Given an array of integers greater[] in which every value of array represents how many elements are greater to its 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 the above idea
// 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 functionint main()
{ int Arr[] = { 6, 3, 2, 1, 0, 1, 0 };
int n = sizeof(Arr) / sizeof(Arr[0]);
originalArray(Arr, n);
return 0;
} |
// 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 program original array from an# array that stores counts of greater# elements on rightdef 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 codearr = [6, 3, 2, 1, 0, 1, 0]
n = len(arr)
originalArray(arr, n)# This code is contributed# by Mohit Kumar |
// 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 */ |
<script>// JavaScript program to generate original array // from an array that stores counts of // greater elements on right. function originalArray(greater,n)
{
// Array that is used to include every
// element only once
let temp = [];
for (let i = 0; i <= n; i++)
temp.push(i);
// Traverse the array element
let arr = new Array(n);
for (let i = 0; i < n; i++)
{
// find the K-th (n-greater[i]-i)
// smallest element in Include_Array
let k = n - greater[i] - i;
arr[i] = temp[k];
// remove current k-th element
// from Include array
temp.splice(k,1);
}
// print resultant array
for (let i = 0; i < n; i++)
document.write(arr[i] + " ");
}
// Driver code
let Arr=[6, 3, 2, 1, 0, 1, 0 ];
let n = Arr.length;
originalArray(Arr, n);
// This code is contributed by patel2127</script> |
1 4 5 6 7 2 3
Time Complexity: (n2) (Erase operation takes O(n) in vector).
Auxiliary Space: O(n), extra space is required for temp and res arrays.