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Generate original array from an array that store the counts of greater elements on right
  • Difficulty Level : Medium
  • Last Updated : 27 Feb, 2019

Given an array of integers greater[] in which every value of array represent how many elements are greater to it’s on right side in an unknown array arr[]. Our task is to generate original array arr[]. It may be assumed that the original array contains elements in range from 1 to n and all elements are unique
Examples:

Input  : greater[] = { 6, 3, 2, 1, 0, 0, 0 }
Output : arr[] = [ 1, 4, 5, 6, 7, 3, 2 ]
 
Input  : greater[] = { 0, 0, 0, 0, 0 }
Output : arr[] = [ 5, 4, 3, 2, 1 ]  

We consider an array of elements temp[] = {1, 2, 3, 4, .. n}. We know value of greater[0] indicates count of elements greater than arr[0]. We can observe that (n – greater[0])-th element of temp[] can be put at arr[0]. So we put this at arr[0] and remove this from temp[]. We repeat above process for remaining elements. For every element greater[i], we put (n – greater[i] – i)-th element of temp[] in arr[i] and remove it from temp[].

Below is the implementation of above idea

C++




// C++ program to generate original array 
// from an array that stores counts of 
// greater elements on right.
#include <bits/stdc++.h>
using namespace std;
  
void originalArray(int greater[], int n)
{
    // Array that is used to include every
    // element only once
    vector<int> temp;
    for (int i = 0; i <= n; i++)
        temp.push_back(i);
  
    // Traverse the array element
    int arr[n];
    for (int i = 0; i < n; i++) {
  
        // find the K-th (n-greater[i]-i) 
        // smallest element in Include_Array
        int k = n - greater[i] - i;
  
        arr[i] = temp[k];
  
        // remove current k-th element 
        // from Include array
        temp.erase(temp.begin() + k);
    }
  
    // print resultant array
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
  
// driver program to test above function
int main()
{
    int Arr[] = { 6, 3, 2, 1, 0, 1, 0 };
    int n = sizeof(Arr) / sizeof(Arr[0]);
    originalArray(Arr, n);
    return 0;
}

Java




// Java program to generate original array 
// from an array that stores counts of 
// greater elements on right.
import java.util.Vector;
  
class GFG 
{
      
static void originalArray(int greater[], int n) 
    // Array that is used to include every 
    // element only once 
    Vector<Integer> temp = new Vector<Integer>(); 
    for (int i = 0; i <= n; i++) 
        temp.add(i); 
  
    // Traverse the array element 
    int arr[] = new int[n]; 
    for (int i = 0; i < n; i++)
    
  
        // find the K-th (n-greater[i]-i) 
        // smallest element in Include_Array 
        int k = n - greater[i] - i; 
  
        arr[i] = temp.get(k); 
  
        // remove current k-th element 
        // from Include array 
        temp.remove(k); 
    
  
    // print resultant array 
    for (int i = 0; i < n; i++) 
            System.out.print(arr[i] + " "); 
  
// Driver code
public static void main(String[] args)
{
    int Arr[] = { 6, 3, 2, 1, 0, 1, 0 }; 
    int n = Arr.length; 
    originalArray(Arr, n);
}
  
// This code is contributed by Rajput-Ji

Python3




# Python3 program original array from an
# array that stores counts of greater 
# elements on right
def originalArray(greater, n):
      
    # array that is used to include 
    # every element only once
    temp = []
      
    for i in range(n + 1):
        temp.append(i)
          
    # traverse the array element
    arr = [0 for i in range(n)]
      
    for i in range(n):
  
        # find the Kth (n-greater[i]-i)
        # smallest element in Include_array
        k = n - greater[i] - i
          
        arr[i] = temp[k]
          
        # remove current kth element
        # from include array
        del temp[k]
          
    for i in range(n):
        print(arr[i], end = " ")
          
# Driver code
arr = [6, 3, 2, 1, 0, 1, 0]
n = len(arr)
originalArray(arr, n)
  
# This code is contributed 
# by Mohit Kumar

C#




// C# program to generate original array 
// from an array that stores counts of 
// greater elements on right.
using System;
using System.Collections.Generic; 
  
class GFG 
{
      
static void originalArray(int []greater, int n) 
    // Array that is used to include every 
    // element only once 
    List<int> temp = new List<int>(); 
    for (int i = 0; i <= n; i++) 
        temp.Add(i); 
  
    // Traverse the array element 
    int []arr = new int[n]; 
    for (int i = 0; i < n; i++)
    
  
        // find the K-th (n-greater[i]-i) 
        // smallest element in Include_Array 
        int k = n - greater[i] - i; 
  
        arr[i] = temp[k]; 
  
        // remove current k-th element 
        // from Include array 
        temp.RemoveAt(k); 
    
  
    // print resultant array 
    for (int i = 0; i < n; i++) 
            Console.Write(arr[i] + " "); 
  
// Driver code
public static void Main()
{
    int []Arr = { 6, 3, 2, 1, 0, 1, 0 }; 
    int n = Arr.Length; 
    originalArray(Arr, n);
}
  
/* This code contributed by PrinciRaj1992 */


Output:
1 4 5 6 7 2 3

Time Complexity : (n2) (Erase operation takes O(n) in vector)




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