从累积和数组中查找排列数组

给定一个由N个元素组成的数组arr [] ，其中每个arr [i]是另一个数组P []的子数组P [ 0…i]的累加和，其中P是从1到N的整数的排列。任务是找到数组P [] ，如果不存在这样的P ，则打印-1 。

例子：

Input: arr[] = {2, 3, 6}
Output: 2 1 3

Input: arr[] = {1, 2, 2, 4}
Output: -1

方法：

• 累积数组的第一个元素必须是置换数组的第一个元素，第i^个位置的元素将是arr [i] – arr [i – 1]，因为arr []是置换数组的累加和数组。
• 因此，从累积和数组中找到该数组，然后标记生成的数组中从1到N的每个元素的出现。
• 如果任何元素出现不止一次，则该排列无效，否则打印该排列。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to find the valid permutation
void getPermutation(int a[], int n)
{
  
    // Find the array from the cumulative sum
    vector 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 ans = new List();
    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 present = new List();
  
    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 ontributed by mohit kumar 29

输出：

2 1 3