans[i] = a[i] - a[i - 1]; // To mark the occurrence of an element var present = Array(n+1).fill(false); for (var i = 0; i < ans.length; i++) { // If current element has already // been seen previously if (present[ans[i]]) { document.write( "-1"); return; } // Mark the current element's occurrence else present[ans[i]] = true; } // Print the required permutation for (var i = 0; i < n; i++) document.write( ans[i] + " ");}// Driver codevar a = [2, 3, 6];var n = a.length;getPermutation(a, n);// This code is contributed by rrrtnx.</script> |
Output:
2 1 3
Time Complexity: O(n)
Auxiliary Space: O(n)
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