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 permutation void 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 code int main()
{ int a[] = { 2, 3, 6 };
int n = sizeof (a) / sizeof (a[0]);
getPermutation(a, n);
return 0;
} |
Java
// Java implementation of the approach class GFG
{ // Function to find the valid permutation static 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 code public 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 permutation static 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 code static 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 permutation function 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 code var 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)