从存储右侧较大元素计数的数组生成原始数组

给定一个整数数组greater[]，其中数组的每个值都表示未知数组arr[]中右侧有多少个元素大于其右侧。我们的任务是生成原始数组 arr[]。可以假设原始数组包含从 1 到 n 范围内的元素，并且所有元素都是唯一的
例子：

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 ]  

我们考虑一个元素数组 temp[] = {1, 2, 3, 4, .. n}。我们知道大于 [0] 的值表示大于 arr[0] 的元素计数。我们可以观察到 temp[] 的第 (n – greater[0]) 个元素可以放在 arr[0] 处。所以我们把它放在 arr[0] 并从 temp[] 中删除它。我们对剩余元素重复上述过程。对于每个更大的元素[i]，我们将 temp[] 的第 (n – greater[i] – i) 个元素放入 arr[i] 并将其从 temp[] 中删除。
下面是上述想法的实现

C++

// C++ program to generate original array
// from an array that stores counts of
// greater elements on right.
#include 
using namespace std;
 
void originalArray(int greater[], int n)
{
    // Array that is used to include every
    // element only once
    vector 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 temp = new Vector();
    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 temp = new List();
    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 */

Javascript

 
// 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
 

输出：

1 4 5 6 7 2 3

时间复杂度： （n ² ）（擦除操作在向量中需要 O(n)）