将绝对指数差与 K 的乘积最大化

给定一个由N 个整数组成的数组A[] ，任务是找到K的最大可能值，使得K * |i – j| <= min(A _i , A _j ) ，其中 (0 ? i < j < N)。

Given expression, k * |i – j| <= min(A_i, A_j) can also be written as k = floor( min(A_i, A_j) / |i – j| )

例子：

Input: N = 5, A[ ] = {80, 10, 12, 15, 90} 
Output: 20 
Explanation: 
For i = 0 and j = 4, the maximum possible value of K can be obtained. 
Maximum k = min(A[0], A[4]) / |0 – 4| = min(80, 90) / |-4| = 80/4 = 20

Input: N = 5, A[ ] = {10, 5, 12, 15, 8} 
Output: 12 
Explanation: 
For i = 2 and j = 3, the maximum possible value of K can be obtained. 
Maximum k = min(A[2], A[3]) / |2 – 3| = min(12, 15) / |-1| = 12/1 = 12 

天真的方法：
解决这个问题最简单的方法是从给定的数组中生成所有可能的对，对于每一对，找到 K 的值并不断更新获得的最大K。最后，打印得到的K的最大值。
时间复杂度： O(N ² )
辅助空间： O(1)

有效的方法：
要优化上述方法，请使用两个指针技术。请按照以下步骤操作：

• 初始化三个变量i 、 j和k 。最初设置i = 0和k = 0 。
• 迭代数组，从j = 1到j = N-1 。
• 现在，对于每对A[i]和A[j] ，找到min(A[i], A[j]) / ( j – i )。如果它大于K ，则更新K 。
• 如果A[ j ] >= A[i] / ( j – i ) ，则将指针i更新为j ，以确保在直到i 的所有元素中， A[i]将导致最大K与所有即将到来的A[ j]
• 最后，遍历完数组后打印K的最大值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above appraoch
#include 
using namespace std;
 
// Function returns maximum
// possible value of k
int solve(int A[], int N)
{
 
    // Pointer i make sure that
    // A[i] will result in max k
    int i = 0;
 
    // Stores maximum possible k
    int k = 0;
 
    for (int j = 1; j < N; j++) {
 
        // Possible value of k for
        // current pair (A[i] and A[j])
        int tempK = min(A[i], A[j])
                    / (j - i);
 
        // If current value exceeds k
        if (tempK > k) {
            // Update the value of k
            k = tempK;
        }
 
        // Update pointer i
        if (A[j] >= A[i] / (j - i))
            i = j;
    }
 
    // Return the maxm possible k
    return k;
}
 
// Driver Code
int main()
{
    int A[] = { 10, 5, 12, 15, 8 };
 
    int N = sizeof(A) / sizeof(A[0]);
 
    cout << solve(A, N);
 
    return 0;
}

Java

// Java program to implement
// the above appraoch
class GFG{
 
// Function returns maximum
// possible value of k
static int solve(int A[], int N)
{
     
    // Pointer i make sure that
    // A[i] will result in max k
    int i = 0;
 
    // Stores maximum possible k
    int k = 0;
 
    for(int j = 1; j < N; j++)
    {
         
        // Possible value of k for
        // current pair (A[i] and A[j])
        int tempK = Math.min(A[i], A[j]) /
                              (j - i);
 
        // If current value exceeds k
        if (tempK > k)
        {
             
            // Update the value of k
            k = tempK;
        }
 
        // Update pointer i
        if (A[j] >= A[i] / (j - i))
            i = j;
    }
 
    // Return the maxm possible k
    return k;
}
 
// Driver Code
public static void main(String[] args)
{
    int A[] = { 10, 5, 12, 15, 8 };
    int N = A.length;
     
    System.out.println(solve(A, N));
}
}
 
// This code is contributed by rutvik_56

Python3

# Python3 program to implement
# the above appraoch
 
# Function returns maximum
# possible value of k
def solve(A, N):
 
    # Pointer i make sure that
    # A[i] will result in max k
    i = 0
 
    # Stores maximum possible k
    k = 0
 
    for j in range(1, N):
 
        # Possible value of k for
        # current pair (A[i] and A[j])
        tempK = (min(A[i], A[j]) // (j - i))
                     
        # If current value exceeds k
        if (tempK > k):
             
            # Update the value of k
            k = tempK
         
        # Update pointer i
        if (A[j] >= A[i] // (j - i)):
            i = j
 
    # Return the maxm possible k
    return k
 
# Driver Code
if __name__ == "__main__":
     
    A = [ 10, 5, 12, 15, 8 ]
    N = len(A);
 
    print(solve(A, N))
 
# This code is contributed by chitranayal

C#

// C# program to implement
// the above appraoch
using System;
class GFG{
  
// Function returns maximum
// possible value of k
static int solve(int[] A, int N)
{
      
    // Pointer i make sure that
    // A[i] will result in max k
    int i = 0;
  
    // Stores maximum possible k
    int k = 0;
  
    for(int j = 1; j < N; j++)
    {
          
        // Possible value of k for
        // current pair (A[i] and A[j])
        int tempK = Math.Min(A[i], A[j]) /
                              (j - i);
  
        // If current value exceeds k
        if (tempK > k)
        {
              
            // Update the value of k
            k = tempK;
        }
  
        // Update pointer i
        if (A[j] >= A[i] / (j - i))
            i = j;
    }
  
    // Return the maxm possible k
    return k;
}
  
// Driver Code
public static void Main(string[] args)
{
    int[] A = { 10, 5, 12, 15, 8 };
    int N = A.Length;
      
    Console.Write(solve(A, N));
}
}
  
// This code is contributed by rock_cool

Javascript

 
// Javascript program to implement
// the above appraoch
 
// Function returns maximum
// possible value of k
function solve(A, N)
{
     
    // Pointer i make sure that
    // A[i] will result in max k
    let i = 0;
 
    // Stores maximum possible k
    let k = 0;
 
    for(let j = 1; j < N; j++)
    {
         
        // Possible value of k for
        // current pair (A[i] and A[j])
        let tempK = Math.min(A[i], A[j]) / (j - i);
 
        // If current value exceeds k
        if (tempK > k)
        {
             
            // Update the value of k
            k = tempK;
        }
 
        // Update pointer i
        if (A[j] >= A[i] / (j - i))
            i = j;
    }
 
    // Return the maxm possible k
    return k;
}
 
// Driver code
let A = [ 10, 5, 12, 15, 8 ];
let N = A.length;
 
document.write(solve(A, N));
 
// This code is contributed by divyeshrabadiya07
 

输出：

12

时间复杂度： O(N)
辅助空间： O(1)

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