查找大小为n的集合中所有大小r个子集的特定算法的复杂性

Question

我找到了标题中写的简单算法代码-将其添加到了这篇文章（source）中。
我发现很难计算该算法的空间复杂度和运行时间。以为它可能是O((nCr)*r)，但我认为我们在这里只传递了nCr个项目（并且可能在打印循环中对每个项目执行r个迭代）。

能否在这里解释如何计算空间复杂度和运行时间？

#include <stdio.h> 
void combinationUtil(int arr[], int n, int r, 
                     int index, int data[], int i); 

// The main function that prints all combinations of  
// size r in arr[] of size n. This function mainly 
// uses combinationUtil() 
void printCombination(int arr[], int n, int r) 
{ 
    // A temporary array to store all combination 
    // one by one 
    int data[r]; 

    // Print all combination using temprary array 'data[]' 
    combinationUtil(arr, n, r, 0, data, 0); 
} 

/* arr[]  ---> Input Array 
   n      ---> Size of input array 
   r      ---> Size of a combination to be printed 
   index  ---> Current index in data[] 
   data[] ---> Temporary array to store current combination 
   i      ---> index of current element in arr[]     */
void combinationUtil(int arr[], int n, int r, int index, 
                     int data[], int i) 
{ 
    // Current cobination is ready, print it 
    if (index == r) { 
        for (int j = 0; j < r; j++) 
            printf("%d ", data[j]); 
        printf("\n"); 
        return; 
    } 

    // When no more elements are there to put in data[] 
    if (i >= n) 
        return; 

    // current is included, put next at next location 
    data[index] = arr[i]; 
    combinationUtil(arr, n, r, index + 1, data, i + 1); 

    // current is excluded, replace it with next 
    // (Note that i+1 is passed, but index is not 
    // changed) 
    combinationUtil(arr, n, r, index, data, i + 1); 
}