给定数组中满足给定条件的最小索引

给定大小为N的数组arr []和整数K ，任务是在数组中找到最小的索引，以便：

floor(arr[i] / 1) + floor(arr[i + 1] / 2) + floor(arr[i + 2] / 2) + ….. + floor(arr[n – 1] / n – i ) ≤ K

如果找不到这样的索引，则打印-1 。

例子：

Input: arr[] = {6, 5, 4, 2}, K = 8
Output: 1
(6 / 1) + (5 / 2) + (4 / 3) + (2 / 4) = 6 + 2 + 1 + 0 = 9 which is > K
(5 / 1) + (4 / 2) + (2 / 3) = 5 + 2 + 0 = 7 < K
Hence i = 1 is the required index.

Input: arr[] = {5, 4, 2, 3, 9, 1, 8, 7}, K = 7
Output: 5

方法：对于从0开始的每个索引i ，找到给定系列的总和，并且给定总和大于或等于K的第一个索引就是我们所需的索引。如果没有这样的索引，则打印-1 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the required index
int getIndex(int arr[], int n, int K)
{
  
    // Check all the indices, the first index
    // satisfying the conidtion is
    // the required index
    for (int i = 0; i < n; i++) {
  
        // To store the sum of the series
        int sum = 0;
  
        // Denominator for the sum series
        int den = 1;
  
        // Find the sum of the series
        for (int j = i; j < n; j++) {
            sum += (arr[j] / den);
  
            // Increment the denominator
            den++;
  
            // If current sum is greater than K
            // no need to execute loop further
            if (sum > K)
                break;
        }
  
        // Found the first valid index
        if (sum <= K)
            return i;
    }
  
    return -1;
}
  
// Driver code
int main()
{
    int arr[] = { 6, 5, 4, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int K = 8;
    cout << getIndex(arr, n, K);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG {
  
    // Function to return the required index
    static int getIndex(int arr[], int n, int K)
    {
  
        // Check all the indices, the first index
        // satisfying the conidtion is
        // the required index
        for (int i = 0; i < n; i++) {
  
            // To store the sum of the series
            int sum = 0;
  
            // Denominator for the sum series
            int den = 1;
  
            // Find the sum of the series
            for (int j = i; j < n; j++) {
                sum += (arr[j] / den);
  
                // Increment the denominator
                den++;
  
                // If current sum is greater than K
                // no need to execute loop further
                if (sum > K)
                    break;
            }
  
            // Found the first valid index
            if (sum <= K)
                return i;
        }
  
        return -1;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 6, 5, 4, 2 };
        int n = arr.length;
        int K = 8;
        System.out.print(getIndex(arr, n, K));
    }
}

Python

# Python3 implementation of the approach
def getIndex(array, k):
  n = len(array)
  for ind in range(n):
    sum = 0
    div = 1
    for item in array:
      sum += item//div
      div += 1
      if sum > k:
        break
    if sum <= k:
      return ind
    
  return -1
  
# Driver code
arr = [6, 5, 4, 2]
K = 8
print(getIndex(arr, K))
  
arr = [5, 4, 2, 3, 9, 1, 8, 7]
K = 7
print(getIndex(arr, K))

C#

// C# implementation of the approach
using System;
  
class GFG
{
    // Function to return the required index
    static int getIndex(int []arr, int n, int K)
    {
  
        // Check all the indices, the first index
        // satisfying the conidtion is
        // the required index
        for (int i = 0; i < n; i++) 
        {
  
            // To store the sum of the series
            int sum = 0;
  
            // Denominator for the sum series
            int den = 1;
  
            // Find the sum of the series
            for (int j = i; j < n; j++) 
            {
                sum += (arr[j] / den);
  
                // Increment the denominator
                den++;
  
                // If current sum is greater than K
                // no need to execute loop further
                if (sum > K)
                    break;
            }
  
            // Found the first valid index
            if (sum <= K)
                return i;
        }
  
        return -1;
    }
  
    // Driver code
    static public void Main ()
    {
          
        int []arr = { 6, 5, 4, 2 };
        int n = arr.Length;
        int K = 8;
        Console.WriteLine(getIndex(arr, n, K));
    }
}
  
// This Code is contributed by ajit.

PHP

 $K)
                break;
        }
  
        // Found the first valid index
        if ($sum <= $K)
            return $i;
    }
  
    return -1;
}
  
// Driver code
$arr = array( 6, 5, 4, 2 );
$n = sizeof($arr);
$K = 8;
  
echo getIndex($arr, $n, $K);
  
// This code is contributed by Ryuga
?>

输出：

1

时间复杂度： O(N ² )
辅助空间： O(1)